Does a square matrix always commute with its transpose? [duplicate]
$begingroup$
This question already has an answer here:
Commutativity of matrix and its transpose
2 answers
Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.
linear-algebra
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
asked Jan 2 at 23:24
Yibei HeYibei He
3139
$endgroup$
marked as duplicate by amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 23:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Commutativity of matrix and its transpose
2 answers
Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.
linear-algebra
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
asked Jan 2 at 23:24
Yibei HeYibei He
3139
$endgroup$
marked as duplicate by amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 23:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28
$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28
add a comment |
$begingroup$
This question already has an answer here:
Commutativity of matrix and its transpose
2 answers
Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.
linear-algebra
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
asked Jan 2 at 23:24
Yibei HeYibei He
3139
$endgroup$
This question already has an answer here:
Commutativity of matrix and its transpose
2 answers
Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.
This question already has an answer here:
Commutativity of matrix and its transpose
2 answers
linear-algebra
linear-algebra
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
asked Jan 2 at 23:24
Yibei HeYibei He
3139
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
asked Jan 2 at 23:24
Yibei HeYibei He
3139
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
edited Jan 7 at 21:46
Rohit Pandey
1,4171023
1,4171023
asked Jan 2 at 23:24
Yibei HeYibei He
3139
asked Jan 2 at 23:24
Yibei HeYibei He
3139
asked Jan 2 at 23:24
Yibei HeYibei He
3139
3139
marked as duplicate by amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 23:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by amWhy linear-algebra
Users with the linear-algebra badge can single-handedly close linear-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
Jan 2 at 23:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28
$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28
add a comment |
$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28
$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28
$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28
$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28
$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28
$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Counterexample:
$$
M = pmatrix{0&1\0&0}
$$
More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
add a comment |
$begingroup$
Here is a simple counter example: Let
$
M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
$
then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.
A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
A general version of the Spectral Theorem states that $M$ is normal if and only
if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
$M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)
answered Jan 2 at 23:36
12131213
1967
$endgroup$
$begingroup$
This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
$endgroup$
– Omnomnomnom
Jan 3 at 0:02
$begingroup$
Corrected, thanks!
$endgroup$
– 1213
Jan 3 at 14:28
add a comment |
$begingroup$
No. Here is a counter example (python code):
import numpy as np
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
np.dot(m,m.T)[0,0]
np.dot(m.T,m)[0,0]
The first entry is 14 and the second is 3.21/
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Counterexample:
$$
M = pmatrix{0&1\0&0}
$$
More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
add a comment |
$begingroup$
Counterexample:
$$
M = pmatrix{0&1\0&0}
$$
More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
add a comment |
$begingroup$
Counterexample:
$$
M = pmatrix{0&1\0&0}
$$
More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
$endgroup$
Counterexample:
$$
M = pmatrix{0&1\0&0}
$$
More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
answered Jan 2 at 23:29
OmnomnomnomOmnomnomnom
128k791185
128k791185
add a comment |
add a comment |
$begingroup$
Here is a simple counter example: Let
$
M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
$
then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.
A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
A general version of the Spectral Theorem states that $M$ is normal if and only
if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
$M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)
answered Jan 2 at 23:36
12131213
1967
$endgroup$
$begingroup$
This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
$endgroup$
– Omnomnomnom
Jan 3 at 0:02
$begingroup$
Corrected, thanks!
$endgroup$
– 1213
Jan 3 at 14:28
add a comment |
$begingroup$
Here is a simple counter example: Let
$
M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
$
then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.
A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
A general version of the Spectral Theorem states that $M$ is normal if and only
if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
$M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)
answered Jan 2 at 23:36
12131213
1967
$endgroup$
$begingroup$
This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
$endgroup$
– Omnomnomnom
Jan 3 at 0:02
$begingroup$
Corrected, thanks!
$endgroup$
– 1213
Jan 3 at 14:28
add a comment |
$begingroup$
Here is a simple counter example: Let
$
M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
$
then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.
A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
A general version of the Spectral Theorem states that $M$ is normal if and only
if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
$M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)
answered Jan 2 at 23:36
12131213
1967
$endgroup$
Here is a simple counter example: Let
$
M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
$
then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.
A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
A general version of the Spectral Theorem states that $M$ is normal if and only
if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
$M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)
answered Jan 2 at 23:36
12131213
1967
edited Jan 3 at 14:27
answered Jan 2 at 23:36
12131213
1967
answered Jan 2 at 23:36
12131213
1967
answered Jan 2 at 23:36
12131213
1967
1967
$begingroup$
This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
$endgroup$
– Omnomnomnom
Jan 3 at 0:02
$begingroup$
Corrected, thanks!
$endgroup$
– 1213
Jan 3 at 14:28
add a comment |
$begingroup$
This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
$endgroup$
– Omnomnomnom
Jan 3 at 0:02
$begingroup$
Corrected, thanks!
$endgroup$
– 1213
Jan 3 at 14:28
$begingroup$
This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
$endgroup$
– Omnomnomnom
Jan 3 at 0:02
$begingroup$
This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
$endgroup$
– Omnomnomnom
Jan 3 at 0:02
$begingroup$
Corrected, thanks!
$endgroup$
– 1213
Jan 3 at 14:28
$begingroup$
Corrected, thanks!
$endgroup$
– 1213
Jan 3 at 14:28
add a comment |
$begingroup$
No. Here is a counter example (python code):
import numpy as np
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
np.dot(m,m.T)[0,0]
np.dot(m.T,m)[0,0]
The first entry is 14 and the second is 3.21/
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
$endgroup$
add a comment |
$begingroup$
No. Here is a counter example (python code):
import numpy as np
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
np.dot(m,m.T)[0,0]
np.dot(m.T,m)[0,0]
The first entry is 14 and the second is 3.21/
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
$endgroup$
add a comment |
$begingroup$
No. Here is a counter example (python code):
import numpy as np
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
np.dot(m,m.T)[0,0]
np.dot(m.T,m)[0,0]
The first entry is 14 and the second is 3.21/
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
$endgroup$
No. Here is a counter example (python code):
import numpy as np
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
np.dot(m,m.T)[0,0]
np.dot(m.T,m)[0,0]
The first entry is 14 and the second is 3.21/
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
answered Jan 2 at 23:29
Rohit PandeyRohit Pandey
1,4171023
1,4171023
add a comment |
add a comment |
$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28
$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28