Mixing
Square root of matrix
$begingroup$
Let $A^2$ be a symmetric matrice such that the spectral theorem allows us to write
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T$$
Suppose $forall i in [![ 1, n]!], lambda_i geq 0$.
I want to know if the following proposition is true:
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T implies A = Poperatorname{Diag}(sqrt{lambda_1} dots sqrt{lambda_n})P^T$$
matrices diagonalization symmetric-matrices
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
$endgroup$
add a comment |
$begingroup$
Let $A^2$ be a symmetric matrice such that the spectral theorem allows us to write
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T$$
Suppose $forall i in [![ 1, n]!], lambda_i geq 0$.
I want to know if the following proposition is true:
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T implies A = Poperatorname{Diag}(sqrt{lambda_1} dots sqrt{lambda_n})P^T$$
matrices diagonalization symmetric-matrices
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
$endgroup$
$begingroup$
yes, it is indeed true
$endgroup$
– pointguard0
Jan 30 at 22:30
$begingroup$
Please notice that my answer was initially only half-true; I presented the correct one now.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
$begingroup$
Let $A^2$ be a symmetric matrice such that the spectral theorem allows us to write
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T$$
Suppose $forall i in [![ 1, n]!], lambda_i geq 0$.
I want to know if the following proposition is true:
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T implies A = Poperatorname{Diag}(sqrt{lambda_1} dots sqrt{lambda_n})P^T$$
matrices diagonalization symmetric-matrices
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
$endgroup$
Let $A^2$ be a symmetric matrice such that the spectral theorem allows us to write
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T$$
Suppose $forall i in [![ 1, n]!], lambda_i geq 0$.
I want to know if the following proposition is true:
$$A^2 = Poperatorname{Diag}(lambda_1 dots lambda_n)P^T implies A = Poperatorname{Diag}(sqrt{lambda_1} dots sqrt{lambda_n})P^T$$
matrices diagonalization symmetric-matrices
matrices diagonalization symmetric-matrices
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
edited Jan 30 at 22:44
Euler Pythagoras
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
asked Jan 30 at 22:25
Euler PythagorasEuler Pythagoras
54212
54212
$begingroup$
yes, it is indeed true
$endgroup$
– pointguard0
Jan 30 at 22:30
$begingroup$
Please notice that my answer was initially only half-true; I presented the correct one now.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
$begingroup$
yes, it is indeed true
$endgroup$
– pointguard0
Jan 30 at 22:30
$begingroup$
Please notice that my answer was initially only half-true; I presented the correct one now.
$endgroup$
– Blazej
Jan 31 at 9:16
$begingroup$
yes, it is indeed true
$endgroup$
– pointguard0
Jan 30 at 22:30
$begingroup$
yes, it is indeed true
$endgroup$
– pointguard0
Jan 30 at 22:30
$begingroup$
Please notice that my answer was initially only half-true; I presented the correct one now.
$endgroup$
– Blazej
Jan 31 at 9:16
$begingroup$
Please notice that my answer was initially only half-true; I presented the correct one now.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is partially wrong; see discussion below
This is false. If you know that $A^2= mathrm{diag}(x_1,...,x_n)$ in some basis, then in this basis $A=mathrm{diag}( pm sqrt{x_1},....,pm sqrt{x_n})$. Thus in general knowing $A^2$ there are $2^n$ choices for (symmetric) $A$. Of course taking square roots becomes unique if you restrict attention to positive matrices.
Correction It was pointed out to me in the comments that answer above is wrong without additional assumptions. Let's take a closer look. We know that $A$ commutes with $A^2$, so (since both are diagonalizable), there exists a common eigenbasis. If $A^2$ has nondegenerate spectrum then this already implies that $A$ is diagonal in the same basis as $A^2$. Otherwise we can group eigenvalues into blocks, so that (perhaps after reshuffling rows and columns) $A^2= mathrm{diag}(y_1,...,y_1,y_2,...,y_2,...,y_n,...y_k)$, where $y_i neq y_j$ for $i neq j$ and $y_i geq 0$ is repeated $g_i$ times. Now $A$ has to leave each eigenspace of $A^2$ invariant, so it is a direct sum of $g_i times g_i$ matrices of the form $sqrt{y_i} S_i$, where $S_i^2 = 1$, $S_i^T = S_i$. For $g_i=1$ the only matrices of this form are $pm 1$ (reproducing the previous answer), but in general this is not the only possibility. For example for $g_1 = 2$ there is the Pauli $x$ matrix $sigma_x = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$.
answered Jan 30 at 22:33
BlazejBlazej
1,632620
$endgroup$
$begingroup$
Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety).
$endgroup$
– loup blanc
Jan 30 at 23:53
$begingroup$
@loupblanc thank you, I corrected mu answer.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is partially wrong; see discussion below
This is false. If you know that $A^2= mathrm{diag}(x_1,...,x_n)$ in some basis, then in this basis $A=mathrm{diag}( pm sqrt{x_1},....,pm sqrt{x_n})$. Thus in general knowing $A^2$ there are $2^n$ choices for (symmetric) $A$. Of course taking square roots becomes unique if you restrict attention to positive matrices.
Correction It was pointed out to me in the comments that answer above is wrong without additional assumptions. Let's take a closer look. We know that $A$ commutes with $A^2$, so (since both are diagonalizable), there exists a common eigenbasis. If $A^2$ has nondegenerate spectrum then this already implies that $A$ is diagonal in the same basis as $A^2$. Otherwise we can group eigenvalues into blocks, so that (perhaps after reshuffling rows and columns) $A^2= mathrm{diag}(y_1,...,y_1,y_2,...,y_2,...,y_n,...y_k)$, where $y_i neq y_j$ for $i neq j$ and $y_i geq 0$ is repeated $g_i$ times. Now $A$ has to leave each eigenspace of $A^2$ invariant, so it is a direct sum of $g_i times g_i$ matrices of the form $sqrt{y_i} S_i$, where $S_i^2 = 1$, $S_i^T = S_i$. For $g_i=1$ the only matrices of this form are $pm 1$ (reproducing the previous answer), but in general this is not the only possibility. For example for $g_1 = 2$ there is the Pauli $x$ matrix $sigma_x = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$.
answered Jan 30 at 22:33
BlazejBlazej
1,632620
$endgroup$
$begingroup$
Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety).
$endgroup$
– loup blanc
Jan 30 at 23:53
$begingroup$
@loupblanc thank you, I corrected mu answer.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
$begingroup$
This is partially wrong; see discussion below
This is false. If you know that $A^2= mathrm{diag}(x_1,...,x_n)$ in some basis, then in this basis $A=mathrm{diag}( pm sqrt{x_1},....,pm sqrt{x_n})$. Thus in general knowing $A^2$ there are $2^n$ choices for (symmetric) $A$. Of course taking square roots becomes unique if you restrict attention to positive matrices.
Correction It was pointed out to me in the comments that answer above is wrong without additional assumptions. Let's take a closer look. We know that $A$ commutes with $A^2$, so (since both are diagonalizable), there exists a common eigenbasis. If $A^2$ has nondegenerate spectrum then this already implies that $A$ is diagonal in the same basis as $A^2$. Otherwise we can group eigenvalues into blocks, so that (perhaps after reshuffling rows and columns) $A^2= mathrm{diag}(y_1,...,y_1,y_2,...,y_2,...,y_n,...y_k)$, where $y_i neq y_j$ for $i neq j$ and $y_i geq 0$ is repeated $g_i$ times. Now $A$ has to leave each eigenspace of $A^2$ invariant, so it is a direct sum of $g_i times g_i$ matrices of the form $sqrt{y_i} S_i$, where $S_i^2 = 1$, $S_i^T = S_i$. For $g_i=1$ the only matrices of this form are $pm 1$ (reproducing the previous answer), but in general this is not the only possibility. For example for $g_1 = 2$ there is the Pauli $x$ matrix $sigma_x = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$.
answered Jan 30 at 22:33
BlazejBlazej
1,632620
$endgroup$
$begingroup$
Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety).
$endgroup$
– loup blanc
Jan 30 at 23:53
$begingroup$
@loupblanc thank you, I corrected mu answer.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
$begingroup$
This is partially wrong; see discussion below
This is false. If you know that $A^2= mathrm{diag}(x_1,...,x_n)$ in some basis, then in this basis $A=mathrm{diag}( pm sqrt{x_1},....,pm sqrt{x_n})$. Thus in general knowing $A^2$ there are $2^n$ choices for (symmetric) $A$. Of course taking square roots becomes unique if you restrict attention to positive matrices.
Correction It was pointed out to me in the comments that answer above is wrong without additional assumptions. Let's take a closer look. We know that $A$ commutes with $A^2$, so (since both are diagonalizable), there exists a common eigenbasis. If $A^2$ has nondegenerate spectrum then this already implies that $A$ is diagonal in the same basis as $A^2$. Otherwise we can group eigenvalues into blocks, so that (perhaps after reshuffling rows and columns) $A^2= mathrm{diag}(y_1,...,y_1,y_2,...,y_2,...,y_n,...y_k)$, where $y_i neq y_j$ for $i neq j$ and $y_i geq 0$ is repeated $g_i$ times. Now $A$ has to leave each eigenspace of $A^2$ invariant, so it is a direct sum of $g_i times g_i$ matrices of the form $sqrt{y_i} S_i$, where $S_i^2 = 1$, $S_i^T = S_i$. For $g_i=1$ the only matrices of this form are $pm 1$ (reproducing the previous answer), but in general this is not the only possibility. For example for $g_1 = 2$ there is the Pauli $x$ matrix $sigma_x = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$.
answered Jan 30 at 22:33
BlazejBlazej
1,632620
$endgroup$
This is partially wrong; see discussion below
This is false. If you know that $A^2= mathrm{diag}(x_1,...,x_n)$ in some basis, then in this basis $A=mathrm{diag}( pm sqrt{x_1},....,pm sqrt{x_n})$. Thus in general knowing $A^2$ there are $2^n$ choices for (symmetric) $A$. Of course taking square roots becomes unique if you restrict attention to positive matrices.
Correction It was pointed out to me in the comments that answer above is wrong without additional assumptions. Let's take a closer look. We know that $A$ commutes with $A^2$, so (since both are diagonalizable), there exists a common eigenbasis. If $A^2$ has nondegenerate spectrum then this already implies that $A$ is diagonal in the same basis as $A^2$. Otherwise we can group eigenvalues into blocks, so that (perhaps after reshuffling rows and columns) $A^2= mathrm{diag}(y_1,...,y_1,y_2,...,y_2,...,y_n,...y_k)$, where $y_i neq y_j$ for $i neq j$ and $y_i geq 0$ is repeated $g_i$ times. Now $A$ has to leave each eigenspace of $A^2$ invariant, so it is a direct sum of $g_i times g_i$ matrices of the form $sqrt{y_i} S_i$, where $S_i^2 = 1$, $S_i^T = S_i$. For $g_i=1$ the only matrices of this form are $pm 1$ (reproducing the previous answer), but in general this is not the only possibility. For example for $g_1 = 2$ there is the Pauli $x$ matrix $sigma_x = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}$.
answered Jan 30 at 22:33
BlazejBlazej
1,632620
edited Jan 31 at 9:15
answered Jan 30 at 22:33
BlazejBlazej
1,632620
answered Jan 30 at 22:33
BlazejBlazej
1,632620
answered Jan 30 at 22:33
BlazejBlazej
1,632620
1,632620
$begingroup$
Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety).
$endgroup$
– loup blanc
Jan 30 at 23:53
$begingroup$
@loupblanc thank you, I corrected mu answer.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
$begingroup$
Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety).
$endgroup$
– loup blanc
Jan 30 at 23:53
$begingroup$
@loupblanc thank you, I corrected mu answer.
$endgroup$
– Blazej
Jan 31 at 9:16
$begingroup$
Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety).
$endgroup$
– loup blanc
Jan 30 at 23:53
$begingroup$
Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety).
$endgroup$
– loup blanc
Jan 30 at 23:53
$begingroup$
@loupblanc thank you, I corrected mu answer.
$endgroup$
– Blazej
Jan 31 at 9:16
$begingroup$
@loupblanc thank you, I corrected mu answer.
$endgroup$
– Blazej
Jan 31 at 9:16
add a comment |
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$begingroup$
yes, it is indeed true
$endgroup$
– pointguard0
Jan 30 at 22:30
$begingroup$
Please notice that my answer was initially only half-true; I presented the correct one now.
$endgroup$
– Blazej
Jan 31 at 9:16