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.



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?


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