# Singular Values & Linear Transformations

1. Apr 8, 2009

### grawil

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. $$T^{-1}$$ exists) and relates matrices A and B:

$$B = T A$$

or

$$A = T^{-1} B$$

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:

$$A = U_A \Lambda_A V_A^T$$

$$B = U_B \Lambda_B V_B^T$$

where U is (m x m), $$\Lambda$$ is (m x n) and V is (n x n). Substitution gives

$$T A = U_B \Lambda_B V_B^T$$

$$T U_A \Lambda_A V_A^T = U_B \Lambda_B V_B^T$$

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

$$T U_A = U_B$$

$$\Lambda_A = \Lambda_B$$

$$V_A = V_B$$

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.

2. Apr 9, 2009

### grawil

After a bit more thought... the above is only true if T is orthogonal/unitary. I still think it is possible to relate the singular values of A and TA for the slightly more general case where T is full-rank (i.e. non-singular) but I'm not entirely sure how. Obviously, if T is not of full rank then all bets are off.