Eigenvalues of two symmetric $4times 4$ matrices: why is one negative of the other?Congruence of invertible skew symmetric matricesEigenvalues of a general block hermitian matrixEigenvalues of Overlapping block diagonal matricesBuilding matrices from eigenvalueseigenvalues for certain hermitian matricesEigenvalues and eigenspaces in a symmetric matrixThe matrix of an endomorphismProve that the span of $M_1, M_2, M_3$ is the set of all symmetric $2times2$ matrices.Looking for properties of, or formulae for eigenvalues of a symmetric matrix reminiscent of Toeplitz matricesDo hermitian matrices commute when they occupy they same elements but have different values?

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Eigenvalues of two symmetric $4times 4$ matrices: why is one negative of the other?


Congruence of invertible skew symmetric matricesEigenvalues of a general block hermitian matrixEigenvalues of Overlapping block diagonal matricesBuilding matrices from eigenvalueseigenvalues for certain hermitian matricesEigenvalues and eigenspaces in a symmetric matrixThe matrix of an endomorphismProve that the span of $M_1, M_2, M_3$ is the set of all symmetric $2times2$ matrices.Looking for properties of, or formulae for eigenvalues of a symmetric matrix reminiscent of Toeplitz matricesDo hermitian matrices commute when they occupy they same elements but have different values?













3












$begingroup$


Consider the following symmetric matrix:



$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



and a very similar matrix:



$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ instead, then they do not share the same eigenvalues.




Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.










share|cite|improve this question











$endgroup$











  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    1 hour ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    1 hour ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    1 hour ago















3












$begingroup$


Consider the following symmetric matrix:



$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



and a very similar matrix:



$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ instead, then they do not share the same eigenvalues.




Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.










share|cite|improve this question











$endgroup$











  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    1 hour ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    1 hour ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    1 hour ago













3












3








3


1



$begingroup$


Consider the following symmetric matrix:



$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



and a very similar matrix:



$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ instead, then they do not share the same eigenvalues.




Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.










share|cite|improve this question











$endgroup$




Consider the following symmetric matrix:



$$
M_0 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & 4 & 3 \
2 & 4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



and a very similar matrix:



$$
M_1 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & -4 & 3 \
2 & -4 & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?



I also tried playing around with the values a little; for example, if the center block is $beginpmatrix1 & pm 4 \ pm 4 & 1endpmatrix$ instead, then they do not share the same eigenvalues.




Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.



$$
M_2 =
beginpmatrix
0 & 1 & 2 & 0 \
1 & 0 & e^ix & 3 \
2 & e^-ix & 0 & 1 \
0 & 3 & 1 & 0
endpmatrix
$$



ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.



Thanks.







linear-algebra matrices eigenvalues-eigenvectors symmetric-matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 27 mins ago









M. Vinay

7,33322136




7,33322136










asked 2 hours ago









TroyTroy

4281519




4281519











  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    1 hour ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    1 hour ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    1 hour ago
















  • $begingroup$
    It's because of all the conveniently placed zeroes.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
    $endgroup$
    – Troy
    1 hour ago






  • 1




    $begingroup$
    In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
    $endgroup$
    – M. Vinay
    1 hour ago






  • 2




    $begingroup$
    In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy that certainly narrows down the search for me, thanks for the input!
    $endgroup$
    – Troy
    1 hour ago















$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
1 hour ago




$begingroup$
It's because of all the conveniently placed zeroes.
$endgroup$
– M. Vinay
1 hour ago












$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
1 hour ago




$begingroup$
@M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind?
$endgroup$
– Troy
1 hour ago




1




1




$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
1 hour ago




$begingroup$
In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say.
$endgroup$
– M. Vinay
1 hour ago




2




2




$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
1 hour ago




$begingroup$
In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix.
$endgroup$
– leonbloy
1 hour ago












$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
1 hour ago




$begingroup$
@leonbloy that certainly narrows down the search for me, thanks for the input!
$endgroup$
– Troy
1 hour ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

$$-M_1=D^-1M_0D$$
where $D=D^-1$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$

and
$$-pmatrix-a_11&a_12&a_13&-a_14\
a_21&-a_22&-a_23&a_24\
a_31&-a_32&-a_33&a_34\
-a_41&a_42&a_43&-a_44$$

are conjugate, for precisely the same reason.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
    $endgroup$
    – Troy
    1 hour ago



















2












$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
beginalign*
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
endalign*

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    1 hour ago


















1












$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
chi_M_a(t)
&= t^4 - left(a^2 + 15right) t^2 - 10 , a t + 25 \
chi_M_-a(t)
&= t^4 - left(a^2 + 15right) t^2 + 10 , a t + 25
endalign*

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
0
&= chi_M_a(t) \
&= lambda^4 - left(a^2 + 15right) lambda^2 - 10 , a lambda + 25\
&= (-lambda)^4 - left(a^2 + 15right) (-lambda)^2 + 10 , a (-lambda) + 25 \
&= chi_M_-a(-lambda)
endalign*

This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
chi_M_a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 - 10 , a t + 25 \
chi_M_-a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 + 10 , a t + 25
endalign*

A similiar argument then shows that $M_a+bi$ and $M_-a+bi$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    1 hour ago











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

$$-M_1=D^-1M_0D$$
where $D=D^-1$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$

and
$$-pmatrix-a_11&a_12&a_13&-a_14\
a_21&-a_22&-a_23&a_24\
a_31&-a_32&-a_33&a_34\
-a_41&a_42&a_43&-a_44$$

are conjugate, for precisely the same reason.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
    $endgroup$
    – Troy
    1 hour ago
















4












$begingroup$

$$-M_1=D^-1M_0D$$
where $D=D^-1$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$

and
$$-pmatrix-a_11&a_12&a_13&-a_14\
a_21&-a_22&-a_23&a_24\
a_31&-a_32&-a_33&a_34\
-a_41&a_42&a_43&-a_44$$

are conjugate, for precisely the same reason.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
    $endgroup$
    – Troy
    1 hour ago














4












4








4





$begingroup$

$$-M_1=D^-1M_0D$$
where $D=D^-1$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$

and
$$-pmatrix-a_11&a_12&a_13&-a_14\
a_21&-a_22&-a_23&a_24\
a_31&-a_32&-a_33&a_34\
-a_41&a_42&a_43&-a_44$$

are conjugate, for precisely the same reason.






share|cite|improve this answer











$endgroup$



$$-M_1=D^-1M_0D$$
where $D=D^-1$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$pmatrixa_11&a_12&a_13&a_14\
a_21&a_22&a_23&a_24\
a_31&a_32&a_33&a_34\
a_41&a_42&a_43&a_44$$

and
$$-pmatrix-a_11&a_12&a_13&-a_14\
a_21&-a_22&-a_23&a_24\
a_31&-a_32&-a_33&a_34\
-a_41&a_42&a_43&-a_44$$

are conjugate, for precisely the same reason.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 48 mins ago

























answered 1 hour ago









Lord Shark the UnknownLord Shark the Unknown

108k1162135




108k1162135







  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
    $endgroup$
    – Troy
    1 hour ago













  • 1




    $begingroup$
    Of course, signature matrix. This is the answer.
    $endgroup$
    – M. Vinay
    1 hour ago










  • $begingroup$
    okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
    $endgroup$
    – Troy
    1 hour ago








1




1




$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
1 hour ago




$begingroup$
Of course, signature matrix. This is the answer.
$endgroup$
– M. Vinay
1 hour ago












$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
1 hour ago





$begingroup$
okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long)
$endgroup$
– Troy
1 hour ago












2












$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
beginalign*
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
endalign*

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    1 hour ago















2












$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
beginalign*
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
endalign*

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    1 hour ago













2












2








2





$begingroup$

This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
beginalign*
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
endalign*

And the cases of the third and fourth rows are obviously similar.






share|cite|improve this answer











$endgroup$



This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.



Let $$M_1 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix, quad M_2 = beginbmatrix0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0endbmatrix$$



Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = beginbmatrixx_1 & x_2 & x_3 & x_4endbmatrix^T$.
Then we can show that
$beginbmatrixx_1 & -x_2 & -x_3 & x_4endbmatrix^T$
is an eigenvector corresponding to eigenvalue $-lambda$ for $M_2$.



For,
beginalign*
a_2 x_2 + a_3 x_3 = lambda x_1 & implies a_2 (-x_2) + a_3(-x_3) = -lambda x_1\
b_1 x_1 + b_3 x_3 + b_4 x_4 = lambda x_2 & implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-lambda)(-x_2).
endalign*

And the cases of the third and fourth rows are obviously similar.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









M. VinayM. Vinay

7,33322136




7,33322136











  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    1 hour ago
















  • $begingroup$
    oh this is promising. let me mull on this a little before I accept. thanks!
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    The would imply that the property has no obvious generalization for larger sizes, no?
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
    $endgroup$
    – M. Vinay
    1 hour ago















$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
1 hour ago




$begingroup$
oh this is promising. let me mull on this a little before I accept. thanks!
$endgroup$
– Troy
1 hour ago












$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
1 hour ago




$begingroup$
The would imply that the property has no obvious generalization for larger sizes, no?
$endgroup$
– leonbloy
1 hour ago












$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
1 hour ago




$begingroup$
@leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter.
$endgroup$
– M. Vinay
1 hour ago











1












$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
chi_M_a(t)
&= t^4 - left(a^2 + 15right) t^2 - 10 , a t + 25 \
chi_M_-a(t)
&= t^4 - left(a^2 + 15right) t^2 + 10 , a t + 25
endalign*

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
0
&= chi_M_a(t) \
&= lambda^4 - left(a^2 + 15right) lambda^2 - 10 , a lambda + 25\
&= (-lambda)^4 - left(a^2 + 15right) (-lambda)^2 + 10 , a (-lambda) + 25 \
&= chi_M_-a(-lambda)
endalign*

This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
chi_M_a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 - 10 , a t + 25 \
chi_M_-a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 + 10 , a t + 25
endalign*

A similiar argument then shows that $M_a+bi$ and $M_-a+bi$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    1 hour ago















1












$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
chi_M_a(t)
&= t^4 - left(a^2 + 15right) t^2 - 10 , a t + 25 \
chi_M_-a(t)
&= t^4 - left(a^2 + 15right) t^2 + 10 , a t + 25
endalign*

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
0
&= chi_M_a(t) \
&= lambda^4 - left(a^2 + 15right) lambda^2 - 10 , a lambda + 25\
&= (-lambda)^4 - left(a^2 + 15right) (-lambda)^2 + 10 , a (-lambda) + 25 \
&= chi_M_-a(-lambda)
endalign*

This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
chi_M_a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 - 10 , a t + 25 \
chi_M_-a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 + 10 , a t + 25
endalign*

A similiar argument then shows that $M_a+bi$ and $M_-a+bi$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    1 hour ago













1












1








1





$begingroup$

I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
chi_M_a(t)
&= t^4 - left(a^2 + 15right) t^2 - 10 , a t + 25 \
chi_M_-a(t)
&= t^4 - left(a^2 + 15right) t^2 + 10 , a t + 25
endalign*

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
0
&= chi_M_a(t) \
&= lambda^4 - left(a^2 + 15right) lambda^2 - 10 , a lambda + 25\
&= (-lambda)^4 - left(a^2 + 15right) (-lambda)^2 + 10 , a (-lambda) + 25 \
&= chi_M_-a(-lambda)
endalign*

This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
chi_M_a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 - 10 , a t + 25 \
chi_M_-a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 + 10 , a t + 25
endalign*

A similiar argument then shows that $M_a+bi$ and $M_-a+bi$ have eigenvalues related by negation.






share|cite|improve this answer











$endgroup$



I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.



Consider the matrix
$$
M_a =
left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a & 3 \
2 & a & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

The characteristic polynomials of $M_a$ and $M_-a$ are
beginalign*
chi_M_a(t)
&= t^4 - left(a^2 + 15right) t^2 - 10 , a t + 25 \
chi_M_-a(t)
&= t^4 - left(a^2 + 15right) t^2 + 10 , a t + 25
endalign*

Now, note that $lambda$ is an eigenvalue of $M_a$ if and only if
beginalign*
0
&= chi_M_a(t) \
&= lambda^4 - left(a^2 + 15right) lambda^2 - 10 , a lambda + 25\
&= (-lambda)^4 - left(a^2 + 15right) (-lambda)^2 + 10 , a (-lambda) + 25 \
&= chi_M_-a(-lambda)
endalign*

This proves that $M_a$ and $M_-a$ have eigenvalues related by negation.



Now, suppose that $M$ instead takes the form
$$
M_a+bi=left[beginarrayrrrr
0 & 1 & 2 & 0 \
1 & 0 & a + i , b & 3 \
2 & a - i , b & 0 & 1 \
0 & 3 & 1 & 0
endarrayright]
$$

In this case, the characteristic polynomials of $M_a+bi$ and $M_-a+bi$ are
beginalign*
chi_M_a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 - 10 , a t + 25 \
chi_M_-a+bi(t)
&= t^4 + left(-a^2 - b^2 - 15right) t^2 + 10 , a t + 25
endalign*

A similiar argument then shows that $M_a+bi$ and $M_-a+bi$ have eigenvalues related by negation.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









Brian FitzpatrickBrian Fitzpatrick

21.8k42959




21.8k42959











  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    1 hour ago
















  • $begingroup$
    thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
    $endgroup$
    – Troy
    1 hour ago










  • $begingroup$
    This does not explain if the property depends on having those non-zero elements.
    $endgroup$
    – leonbloy
    1 hour ago










  • $begingroup$
    @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
    $endgroup$
    – Brian Fitzpatrick
    1 hour ago















$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
1 hour ago




$begingroup$
thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless.
$endgroup$
– Troy
1 hour ago












$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
1 hour ago




$begingroup$
This does not explain if the property depends on having those non-zero elements.
$endgroup$
– leonbloy
1 hour ago












$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
1 hour ago




$begingroup$
@leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here.
$endgroup$
– Brian Fitzpatrick
1 hour ago

















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Францішак Багушэвіч Змест Сям'я | Біяграфія | Творчасць | Мова Багушэвіча | Ацэнкі дзейнасці | Цікавыя факты | Спадчына | Выбраная бібліяграфія | Ушанаванне памяці | У філатэліі | Зноскі | Літаратура | Спасылкі | НавігацыяЛяхоўскі У. Рупіўся дзеля Бога і людзей: Жыццёвы шлях Лявона Вітан-Дубейкаўскага // Вольскі і Памідораў з песняй пра немца Адвакат, паэт, народны заступнік Ашмянскі веснікВ Минске появится площадь Богушевича и улица Сырокомли, Белорусская деловая газета, 19 июля 2001 г.Айцец беларускай нацыянальнай ідэі паўстаў у бронзе Сяргей Аляксандравіч Адашкевіч (1918, Мінск). 80-я гады. Бюст «Францішак Багушэвіч».Яўген Мікалаевіч Ціхановіч. «Партрэт Францішка Багушэвіча»Мікола Мікалаевіч Купава. «Партрэт зачынальніка новай беларускай літаратуры Францішка Багушэвіча»Уладзімір Іванавіч Мелехаў. На помніку «Змагарам за родную мову» Барэльеф «Францішак Багушэвіч»Памяць пра Багушэвіча на Віленшчыне Страчаная сталіца. Беларускія шыльды на вуліцах Вільні«Krynica». Ideologia i przywódcy białoruskiego katolicyzmuФранцішак БагушэвічТворы на knihi.comТворы Францішка Багушэвіча на bellib.byСодаль Уладзімір. Францішак Багушэвіч на Лідчыне;Луцкевіч Антон. Жыцьцё і творчасьць Фр. Багушэвіча ў успамінах ягоных сучасьнікаў // Запісы Беларускага Навуковага таварыства. Вільня, 1938. Сшытак 1. С. 16-34.Большая российская1188761710000 0000 5537 633Xn9209310021619551927869394п

Беларусь Змест Назва Гісторыя Геаграфія Сімволіка Дзяржаўны лад Палітычныя партыі Міжнароднае становішча і знешняя палітыка Адміністрацыйны падзел Насельніцтва Эканоміка Культура і грамадства Сацыяльная сфера Узброеныя сілы Заўвагі Літаратура Спасылкі НавігацыяHGЯOiТоп-2011 г. (па версіі ej.by)Топ-2013 г. (па версіі ej.by)Топ-2016 г. (па версіі ej.by)Топ-2017 г. (па версіі ej.by)Нацыянальны статыстычны камітэт Рэспублікі БеларусьШчыльнасць насельніцтва па краінахhttp://naviny.by/rubrics/society/2011/09/16/ic_articles_116_175144/А. Калечыц, У. Ксяндзоў. Спробы засялення краю неандэртальскім чалавекам.І ў Менску былі мамантыА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіГ. Штыхаў. Балты і славяне ў VI—VIII стст.М. Клімаў. Полацкае княства ў IX—XI стст.Г. Штыхаў, В. Ляўко. Палітычная гісторыя Полацкай зямліГ. Штыхаў. Дзяржаўны лад у землях-княствахГ. Штыхаў. Дзяржаўны лад у землях-княствахБеларускія землі ў складзе Вялікага Княства ЛітоўскагаЛюблінская унія 1569 г."The Early Stages of Independence"Zapomniane prawdy25 гадоў таму было аб'яўлена, што Язэп Пілсудскі — беларус (фота)Наша вадаДакументы ЧАЭС: Забруджванне тэрыторыі Беларусі « ЧАЭС Зона адчужэнняСведения о политических партиях, зарегистрированных в Республике Беларусь // Министерство юстиции Республики БеларусьСтатыстычны бюлетэнь „Полаўзроставая структура насельніцтва Рэспублікі Беларусь на 1 студзеня 2012 года і сярэднегадовая колькасць насельніцтва за 2011 год“Индекс человеческого развития Беларуси — не было бы нижеБеларусь занимает первое место в СНГ по индексу развития с учетом гендерного факцёраНацыянальны статыстычны камітэт Рэспублікі БеларусьКанстытуцыя РБ. Артыкул 17Трансфармацыйныя задачы БеларусіВыйсце з крызісу — далейшае рэфармаванне Беларускі рубель — сусветны лідар па дэвальвацыяхПра змену коштаў у кастрычніку 2011 г.Бядней за беларусаў у СНД толькі таджыкіСярэдні заробак у верасні дасягнуў 2,26 мільёна рублёўЭканомікаГаласуем за ТОП-100 беларускай прозыСучасныя беларускія мастакіАрхитектура Беларуси BELARUS.BYА. Каханоўскі. Культура Беларусі ўсярэдзіне XVII—XVIII ст.Анталогія беларускай народнай песні, гуказапісы спеваўБеларускія Музычныя IнструментыБеларускі рок, які мы страцілі. Топ-10 гуртоў«Мясцовы час» — нязгаслая легенда беларускай рок-музыкіСЯРГЕЙ БУДКІН. МЫ НЯ ЗНАЕМ СВАЁЙ МУЗЫКІМ. А. Каладзінскі. НАРОДНЫ ТЭАТРМагнацкія культурныя цэнтрыПублічная дыскусія «Беларуская новая пьеса: без беларускай мовы ці беларуская?»Беларускія драматургі па-ранейшаму лепш ставяцца за мяжой, чым на радзіме«Працэс незалежнага кіно пайшоў, і дзяржаву турбуе яго непадкантрольнасць»Беларускія філосафы ў пошуках прасторыВсе идём в библиотекуАрхіваванаАб Нацыянальнай праграме даследавання і выкарыстання касмічнай прасторы ў мірных мэтах на 2008—2012 гадыУ космас — разам.У суседнім з Барысаўскім раёне пабудуюць Камандна-вымяральны пунктСвяты і абрады беларусаў«Мірныя бульбашы з малой краіны» — 5 непраўдзівых стэрэатыпаў пра БеларусьМ. Раманюк. Беларускае народнае адзеннеУ Беларусі скарачаецца колькасць злачынстваўЛукашэнка незадаволены мінскімі ўладамі Крадзяжы складаюць у Мінску каля 70% злачынстваў Узровень злачыннасці ў Мінскай вобласці — адзін з самых высокіх у краіне Генпракуратура аналізуе стан са злачыннасцю ў Беларусі па каэфіцыенце злачыннасці У Беларусі стабілізавалася крымінагеннае становішча, лічыць генпракурорЗамежнікі сталі здзяйсняць у Беларусі больш злачынстваўМУС Беларусі турбуе рост рэцыдыўнай злачыннасціЯ з ЖЭСа. Дазволіце вас абкрасці! Рэйтынг усіх службаў і падраздзяленняў ГУУС Мінгарвыканкама вырасАб КДБ РБГісторыя Аператыўна-аналітычнага цэнтра РБГісторыя ДКФРТаможняagentura.ruБеларусьBelarus.by — Афіцыйны сайт Рэспублікі БеларусьСайт урада БеларусіRadzima.org — Збор архітэктурных помнікаў, гісторыя Беларусі«Глобус Беларуси»Гербы и флаги БеларусиАсаблівасці каменнага веку на БеларусіА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіУ. Ксяндзоў. Сярэдні каменны век (мезаліт). Засяленне краю плямёнамі паляўнічых, рыбакоў і збіральнікаўА. Калечыц, М. Чарняўскі. Плямёны на тэрыторыі Беларусі ў новым каменным веку (неаліце)А. Калечыц, У. Ксяндзоў, М. Чарняўскі. Гаспадарчыя заняткі ў каменным векуЭ. Зайкоўскі. Духоўная культура ў каменным векуАсаблівасці бронзавага веку на БеларусіФарміраванне супольнасцей ранняга перыяду бронзавага векуФотографии БеларусиРоля беларускіх зямель ва ўтварэнні і ўмацаванні ВКЛВ. Фадзеева. З гісторыі развіцця беларускай народнай вышыўкіDMOZGran catalanaБольшая российскаяBritannica (анлайн)Швейцарскі гістарычны15325917611952699xDA123282154079143-90000 0001 2171 2080n9112870100577502ge128882171858027501086026362074122714179пппппп

ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result