Singular values of a matrix times a diagonal matrix

In summary: However, your singular values and vectors are again the eigenvalues and eigenvectors ofA=A^*=\begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ \sigma_1 & 0 & 0 & \cdots & 0 \end{bmatrix}which is singular and has the obvious EVD.In summary, when dealing with singular circulant matrices, it is more efficient to use an eigenvalue decomposition
  • #1
jdevita
1
0
Hi,

I have been struggling with this problem for a while, and I have not found the answer in textbooks or google. Any help would be very much appreciated.

Suppose I know the singular value decomposition of matrix B, which is a singular, circulant matrix. That is, I know [tex]u_i[/tex], [tex]v_i[/tex], and [tex]\sigma_i[/tex], such that [tex]BB^*v_i = \sigma_i^2v_i[/tex] and [tex]B^*Bu_i = \sigma_i^2u_i[/tex]. Where [tex]B^*[/tex] is the conjugate transpose.

Now let A = DB, where D is a diagonal matrix. Is there any way to determine the singular values and vectors of A from the singular values and vectors of B?

Thank you,
Jason
 
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  • #2
You should take advantage of the special and desirable properties of circulant matrices, namely, use an eigenvalue decomposition (EVD) on B instead of a SVD. Every circulant nxn matrix B has the EVD

[tex]B=W\Lambda W^H[/tex]

where [tex]\Lambda[/tex] is a diagonal matrix of eigenvalues and where the columns of W contain the eigenvectors. W contains the complete basis set for the complex discrete Fourier transform of length n, regardless of details of B. Since

[tex]W^H = W^{-1}[/tex]

it is easy to show that

[tex]B^{-1}=W\Lambda^{-1} W^H[/tex]

and B is singular if has one or more zero eigenvalue. To expand on your result, note that

[tex]BB^H=W\Lambda \Lambda^HW^H=W|\Lambda|^2W^H[/tex]

applied to one of the eigenvectors [tex]w_i[/tex] gives

[tex]BB^Hw_i=|\lambda_i|^2w_i[/tex].

Your singlular values squared [tex]{\sigma_i}^2[/tex] are known to be the eigenvalues of [tex]BB^H[/tex], and comparison to the above shows that they are in fact the eigenvalues squared of B.

Multiplying B by a diagonal matrix D removes the circulant symmetry, and I don't see a simple relation between the expansion of A and that of B.
 
Last edited:

What are singular values of a matrix times a diagonal matrix?

Singular values of a matrix times a diagonal matrix are the square roots of the eigenvalues of the product of the matrix and its conjugate transpose. They represent the scaling factors for the orthogonal basis vectors in the singular value decomposition of the matrix.

How are singular values of a matrix times a diagonal matrix calculated?

The singular values of a matrix times a diagonal matrix can be calculated by first finding the singular value decomposition (SVD) of the original matrix. Then, the singular values of the resulting diagonal matrix are simply the diagonal entries of the matrix.

What is the relationship between singular values of a matrix and its diagonal matrix?

The singular values of a matrix and its diagonal matrix are closely related. The singular values of the diagonal matrix are the diagonal entries of the matrix, and the singular values of the original matrix are the square roots of the eigenvalues of the product of the matrix and its conjugate transpose.

Why are singular values of a matrix times a diagonal matrix important?

Singular values of a matrix times a diagonal matrix have several important applications in linear algebra and data analysis. They can be used to determine the rank of a matrix, identify important features in data, and perform dimensionality reduction techniques such as principal component analysis.

Can the singular values of a matrix times a diagonal matrix be negative?

No, the singular values of a matrix times a diagonal matrix are always non-negative. This is because they are the square roots of eigenvalues, which are always non-negative. In some cases, the singular values may be zero if the matrix is singular.

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