Mixing
$mathbf{A}mathbf{X} = mathbf{X}mathbf{M}$ implies eigenvalues of $mathbf{M}$ are eigenvalues of $mathbf{A}$
$begingroup$
Let $mathbf{A} in mathbb{C}^{n times n}$ and $mathbf{X} in mathbb{C}^{n times p}$ with $Rank(mathbf{X})=p$ for $1 leq p < n$. I want to show that if $mathbf{A}mathbf{X} = mathbf{X}mathbf{M}$, then the eigenvalues of $mathbf{M}$ are eigenvalues of $mathbf{A}$.
I have been trying to use QR and Schur decompositions to get some sort of similarity transformation that would prove this statement. However, I am having trouble with this. I also am not sure where $mathbf{X}$ having full column rank fits into this.
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
$endgroup$
add a comment |
$begingroup$
Let $mathbf{A} in mathbb{C}^{n times n}$ and $mathbf{X} in mathbb{C}^{n times p}$ with $Rank(mathbf{X})=p$ for $1 leq p < n$. I want to show that if $mathbf{A}mathbf{X} = mathbf{X}mathbf{M}$, then the eigenvalues of $mathbf{M}$ are eigenvalues of $mathbf{A}$.
I have been trying to use QR and Schur decompositions to get some sort of similarity transformation that would prove this statement. However, I am having trouble with this. I also am not sure where $mathbf{X}$ having full column rank fits into this.
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
$endgroup$
add a comment |
$begingroup$
Let $mathbf{A} in mathbb{C}^{n times n}$ and $mathbf{X} in mathbb{C}^{n times p}$ with $Rank(mathbf{X})=p$ for $1 leq p < n$. I want to show that if $mathbf{A}mathbf{X} = mathbf{X}mathbf{M}$, then the eigenvalues of $mathbf{M}$ are eigenvalues of $mathbf{A}$.
I have been trying to use QR and Schur decompositions to get some sort of similarity transformation that would prove this statement. However, I am having trouble with this. I also am not sure where $mathbf{X}$ having full column rank fits into this.
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
$endgroup$
Let $mathbf{A} in mathbb{C}^{n times n}$ and $mathbf{X} in mathbb{C}^{n times p}$ with $Rank(mathbf{X})=p$ for $1 leq p < n$. I want to show that if $mathbf{A}mathbf{X} = mathbf{X}mathbf{M}$, then the eigenvalues of $mathbf{M}$ are eigenvalues of $mathbf{A}$.
I have been trying to use QR and Schur decompositions to get some sort of similarity transformation that would prove this statement. However, I am having trouble with this. I also am not sure where $mathbf{X}$ having full column rank fits into this.
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
asked Jan 28 at 1:22
Andreu PayneAndreu Payne
354
354
add a comment |
add a comment |
1 Answer
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$begingroup$
If $lambda$ is an eigenvalue of $M$ and $Y neq 0$ is an eigenvector corresponding to $lambda$, then
$$A(XY) = (AX)Y = (XM)Y = X(MY) = lambda(XY);$$
Since $X$ has full rank, $XY neq 0$, hence $lambda$ is an eigenvalue of $A$.
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $lambda$ is an eigenvalue of $M$ and $Y neq 0$ is an eigenvector corresponding to $lambda$, then
$$A(XY) = (AX)Y = (XM)Y = X(MY) = lambda(XY);$$
Since $X$ has full rank, $XY neq 0$, hence $lambda$ is an eigenvalue of $A$.
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
$endgroup$
add a comment |
$begingroup$
If $lambda$ is an eigenvalue of $M$ and $Y neq 0$ is an eigenvector corresponding to $lambda$, then
$$A(XY) = (AX)Y = (XM)Y = X(MY) = lambda(XY);$$
Since $X$ has full rank, $XY neq 0$, hence $lambda$ is an eigenvalue of $A$.
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
$endgroup$
add a comment |
$begingroup$
If $lambda$ is an eigenvalue of $M$ and $Y neq 0$ is an eigenvector corresponding to $lambda$, then
$$A(XY) = (AX)Y = (XM)Y = X(MY) = lambda(XY);$$
Since $X$ has full rank, $XY neq 0$, hence $lambda$ is an eigenvalue of $A$.
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
$endgroup$
If $lambda$ is an eigenvalue of $M$ and $Y neq 0$ is an eigenvector corresponding to $lambda$, then
$$A(XY) = (AX)Y = (XM)Y = X(MY) = lambda(XY);$$
Since $X$ has full rank, $XY neq 0$, hence $lambda$ is an eigenvalue of $A$.
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
answered Jan 28 at 1:31
Catalin ZaraCatalin Zara
3,752514
3,752514
add a comment |
add a comment |
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