non-singular matrix. Show if $a^TAa \neq -1$

Question

Let $A$ a nonsingular $n\times n$ matrix and $a$ $\in \mathbb{R}^n$. Show that if $a^TAa \neq -1$, then

$(A+aa^T)^{-1}= A^{-1} -\frac{1}{1+a^TAa}A^{-1}aa^TA^{-1}$

I just don't see how I can apply any invertible matrixes properties here. What approach should I take?

If I want to show $X^{-1}$ is $Y$, I try multiplying out $XY$ and $YX$ to see if I get the identity. Actually, I only need to compute one of $XY$ or $YX$, because with matrices, a left inverse is always a right inverse. — Stephen Montgomery-Smith, Sep 06 '20 at 00:14
@JCAA I changed my comment before you posted yours to incorporate what you said. You are correct that there must be some mistake in the OPs question. — Stephen Montgomery-Smith, Sep 06 '20 at 00:22

score 2 · Answer 1 · answered Sep 06 '20 at 00:14

2

Hint: Try multiplying out $$(A+aa^T)\left(A^{-1} -\frac{1}{1+a^TAa}A^{-1}aa^TA^{-1}\right)$$ and see what you get. Note that you may have to appropriately group terms to simplify this to get the identity matrix.

answered Sep 06 '20 at 00:14

JimmyK4542

55,969

score 1 · Answer 2 · answered Sep 06 '20 at 00:19

1

Hint Try using the following the Woodbury matrix identity, https://en.wikipedia.org/wiki/Woodbury_matrix_identity.

\begin{align} (A+UCV)^{-1} = A^{-1}-A^{-1}U(C^{-1}+VA^{-1}U)^{-1}VA^{-1} \end{align}

answered Sep 06 '20 at 00:19

dgadjov

1,349

score 1 · Answer 3 · answered Sep 06 '20 at 01:05

1

This is called the Sherman–Morrison formula.

Check the statement and the proof in the linked Wikipedia article.

answered Sep 06 '20 at 01:05

non-singular matrix. Show if $a^TAa \neq -1$

3 Answers3