A short one on matrix multiplication

[Päällikkö]

Let A and B be real invertible n x n matrices so that B = (A^T)^-1.
Show that B_m = (I - B₁A₁^T)(A_m^T)⁺, where
B₁ = [b₁, ..., b_m],
B_m = [b_m+1, ..., b_n],
A₁ = [a₁, ..., a_m],
A_m = [a_m+1, ..., a_n].


Any pointers on how one would go about proving the above? I'm fresh out of ideas.

I'm not sure if the above statement is even actually true. Using random matrices in matlab, I have not found any counterexamples, though.

 

[Dick]

What does the '+' symbol exponent on a column vector mean?

 

[Päällikkö]

I'm not sure if that's a hint or a question: the pseudoinverse.

 

[Dick]

It's a question - just trying to understand the problem.

 

[Hurkyl]

So, as block matrices,

A = [ A₁ | A_m ]
B = [ B₁ | B_m ]

Have you tried just multiplying things out in block form?

 

[Päällikkö]

How do you get to a situation where you get to use the matrices A and B?

Do you mean something like
(I - B₁A₁^T)(A_m^T)⁺ = (I - BF(AF)^T)((AG)^T)⁺,
where F = [I O]^T (where I is m x m and F is n x m) and G = [O I]^T (where I is (n-m) x (n-m) and G is n x (n-m)).

I doubt you meant that, could you elaborate? Anyways, the above I've tried to no avail.

EDIT: or do you mean starting from the equation B = [B₁ B_m] = (A^T)^-1 = ([A₁ A_m]^T)^-1 and trying to solve for B_m? That I haven't actually tried but it seems a bit like a long shot. I'll see if I can get it going.

EDIT: No, can't get it going.

 

[Dick]

Ok, try getting it going this way. Look at matrices as blocks, as Hurkyl suggests. Your premise shows B.A^T=1. The partitioning splits this into B1.A1^T+Bm.Am^T=1. So Bm.Am^T=1-B1.A1^T. Now you just have to show that (Am^T)^+ is a good enough inverse to Am^T that you can multiply both sides by it and get what you want. I'm still scratching my head over how to write this down formally, but maybe you can figure it out.

 

[Dick]

Yeah, and the last step is easy as well. Am^T has linearly independent rows. Writing out (Am^T)^+ explicitly (as in the wikipedia entry on pseudoinverses) and then multiplying Am^T by it gives an n-mxn-m unit matrix.

 

[Päällikkö]

Ah indeed, partitioning it in that manner makes it all clear. Thank you very much.