Mixing
Polar decomposition of real matrices
$begingroup$
Can we decompose real matrices in a way that is analogous to polar decomposition in the complex case?
What I mean is: given an invertible real matrix $M$, can we always write:
$$
M = OP,
$$
maybe uniquely, where $O$ is orthogonal, and $P$ is symmetric positive-definite?
Thanks.
linear-algebra matrix-decomposition
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
$endgroup$
add a comment |
$begingroup$
Can we decompose real matrices in a way that is analogous to polar decomposition in the complex case?
What I mean is: given an invertible real matrix $M$, can we always write:
$$
M = OP,
$$
maybe uniquely, where $O$ is orthogonal, and $P$ is symmetric positive-definite?
Thanks.
linear-algebra matrix-decomposition
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
$endgroup$
$begingroup$
en.wikipedia.org/wiki/Polar_decomposition
$endgroup$
– Bill Cook
Sep 29 '15 at 11:56
$begingroup$
@BillCook More or less... oh wait, I forgot symmetry!
$endgroup$
– geodude
Sep 29 '15 at 11:57
$begingroup$
The answer is yes. This is a consequence of the spectral theorem, whereby $M^TM$ can be orthogonally diagonalized.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 14:37
$begingroup$
@Omnomnomnom Wow. Can you explain more? (Or make it into an answer?)
$endgroup$
– geodude
Sep 29 '15 at 14:39
add a comment |
$begingroup$
Can we decompose real matrices in a way that is analogous to polar decomposition in the complex case?
What I mean is: given an invertible real matrix $M$, can we always write:
$$
M = OP,
$$
maybe uniquely, where $O$ is orthogonal, and $P$ is symmetric positive-definite?
Thanks.
linear-algebra matrix-decomposition
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
$endgroup$
Can we decompose real matrices in a way that is analogous to polar decomposition in the complex case?
What I mean is: given an invertible real matrix $M$, can we always write:
$$
M = OP,
$$
maybe uniquely, where $O$ is orthogonal, and $P$ is symmetric positive-definite?
Thanks.
linear-algebra matrix-decomposition
linear-algebra matrix-decomposition
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
edited Sep 29 '15 at 11:58
geodude
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
asked Sep 29 '15 at 11:52
geodudegeodude
4,1261343
4,1261343
$begingroup$
en.wikipedia.org/wiki/Polar_decomposition
$endgroup$
– Bill Cook
Sep 29 '15 at 11:56
$begingroup$
@BillCook More or less... oh wait, I forgot symmetry!
$endgroup$
– geodude
Sep 29 '15 at 11:57
$begingroup$
The answer is yes. This is a consequence of the spectral theorem, whereby $M^TM$ can be orthogonally diagonalized.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 14:37
$begingroup$
@Omnomnomnom Wow. Can you explain more? (Or make it into an answer?)
$endgroup$
– geodude
Sep 29 '15 at 14:39
add a comment |
$begingroup$
en.wikipedia.org/wiki/Polar_decomposition
$endgroup$
– Bill Cook
Sep 29 '15 at 11:56
$begingroup$
@BillCook More or less... oh wait, I forgot symmetry!
$endgroup$
– geodude
Sep 29 '15 at 11:57
$begingroup$
The answer is yes. This is a consequence of the spectral theorem, whereby $M^TM$ can be orthogonally diagonalized.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 14:37
$begingroup$
@Omnomnomnom Wow. Can you explain more? (Or make it into an answer?)
$endgroup$
– geodude
Sep 29 '15 at 14:39
$begingroup$
en.wikipedia.org/wiki/Polar_decomposition
$endgroup$
– Bill Cook
Sep 29 '15 at 11:56
$begingroup$
en.wikipedia.org/wiki/Polar_decomposition
$endgroup$
– Bill Cook
Sep 29 '15 at 11:56
$begingroup$
@BillCook More or less... oh wait, I forgot symmetry!
$endgroup$
– geodude
Sep 29 '15 at 11:57
$begingroup$
@BillCook More or less... oh wait, I forgot symmetry!
$endgroup$
– geodude
Sep 29 '15 at 11:57
$begingroup$
The answer is yes. This is a consequence of the spectral theorem, whereby $M^TM$ can be orthogonally diagonalized.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 14:37
$begingroup$
The answer is yes. This is a consequence of the spectral theorem, whereby $M^TM$ can be orthogonally diagonalized.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 14:37
$begingroup$
@Omnomnomnom Wow. Can you explain more? (Or make it into an answer?)
$endgroup$
– geodude
Sep 29 '15 at 14:39
$begingroup$
@Omnomnomnom Wow. Can you explain more? (Or make it into an answer?)
$endgroup$
– geodude
Sep 29 '15 at 14:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer is yes. In fact, we don't need the spectral theorem to prove it.
Suppose that $M$ is a real invertible matrix. Then $M^TM$ is positive definite and has the unique positive semidefinite square root $P = sqrt{M^TM}$. Now, note that $P$ has the property $|Px| = |Mx|$ for all vectors $x$.
If $M$ is invertible, then $P$ is invertible as well, and we have $M = (MP^{-1})P$. We note that $MP^{-1}$ is orthogonal since for all $y = Px$, we have
$$
|MP^{-1}y| = |y|
$$
Thus, we have a polar decomposition with $O =( MP^{-1})$ .
We note that this decomposition is unique. In particular, suppose $M = OP$ for orthogonal $O$ and positive $P$. then
$$
M^TM = PO^TOP = P^2
$$
And so, by the uniqueness of positive definite square roots, $P$ is uniquely determined. Then we can rearrange $M = OP$ to find $O = MP^{-1}$ is also unique determined.
Things get a bit trickier when $M$ is not invertible, but we can still guarantee a (non-unique) polar decomposition.
edited Jan 16 at 16:08
SRJ
1,8081620
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
This can also be derived easily from the singular value decomposition.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 15:01
$begingroup$
Wait...how do you prove that $PM^{-1}P=M$?
$endgroup$
– geodude
Sep 30 '15 at 12:36
$begingroup$
I think I meant to write $MP^{-1}$... the proof is pretty much the same, though. The only other thing we need to show is that $M$ invertible $iff$ $M^TM$ invertible $iff$ $P$ invertible, which isn't too hard.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:10
$begingroup$
See my latest edit.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:12
$begingroup$
Oh yeah, like this it works! Thanks.
$endgroup$
– geodude
Sep 30 '15 at 13:29
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
The answer is yes. In fact, we don't need the spectral theorem to prove it.
Suppose that $M$ is a real invertible matrix. Then $M^TM$ is positive definite and has the unique positive semidefinite square root $P = sqrt{M^TM}$. Now, note that $P$ has the property $|Px| = |Mx|$ for all vectors $x$.
If $M$ is invertible, then $P$ is invertible as well, and we have $M = (MP^{-1})P$. We note that $MP^{-1}$ is orthogonal since for all $y = Px$, we have
$$
|MP^{-1}y| = |y|
$$
Thus, we have a polar decomposition with $O =( MP^{-1})$ .
We note that this decomposition is unique. In particular, suppose $M = OP$ for orthogonal $O$ and positive $P$. then
$$
M^TM = PO^TOP = P^2
$$
And so, by the uniqueness of positive definite square roots, $P$ is uniquely determined. Then we can rearrange $M = OP$ to find $O = MP^{-1}$ is also unique determined.
Things get a bit trickier when $M$ is not invertible, but we can still guarantee a (non-unique) polar decomposition.
edited Jan 16 at 16:08
SRJ
1,8081620
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
This can also be derived easily from the singular value decomposition.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 15:01
$begingroup$
Wait...how do you prove that $PM^{-1}P=M$?
$endgroup$
– geodude
Sep 30 '15 at 12:36
$begingroup$
I think I meant to write $MP^{-1}$... the proof is pretty much the same, though. The only other thing we need to show is that $M$ invertible $iff$ $M^TM$ invertible $iff$ $P$ invertible, which isn't too hard.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:10
$begingroup$
See my latest edit.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:12
$begingroup$
Oh yeah, like this it works! Thanks.
$endgroup$
– geodude
Sep 30 '15 at 13:29
add a comment |
$begingroup$
The answer is yes. In fact, we don't need the spectral theorem to prove it.
Suppose that $M$ is a real invertible matrix. Then $M^TM$ is positive definite and has the unique positive semidefinite square root $P = sqrt{M^TM}$. Now, note that $P$ has the property $|Px| = |Mx|$ for all vectors $x$.
If $M$ is invertible, then $P$ is invertible as well, and we have $M = (MP^{-1})P$. We note that $MP^{-1}$ is orthogonal since for all $y = Px$, we have
$$
|MP^{-1}y| = |y|
$$
Thus, we have a polar decomposition with $O =( MP^{-1})$ .
We note that this decomposition is unique. In particular, suppose $M = OP$ for orthogonal $O$ and positive $P$. then
$$
M^TM = PO^TOP = P^2
$$
And so, by the uniqueness of positive definite square roots, $P$ is uniquely determined. Then we can rearrange $M = OP$ to find $O = MP^{-1}$ is also unique determined.
Things get a bit trickier when $M$ is not invertible, but we can still guarantee a (non-unique) polar decomposition.
edited Jan 16 at 16:08
SRJ
1,8081620
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
This can also be derived easily from the singular value decomposition.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 15:01
$begingroup$
Wait...how do you prove that $PM^{-1}P=M$?
$endgroup$
– geodude
Sep 30 '15 at 12:36
$begingroup$
I think I meant to write $MP^{-1}$... the proof is pretty much the same, though. The only other thing we need to show is that $M$ invertible $iff$ $M^TM$ invertible $iff$ $P$ invertible, which isn't too hard.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:10
$begingroup$
See my latest edit.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:12
$begingroup$
Oh yeah, like this it works! Thanks.
$endgroup$
– geodude
Sep 30 '15 at 13:29
add a comment |
$begingroup$
The answer is yes. In fact, we don't need the spectral theorem to prove it.
Suppose that $M$ is a real invertible matrix. Then $M^TM$ is positive definite and has the unique positive semidefinite square root $P = sqrt{M^TM}$. Now, note that $P$ has the property $|Px| = |Mx|$ for all vectors $x$.
If $M$ is invertible, then $P$ is invertible as well, and we have $M = (MP^{-1})P$. We note that $MP^{-1}$ is orthogonal since for all $y = Px$, we have
$$
|MP^{-1}y| = |y|
$$
Thus, we have a polar decomposition with $O =( MP^{-1})$ .
We note that this decomposition is unique. In particular, suppose $M = OP$ for orthogonal $O$ and positive $P$. then
$$
M^TM = PO^TOP = P^2
$$
And so, by the uniqueness of positive definite square roots, $P$ is uniquely determined. Then we can rearrange $M = OP$ to find $O = MP^{-1}$ is also unique determined.
Things get a bit trickier when $M$ is not invertible, but we can still guarantee a (non-unique) polar decomposition.
edited Jan 16 at 16:08
SRJ
1,8081620
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
The answer is yes. In fact, we don't need the spectral theorem to prove it.
Suppose that $M$ is a real invertible matrix. Then $M^TM$ is positive definite and has the unique positive semidefinite square root $P = sqrt{M^TM}$. Now, note that $P$ has the property $|Px| = |Mx|$ for all vectors $x$.
If $M$ is invertible, then $P$ is invertible as well, and we have $M = (MP^{-1})P$. We note that $MP^{-1}$ is orthogonal since for all $y = Px$, we have
$$
|MP^{-1}y| = |y|
$$
Thus, we have a polar decomposition with $O =( MP^{-1})$ .
We note that this decomposition is unique. In particular, suppose $M = OP$ for orthogonal $O$ and positive $P$. then
$$
M^TM = PO^TOP = P^2
$$
And so, by the uniqueness of positive definite square roots, $P$ is uniquely determined. Then we can rearrange $M = OP$ to find $O = MP^{-1}$ is also unique determined.
Things get a bit trickier when $M$ is not invertible, but we can still guarantee a (non-unique) polar decomposition.
edited Jan 16 at 16:08
SRJ
1,8081620
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
edited Jan 16 at 16:08
SRJ
1,8081620
edited Jan 16 at 16:08
SRJ
1,8081620
edited Jan 16 at 16:08
SRJ
1,8081620
1,8081620
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
answered Sep 29 '15 at 15:00
OmnomnomnomOmnomnomnom
128k791184
128k791184
$begingroup$
This can also be derived easily from the singular value decomposition.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 15:01
$begingroup$
Wait...how do you prove that $PM^{-1}P=M$?
$endgroup$
– geodude
Sep 30 '15 at 12:36
$begingroup$
I think I meant to write $MP^{-1}$... the proof is pretty much the same, though. The only other thing we need to show is that $M$ invertible $iff$ $M^TM$ invertible $iff$ $P$ invertible, which isn't too hard.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:10
$begingroup$
See my latest edit.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:12
$begingroup$
Oh yeah, like this it works! Thanks.
$endgroup$
– geodude
Sep 30 '15 at 13:29
add a comment |
$begingroup$
This can also be derived easily from the singular value decomposition.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 15:01
$begingroup$
Wait...how do you prove that $PM^{-1}P=M$?
$endgroup$
– geodude
Sep 30 '15 at 12:36
$begingroup$
I think I meant to write $MP^{-1}$... the proof is pretty much the same, though. The only other thing we need to show is that $M$ invertible $iff$ $M^TM$ invertible $iff$ $P$ invertible, which isn't too hard.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:10
$begingroup$
See my latest edit.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:12
$begingroup$
Oh yeah, like this it works! Thanks.
$endgroup$
– geodude
Sep 30 '15 at 13:29
$begingroup$
This can also be derived easily from the singular value decomposition.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 15:01
$begingroup$
This can also be derived easily from the singular value decomposition.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 15:01
$begingroup$
Wait...how do you prove that $PM^{-1}P=M$?
$endgroup$
– geodude
Sep 30 '15 at 12:36
$begingroup$
Wait...how do you prove that $PM^{-1}P=M$?
$endgroup$
– geodude
Sep 30 '15 at 12:36
$begingroup$
I think I meant to write $MP^{-1}$... the proof is pretty much the same, though. The only other thing we need to show is that $M$ invertible $iff$ $M^TM$ invertible $iff$ $P$ invertible, which isn't too hard.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:10
$begingroup$
I think I meant to write $MP^{-1}$... the proof is pretty much the same, though. The only other thing we need to show is that $M$ invertible $iff$ $M^TM$ invertible $iff$ $P$ invertible, which isn't too hard.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:10
$begingroup$
See my latest edit.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:12
$begingroup$
See my latest edit.
$endgroup$
– Omnomnomnom
Sep 30 '15 at 13:12
$begingroup$
Oh yeah, like this it works! Thanks.
$endgroup$
– geodude
Sep 30 '15 at 13:29
$begingroup$
Oh yeah, like this it works! Thanks.
$endgroup$
– geodude
Sep 30 '15 at 13:29
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Polar_decomposition
$endgroup$
– Bill Cook
Sep 29 '15 at 11:56
$begingroup$
@BillCook More or less... oh wait, I forgot symmetry!
$endgroup$
– geodude
Sep 29 '15 at 11:57
$begingroup$
The answer is yes. This is a consequence of the spectral theorem, whereby $M^TM$ can be orthogonally diagonalized.
$endgroup$
– Omnomnomnom
Sep 29 '15 at 14:37
$begingroup$
@Omnomnomnom Wow. Can you explain more? (Or make it into an answer?)
$endgroup$
– geodude
Sep 29 '15 at 14:39