$XA=A^TX$ prove $X$ symmetric matrix The 2019 Stack Overflow Developer Survey Results Are InJordan block and cyclic vector spacesWhich polynomials are characteristic polynomials of a symmetric matrix?Besides being symmetric, when will a matrix have ONLY real eigenvalues?Symmetric Matrices with trace zerofinding two conditions so that for two polynomials, there exists exactly one matrixSignificance of symmetric characteristic polynomials?Cayley-Hamilton equation for a given matrix $A$ and other matricesFind all $4 times 4$ real matrices such that $A^3=I$Testing if given linear space of symmetric matrices contains matrix with given eigenspectrum?Determinant of a symmetric banded matrix (4 diagonals)
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$XA=A^TX$ prove $X$ symmetric matrix
The 2019 Stack Overflow Developer Survey Results Are InJordan block and cyclic vector spacesWhich polynomials are characteristic polynomials of a symmetric matrix?Besides being symmetric, when will a matrix have ONLY real eigenvalues?Symmetric Matrices with trace zerofinding two conditions so that for two polynomials, there exists exactly one matrixSignificance of symmetric characteristic polynomials?Cayley-Hamilton equation for a given matrix $A$ and other matricesFind all $4 times 4$ real matrices such that $A^3=I$Testing if given linear space of symmetric matrices contains matrix with given eigenspectrum?Determinant of a symmetric banded matrix (4 diagonals)
$begingroup$
Let $A$ be a nonderogatory matrix. This means the characteristic polynomial and the minimal polynomial of $A$ are coincide. Or, equivalently, every matrix $X$ that satisfies $XA=AX$ can be written as a polynomial of $A$. If $AX=XA$ and $XA=A^TX$, prove that $X$ is a symmetric (complex) matrix.
So since $X=f(A)$ for some complex polynomial $f(x)$, from $XA=A^TX$ we can deduce $X^2=X^TX$, but only when $X$ is a real matrix, we can have $X$ is symmetric. I wonder if the original question is wrong, that we should add $X$ being a real matrix? I also check some low order nonderogatory matrices but fail to find a counterexample. Can anyone help me? Thanks!
linear-algebra matrices symmetric-matrices
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
asked Mar 23 at 5:52
IdeleIdele
1,020410
$endgroup$
add a comment |
$begingroup$
Let $A$ be a nonderogatory matrix. This means the characteristic polynomial and the minimal polynomial of $A$ are coincide. Or, equivalently, every matrix $X$ that satisfies $XA=AX$ can be written as a polynomial of $A$. If $AX=XA$ and $XA=A^TX$, prove that $X$ is a symmetric (complex) matrix.
So since $X=f(A)$ for some complex polynomial $f(x)$, from $XA=A^TX$ we can deduce $X^2=X^TX$, but only when $X$ is a real matrix, we can have $X$ is symmetric. I wonder if the original question is wrong, that we should add $X$ being a real matrix? I also check some low order nonderogatory matrices but fail to find a counterexample. Can anyone help me? Thanks!
linear-algebra matrices symmetric-matrices
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
asked Mar 23 at 5:52
IdeleIdele
1,020410
$endgroup$
1
$begingroup$
So you have $AX = XA = A^T X$? I'm asking because the first part is missing in the question title, but is mentioned in the body.
$endgroup$
– M. Vinay
Mar 23 at 8:00
add a comment |
$begingroup$
Let $A$ be a nonderogatory matrix. This means the characteristic polynomial and the minimal polynomial of $A$ are coincide. Or, equivalently, every matrix $X$ that satisfies $XA=AX$ can be written as a polynomial of $A$. If $AX=XA$ and $XA=A^TX$, prove that $X$ is a symmetric (complex) matrix.
So since $X=f(A)$ for some complex polynomial $f(x)$, from $XA=A^TX$ we can deduce $X^2=X^TX$, but only when $X$ is a real matrix, we can have $X$ is symmetric. I wonder if the original question is wrong, that we should add $X$ being a real matrix? I also check some low order nonderogatory matrices but fail to find a counterexample. Can anyone help me? Thanks!
linear-algebra matrices symmetric-matrices
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
asked Mar 23 at 5:52
IdeleIdele
1,020410
$endgroup$
Let $A$ be a nonderogatory matrix. This means the characteristic polynomial and the minimal polynomial of $A$ are coincide. Or, equivalently, every matrix $X$ that satisfies $XA=AX$ can be written as a polynomial of $A$. If $AX=XA$ and $XA=A^TX$, prove that $X$ is a symmetric (complex) matrix.
So since $X=f(A)$ for some complex polynomial $f(x)$, from $XA=A^TX$ we can deduce $X^2=X^TX$, but only when $X$ is a real matrix, we can have $X$ is symmetric. I wonder if the original question is wrong, that we should add $X$ being a real matrix? I also check some low order nonderogatory matrices but fail to find a counterexample. Can anyone help me? Thanks!
linear-algebra matrices symmetric-matrices
linear-algebra matrices symmetric-matrices
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
asked Mar 23 at 5:52
IdeleIdele
1,020410
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
asked Mar 23 at 5:52
IdeleIdele
1,020410
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
edited Mar 23 at 6:33
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked Mar 23 at 5:52
IdeleIdele
1,020410
asked Mar 23 at 5:52
IdeleIdele
1,020410
asked Mar 23 at 5:52
IdeleIdele
1,020410
1,020410
1
$begingroup$
So you have $AX = XA = A^T X$? I'm asking because the first part is missing in the question title, but is mentioned in the body.
$endgroup$
– M. Vinay
Mar 23 at 8:00
add a comment |
1
$begingroup$
So you have $AX = XA = A^T X$? I'm asking because the first part is missing in the question title, but is mentioned in the body.
$endgroup$
– M. Vinay
Mar 23 at 8:00
1
1
$begingroup$
So you have $AX = XA = A^T X$? I'm asking because the first part is missing in the question title, but is mentioned in the body.
$endgroup$
– M. Vinay
Mar 23 at 8:00
$begingroup$
So you have $AX = XA = A^T X$? I'm asking because the first part is missing in the question title, but is mentioned in the body.
$endgroup$
– M. Vinay
Mar 23 at 8:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The hypothesis is that $A=SDS^-1$ with $D$ diagonal and all its diagonal entries distinct. From $AX=XA$, it follows that $X=SES^-1$ with $E$ diagonal.
The equality $XA=A^TX$ can be written as
$$
SEDS^-1=S^-TDS^TSES^-1.
$$
Multiplying on the left by $S^T$ and on the right by $S$, we get
$$
S^TSED=DS^TSE.
$$
For any $k$ such that $E_kkne0$, the entries in the above equality give us
$$
(S^TS)_kj D_jj=(S^T)_kjD_kk.
$$
So, $E_kkne0$ and $kne j$ imply $(S^TS)_kj=0$.
Now, if $E_kkne0$ or $E_jjne0$,
$$
(ES^TS)_kj=E_kk(S^TS)_kj=delta_kjE_kk(S^TS)_kk
$$
$$
(S^TSE)_kj=(S^T)_kjE_jj=delta_kj(S^TS)_kkE_jj
$$
and then $(ES^TS)_kj=(S^TSE)_kj$. If $E_kk=E_jj=0$, we also get $(ES^TS)_kj=(S^TSE)_kj$. So
$$
ES^TS=S^TSE.
$$
Multiply by $S^-1$ on the right, and by $S^-T$ on the left, to get
$$
X^T=S^-TES^T=SES^-1=X.
$$
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Yes, but you proved a special case. Consider any Jordan block, it is a nonderogatory matirx but can't similar to diagonal matrix :(
$endgroup$
– Idele
Apr 5 at 4:41
$begingroup$
You are right, but I think the argument survives even if $D $ is a Jordan form with distinct eigenvalues in each block. I'll check sometime tomorrow.
$endgroup$
– Martin Argerami
Apr 5 at 6:18
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
The hypothesis is that $A=SDS^-1$ with $D$ diagonal and all its diagonal entries distinct. From $AX=XA$, it follows that $X=SES^-1$ with $E$ diagonal.
The equality $XA=A^TX$ can be written as
$$
SEDS^-1=S^-TDS^TSES^-1.
$$
Multiplying on the left by $S^T$ and on the right by $S$, we get
$$
S^TSED=DS^TSE.
$$
For any $k$ such that $E_kkne0$, the entries in the above equality give us
$$
(S^TS)_kj D_jj=(S^T)_kjD_kk.
$$
So, $E_kkne0$ and $kne j$ imply $(S^TS)_kj=0$.
Now, if $E_kkne0$ or $E_jjne0$,
$$
(ES^TS)_kj=E_kk(S^TS)_kj=delta_kjE_kk(S^TS)_kk
$$
$$
(S^TSE)_kj=(S^T)_kjE_jj=delta_kj(S^TS)_kkE_jj
$$
and then $(ES^TS)_kj=(S^TSE)_kj$. If $E_kk=E_jj=0$, we also get $(ES^TS)_kj=(S^TSE)_kj$. So
$$
ES^TS=S^TSE.
$$
Multiply by $S^-1$ on the right, and by $S^-T$ on the left, to get
$$
X^T=S^-TES^T=SES^-1=X.
$$
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Yes, but you proved a special case. Consider any Jordan block, it is a nonderogatory matirx but can't similar to diagonal matrix :(
$endgroup$
– Idele
Apr 5 at 4:41
$begingroup$
You are right, but I think the argument survives even if $D $ is a Jordan form with distinct eigenvalues in each block. I'll check sometime tomorrow.
$endgroup$
– Martin Argerami
Apr 5 at 6:18
add a comment |
$begingroup$
The hypothesis is that $A=SDS^-1$ with $D$ diagonal and all its diagonal entries distinct. From $AX=XA$, it follows that $X=SES^-1$ with $E$ diagonal.
The equality $XA=A^TX$ can be written as
$$
SEDS^-1=S^-TDS^TSES^-1.
$$
Multiplying on the left by $S^T$ and on the right by $S$, we get
$$
S^TSED=DS^TSE.
$$
For any $k$ such that $E_kkne0$, the entries in the above equality give us
$$
(S^TS)_kj D_jj=(S^T)_kjD_kk.
$$
So, $E_kkne0$ and $kne j$ imply $(S^TS)_kj=0$.
Now, if $E_kkne0$ or $E_jjne0$,
$$
(ES^TS)_kj=E_kk(S^TS)_kj=delta_kjE_kk(S^TS)_kk
$$
$$
(S^TSE)_kj=(S^T)_kjE_jj=delta_kj(S^TS)_kkE_jj
$$
and then $(ES^TS)_kj=(S^TSE)_kj$. If $E_kk=E_jj=0$, we also get $(ES^TS)_kj=(S^TSE)_kj$. So
$$
ES^TS=S^TSE.
$$
Multiply by $S^-1$ on the right, and by $S^-T$ on the left, to get
$$
X^T=S^-TES^T=SES^-1=X.
$$
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Yes, but you proved a special case. Consider any Jordan block, it is a nonderogatory matirx but can't similar to diagonal matrix :(
$endgroup$
– Idele
Apr 5 at 4:41
$begingroup$
You are right, but I think the argument survives even if $D $ is a Jordan form with distinct eigenvalues in each block. I'll check sometime tomorrow.
$endgroup$
– Martin Argerami
Apr 5 at 6:18
add a comment |
$begingroup$
The hypothesis is that $A=SDS^-1$ with $D$ diagonal and all its diagonal entries distinct. From $AX=XA$, it follows that $X=SES^-1$ with $E$ diagonal.
The equality $XA=A^TX$ can be written as
$$
SEDS^-1=S^-TDS^TSES^-1.
$$
Multiplying on the left by $S^T$ and on the right by $S$, we get
$$
S^TSED=DS^TSE.
$$
For any $k$ such that $E_kkne0$, the entries in the above equality give us
$$
(S^TS)_kj D_jj=(S^T)_kjD_kk.
$$
So, $E_kkne0$ and $kne j$ imply $(S^TS)_kj=0$.
Now, if $E_kkne0$ or $E_jjne0$,
$$
(ES^TS)_kj=E_kk(S^TS)_kj=delta_kjE_kk(S^TS)_kk
$$
$$
(S^TSE)_kj=(S^T)_kjE_jj=delta_kj(S^TS)_kkE_jj
$$
and then $(ES^TS)_kj=(S^TSE)_kj$. If $E_kk=E_jj=0$, we also get $(ES^TS)_kj=(S^TSE)_kj$. So
$$
ES^TS=S^TSE.
$$
Multiply by $S^-1$ on the right, and by $S^-T$ on the left, to get
$$
X^T=S^-TES^T=SES^-1=X.
$$
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
The hypothesis is that $A=SDS^-1$ with $D$ diagonal and all its diagonal entries distinct. From $AX=XA$, it follows that $X=SES^-1$ with $E$ diagonal.
The equality $XA=A^TX$ can be written as
$$
SEDS^-1=S^-TDS^TSES^-1.
$$
Multiplying on the left by $S^T$ and on the right by $S$, we get
$$
S^TSED=DS^TSE.
$$
For any $k$ such that $E_kkne0$, the entries in the above equality give us
$$
(S^TS)_kj D_jj=(S^T)_kjD_kk.
$$
So, $E_kkne0$ and $kne j$ imply $(S^TS)_kj=0$.
Now, if $E_kkne0$ or $E_jjne0$,
$$
(ES^TS)_kj=E_kk(S^TS)_kj=delta_kjE_kk(S^TS)_kk
$$
$$
(S^TSE)_kj=(S^T)_kjE_jj=delta_kj(S^TS)_kkE_jj
$$
and then $(ES^TS)_kj=(S^TSE)_kj$. If $E_kk=E_jj=0$, we also get $(ES^TS)_kj=(S^TSE)_kj$. So
$$
ES^TS=S^TSE.
$$
Multiply by $S^-1$ on the right, and by $S^-T$ on the left, to get
$$
X^T=S^-TES^T=SES^-1=X.
$$
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 24 at 6:56
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
Yes, but you proved a special case. Consider any Jordan block, it is a nonderogatory matirx but can't similar to diagonal matrix :(
$endgroup$
– Idele
Apr 5 at 4:41
$begingroup$
You are right, but I think the argument survives even if $D $ is a Jordan form with distinct eigenvalues in each block. I'll check sometime tomorrow.
$endgroup$
– Martin Argerami
Apr 5 at 6:18
add a comment |
$begingroup$
Yes, but you proved a special case. Consider any Jordan block, it is a nonderogatory matirx but can't similar to diagonal matrix :(
$endgroup$
– Idele
Apr 5 at 4:41
$begingroup$
You are right, but I think the argument survives even if $D $ is a Jordan form with distinct eigenvalues in each block. I'll check sometime tomorrow.
$endgroup$
– Martin Argerami
Apr 5 at 6:18
$begingroup$
Yes, but you proved a special case. Consider any Jordan block, it is a nonderogatory matirx but can't similar to diagonal matrix :(
$endgroup$
– Idele
Apr 5 at 4:41
$begingroup$
Yes, but you proved a special case. Consider any Jordan block, it is a nonderogatory matirx but can't similar to diagonal matrix :(
$endgroup$
– Idele
Apr 5 at 4:41
$begingroup$
You are right, but I think the argument survives even if $D $ is a Jordan form with distinct eigenvalues in each block. I'll check sometime tomorrow.
$endgroup$
– Martin Argerami
Apr 5 at 6:18
$begingroup$
You are right, but I think the argument survives even if $D $ is a Jordan form with distinct eigenvalues in each block. I'll check sometime tomorrow.
$endgroup$
– Martin Argerami
Apr 5 at 6:18
add a comment |
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$begingroup$
So you have $AX = XA = A^T X$? I'm asking because the first part is missing in the question title, but is mentioned in the body.
$endgroup$
– M. Vinay
Mar 23 at 8:00