How to derive the following formula for the inverse of a matrix?

Question

I came across the following theorem:

Let $A$ be a nonsingular square $p \times p$ matrix and $z$ be a p-dimensional column vector. The matrix $(A - z z^T)^{-1}$ is given by

$$(A- zz^T)^{-1} = A^{-1} + \frac{A^{-1}zz^TA^{-1}}{1-z^T A^{-1}z}$$

Now I tried using $A-zz^T$ multiply the matrix on the right side of the above formula and I cannot obtain an identity matrix. I tried:

$$(A^{-1} + \frac{A^{-1}zz^TA^{-1}}{1-z^T A^{-1}z})(A-zz^T) = I - A^{-1}zz^T - \frac{1}{1-z^TA^{-1}z}(A^{-1}zz^T+A^{-1}zz^TA^{-1}zz^T)$$

This is where I got stuck. Can someone help me on this please?

Answer 1

You've almost got it! There's a small typo in your formula: the last term should be $$ {}+ \frac{1}{1-z^TA^{-1}z}(A^{-1}zz^T-A^{-1}zz^TA^{-1}zz^T). $$ And notice this equals $$ \frac{1}{1-z^TA^{-1}z} \big( A^{-1}z ( 1 - z^TA^{-1}z) z^T \big), $$ which should get you where you want to go.

Answer 2

It is a particular case of Sherman-Morrison theorem

(https://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula)

with $u=z$ and $v=-z$.

Answer 3

Remember that matrix multiplication is associative meaning $(AB)C = A(BC)$ so that

$$(A^{-1}zz^T)(A^{-1}zz^T) = A^{-1}z(z^TA^{-1}z)z^T = (z^TA^{-1}z).A^{-1}zz^T$$

Notice that the last equality is due to the fact that $z^TA^{-1}z$ is a scalar. Going back to your equality we now have:

$$ \begin{alignat}{} (A^{-1} + \frac{A^{-1}zz^TA^{-1}}{1-z^T A^{-1}z})(A-zz^T) &&= I - A^{-1}zz^T + \frac{1}{1-z^TA^{-1}z}(A^{-1}zz^T - A^{-1}zz^TA^{-1}zz^T) \\\\&&= I - A^{-1}zz^T + \frac{1}{1-z^TA^{-1}z}(A^{-1}zz^T - (z^TA^{-1}z).A^{-1}zz^T) \\\\&&= I - A^{-1}zz^T + \frac{A^{-1}zz^T}{1-z^TA^{-1}z}(1 - (z^TA^{-1}z)) \\\\&&= I \end{alignat} $$