prove that the matrix A is diagonalizableBlock Diagonal Matrix DiagonalizableNew proof about normal matrix is diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Is the Matrix Diagonalizable if $A^2=4I$Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix
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prove that the matrix A is diagonalizable
Block Diagonal Matrix DiagonalizableNew proof about normal matrix is diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Is the Matrix Diagonalizable if $A^2=4I$Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix
$begingroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 4 hours ago
JoshuaKJoshuaK
264
$endgroup$
add a comment |
$begingroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 4 hours ago
JoshuaKJoshuaK
264
$endgroup$
2
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
4 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
48 mins ago
add a comment |
$begingroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 4 hours ago
JoshuaKJoshuaK
264
$endgroup$
We have :
$A^3-3A^2-A+3I_n = 0 $
how can i prove that A is diagonalizable .
I don't know how to do when A is written this way
linear-algebra matrices eigenvalues-eigenvectors diagonalization
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked 4 hours ago
JoshuaKJoshuaK
264
asked 4 hours ago
JoshuaKJoshuaK
264
edited 4 hours ago
JoshuaK
asked 4 hours ago
JoshuaKJoshuaK
264
asked 4 hours ago
JoshuaKJoshuaK
264
asked 4 hours ago
JoshuaKJoshuaK
264
264
2
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
4 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
48 mins ago
add a comment |
2
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
4 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
48 mins ago
2
2
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
4 hours ago
$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
4 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
48 mins ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
48 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
2
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
2 hours ago
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 4 hours ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 4 hours ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
4 hours ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
2
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
2 hours ago
add a comment |
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
2
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
2 hours ago
add a comment |
$begingroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
$endgroup$
The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
answered 4 hours ago
TheSilverDoeTheSilverDoe
5,324215
5,324215
2
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
2 hours ago
add a comment |
2
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
2 hours ago
2
2
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
2 hours ago
$begingroup$
I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
$endgroup$
– Acccumulation
2 hours ago
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 4 hours ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 4 hours ago
EricEric
513
$endgroup$
add a comment |
$begingroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 4 hours ago
EricEric
513
$endgroup$
Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.
answered 4 hours ago
EricEric
513
answered 4 hours ago
EricEric
513
answered 4 hours ago
EricEric
513
answered 4 hours ago
EricEric
513
513
add a comment |
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 4 hours ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
4 hours ago
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 4 hours ago
GSoferGSofer
8631313
$endgroup$
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
4 hours ago
add a comment |
$begingroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 4 hours ago
GSoferGSofer
8631313
$endgroup$
We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.
answered 4 hours ago
GSoferGSofer
8631313
answered 4 hours ago
GSoferGSofer
8631313
answered 4 hours ago
GSoferGSofer
8631313
answered 4 hours ago
GSoferGSofer
8631313
8631313
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
4 hours ago
add a comment |
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
$endgroup$
– JoshuaK
4 hours ago
add a comment |
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$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
4 hours ago
$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
4 hours ago
$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
48 mins ago