Let 2 $v \times 1$ column matrices X and Y such that the $v \times v$ matrix $A = I +XY^{T}$ to be invertible.Show that $A^{-1}= I - \frac{1}{1+Y^{T}X}XY^{T}$
My effort:
I have to show that $AA^{-1}=I$. So:
$$AA^{-1}=(I +XY^{T})\left(I - \frac{1}{1+Y^{T}X}XY^{T}\right)$$
\begin{align*} AA^{-1}&=(I +XY^{T})\left(I - \frac{1}{1+Y^{T}X}XY^{T}\right)\\ &= I\left(I - \frac{1}{1+Y^{T}X}XY^{T}\right) + XY^{T}\left(I - \frac{1}{1+Y^{T}X}XY^{T}\right)\\ &= I - \frac{1}{1+Y^{T}X}XY^{T} + XY^{T} - \frac{1}{1+Y^{T}X}XY^{T}XY^{T}\\ \end{align*}
How I proceed further ?
Note that $Y^T$ is a $1\times n$ row vector, and $X$ is a $n\times 1$ column vector. Therefore, $Y^TX=b$, where $b$ is a scalar (a number).
\begin{align*} AA^{-1}&=(I +XY^{T})\left(I - \frac{1}{1+Y^{T}X}XY^{T}\right)\\ \\ &=(I +XY^{T})\left(I - \frac{1}{1+b}XY^{T}\right)\\ \\ &= I+XY^T-\frac{1}{1+b}XY^T-\frac{1}{1+b}XY^TXY^T\\ \\ &= I+XY^T-\frac{1}{1+b}XY^T-\frac{1}{1+b}X(Y^TX)Y^T\\ \\ &= I+XY^T-\frac{1}{1+b}XY^T-\frac{1}{1+b}X(b)Y^T\\ \\ &= I+XY^T-\frac{1}{1+b}XY^T-\frac{b}{1+b}XY^T\\ \\ &= I+\left(1-\frac{1}{1+b}-\frac{b}{1+b}\right)XY^T\\ \\ &= I+\left(0\right)XY^T\\ \\ &=I \end{align*}