I'm struggling to grasp what should be a trivial property of singular value decomposition. Say that I have a linear transformation T that is non-singular (i.e. [tex]T^{-1}[/tex] exists) and relates matrices A and B:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]B = T A[/tex]

or

[tex]A = T^{-1} B[/tex]

What I would like to know is how the singular values of A and B are related?

A and B are both dimension (m x n), i.e. T is square (m x m). My thoughts are:

[tex]A = U_A \Lambda_A V_A^T[/tex]

[tex]B = U_B \Lambda_B V_B^T[/tex]

where U is (m x m), [tex]\Lambda[/tex] is (m x n) and V is (n x n). Substitution gives

[tex]T A = U_B \Lambda_B V_B^T[/tex]

[tex]T U_A \Lambda_A V_A^T = U_B \Lambda_B V_B^T[/tex]

but I'm not sure where to go next. From the above, can I simply write:

[tex]T U_A = U_B[/tex]

[tex]\Lambda_A = \Lambda_B[/tex]

[tex]V_A = V_B[/tex]

This implies the singular values are identical (with the caveat that T is non-singular) and that only the basis vectors of A are affected by the transformation... this is what I want to show but I'm unsure if I've actually succeeded with the above.

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# Singular Values & Linear Transformations

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