Skip to main content
formatting; edited tags
Source Link
YCor
  • 63.9k
  • 5
  • 187
  • 285

Is the Singular Value Decompositionsingular value decomposition a measurable function?

Consider$\DeclareMathOperator\svd{svd}$Consider the SVD of rectangular matrices as operators

$$svd:\mathbb C^{n\times m}\to \mathbb U_n\times \mathbb D_{n, m}\times\mathbb U_m$$$$\svd:\mathbb C^{n\times m}\to \mathbb U_n\times \mathbb D_{n, m}\times\mathbb U_m$$

where $\mathbb U_n$ is the space of $n\times n$ unitary matrices and $\mathbb D_{n, m}$ is the space of real nonnegative diagonal rectangular matrices.

We know that this operator is not well defined, since there's more than one SVD for each matrix.

What I want to know is, there exists a specification of $svd$$\svd$ on $\mathbb C^{n\times m}$ that makes it a Borel-measurable function?

I already know that it can't be continuous, and that if I play enough with the signs of the entries, I guess I can make it not even Lebesgue measurable, but what I need here is for just one specification. If I had to try to do it, I guess I'd try to define it locally and then try to glue together the charts, but I feel like it might badly backfire.

(transferred from stackexchange)

Is the Singular Value Decomposition a measurable function?

Consider the SVD of rectangular matrices as operators

$$svd:\mathbb C^{n\times m}\to \mathbb U_n\times \mathbb D_{n, m}\times\mathbb U_m$$

where $\mathbb U_n$ is the space of $n\times n$ unitary matrices and $\mathbb D_{n, m}$ is the space of real nonnegative diagonal rectangular matrices.

We know that this operator is not well defined, since there's more than one SVD for each matrix.

What I want to know is, there exists a specification of $svd$ on $\mathbb C^{n\times m}$ that makes it a Borel-measurable function?

I already know that it can't be continuous, and that if I play enough with the signs of the entries, I guess I can make it not even Lebesgue measurable, but what I need here is for just one specification. If I had to try to do it, I guess I'd try to define it locally and then try to glue together the charts, but I feel like it might badly backfire.

(transferred from stackexchange)

Is the singular value decomposition a measurable function?

$\DeclareMathOperator\svd{svd}$Consider the SVD of rectangular matrices as operators

$$\svd:\mathbb C^{n\times m}\to \mathbb U_n\times \mathbb D_{n, m}\times\mathbb U_m$$

where $\mathbb U_n$ is the space of $n\times n$ unitary matrices and $\mathbb D_{n, m}$ is the space of real nonnegative diagonal rectangular matrices.

We know that this operator is not well defined, since there's more than one SVD for each matrix.

What I want to know is, there exists a specification of $\svd$ on $\mathbb C^{n\times m}$ that makes it a Borel-measurable function?

I already know that it can't be continuous, and that if I play enough with the signs of the entries, I guess I can make it not even Lebesgue measurable, but what I need here is for just one specification. If I had to try to do it, I guess I'd try to define it locally and then try to glue together the charts, but I feel like it might badly backfire.

(transferred from stackexchange)

Source Link
Exodd
  • 201
  • 1
  • 5

Is the Singular Value Decomposition a measurable function?

Consider the SVD of rectangular matrices as operators

$$svd:\mathbb C^{n\times m}\to \mathbb U_n\times \mathbb D_{n, m}\times\mathbb U_m$$

where $\mathbb U_n$ is the space of $n\times n$ unitary matrices and $\mathbb D_{n, m}$ is the space of real nonnegative diagonal rectangular matrices.

We know that this operator is not well defined, since there's more than one SVD for each matrix.

What I want to know is, there exists a specification of $svd$ on $\mathbb C^{n\times m}$ that makes it a Borel-measurable function?

I already know that it can't be continuous, and that if I play enough with the signs of the entries, I guess I can make it not even Lebesgue measurable, but what I need here is for just one specification. If I had to try to do it, I guess I'd try to define it locally and then try to glue together the charts, but I feel like it might badly backfire.

(transferred from stackexchange)