- #1
grawil
- 3
- 0
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.
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.
