Is a Polar Decomposition of Normal Operators Unique?
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Here I am concerned with finite dimensional complex inner product spaces.
We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)
linear-algebra functional-analysis
asked Dec 15 '18 at 4:19
CF8432CF8432
775
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add a comment |
$begingroup$
Here I am concerned with finite dimensional complex inner product spaces.
We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)
linear-algebra functional-analysis
asked Dec 15 '18 at 4:19
CF8432CF8432
775
$endgroup$
add a comment |
$begingroup$
Here I am concerned with finite dimensional complex inner product spaces.
We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)
linear-algebra functional-analysis
asked Dec 15 '18 at 4:19
CF8432CF8432
775
$endgroup$
Here I am concerned with finite dimensional complex inner product spaces.
We know that the (semi)positive operator in such a decomposition is always unique, and so it's clear that the polar decomposition of every invertible operator is unique. But is it necessarily unique for normal operators in general? (We may as well exclude isometries)
linear-algebra functional-analysis
linear-algebra functional-analysis
asked Dec 15 '18 at 4:19
CF8432CF8432
775
asked Dec 15 '18 at 4:19
CF8432CF8432
775
asked Dec 15 '18 at 4:19
CF8432CF8432
775
asked Dec 15 '18 at 4:19
CF8432CF8432
775
asked Dec 15 '18 at 4:19
CF8432CF8432
775
775
add a comment |
add a comment |
1 Answer
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$begingroup$
The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.
For a concrete example: take
$$
A = pmatrix{1&0\0&0\&&ddots\&&&0}
$$
Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
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1 Answer
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1 Answer
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$begingroup$
The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.
For a concrete example: take
$$
A = pmatrix{1&0\0&0\&&ddots\&&&0}
$$
Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.
For a concrete example: take
$$
A = pmatrix{1&0\0&0\&&ddots\&&&0}
$$
Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.
For a concrete example: take
$$
A = pmatrix{1&0\0&0\&&ddots\&&&0}
$$
Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
The polar decomposition will fail to be unique whenever $A$ is non-invertible, whether or not $A$ is normal. You may find it interesting, however, that the two matrices in the polar decomposition commute if and only if $A$ is normal.
For a concrete example: take
$$
A = pmatrix{1&0\0&0\&&ddots\&&&0}
$$
Note that $UA = AU = A$ whenever $U$ is unitary and has $(1,0,dots,0)$ as its first column.
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
answered Dec 15 '18 at 4:26
OmnomnomnomOmnomnomnom
127k790178
127k790178
add a comment |
add a comment |
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