# Matrix Algebra simplify $(A^{T} A)^{-1}A^{T}(B^{-1}A^{T})^{T}B^{T}B^{2}B^{-1}$

I'm sorry, this is probably very basic... I'm trying to review stuff to make sure I dont forget things. The question is simplyfy the below as much as possible: $(A^{T} A)^{-1}A^{T}(B^{-1}A^{T})^{T}B^{T}B^{2}B^{-1}$ You can assume that all matrices inverted in the expression exist.

Ok, I did notice that $A^{-1}$ is not in the expression... so I can't assume it exists.

I don't think i'm too far... and I know the final answer is B (The question comes from Linear Algebra Concepts and Methods, exercise 1.7) $(A^{T} A)^{-1}A^{T}(B^{-1}A^{T})^{T}B^{T}B^{2}B^{-1}=$ $(A^{T} A)^{-1}A^{T}(B^{-1}A^{T})^{T}B^{T}BBB^{-1}=$ $(A^{T} A)^{-1}A^{T}(B^{-1}A^{T})^{T}B^{T}B=$ $A^{-1}(A^{T})^{-1}A^{T}(B^{-1}A^{T})^{T}B^{T}B=$ $A^{-1}(B^{-1}A^{T})^{T}B^{T}B$

I know I can: $A^{-1}(B^{-1}A^{T})^{T}B^{T}B=$$A^{-1}(B^{-1})^{T}AB^{T}B but that doesn't seem to help much... I guess I'm missing something. Thanks! ## 3 Answers Use the following properties: (PQ)^{-1} = Q^{-1}P^{-1} (PQ)^T = Q^TP^T PP^{-1} = I = P^{-1}P (P^T)^T = P Simplify step by step and you should indeed end up with just B. Your last line where you say "I know I can", is not quite correct (you actually can't (in general anyway)!). Hint: Property 3 on this link {(AB)^T} = {B^TA^T} It looks like you just did {(AB)^T} = {A^TB^T} which is not (generally) correct. First of all, let's simplify the transpose in parentheses:$$(B^{-1} A^T)^T = A (B^{-1})^T = A (B^T)^{-1}.$$Note that here we have used the property (AB)^T=B^T A^T: the transpose "distributes" but in the reverse order, just like the inverse. So now you have$$(A^T A)^{-1} A^T A (B^T)^{-1} B^T B^2 B^{-1}.$\$

Can you finish from here?