Is $(I + u\mathbf{v}^\top)^{-1} = I - \frac{u\mathbf{v}^\top}{1+ \mathbf{u}^\top \mathbf{v}}$ in the case that $\mathbf{u}^\top \mathbf{v} \neq -1$?

Question

How can i show that $(I + u\mathbf{v}^\top)^{-1} = I - \frac{u\mathbf{v}^\top}{1+ \mathbf{u}^\top \mathbf{v}}$, in the case that $\mathbf{u}^\top \mathbf{v} \neq -1$?

I is the identity matrix and $u,v \in \mathbb{R} ^{n} $ are column vecotrs.

And if I could show that, would that mean that $(I + u\mathbf{v}^\top) $ would be singular if $\mathbf{u}^\top \mathbf{v} = -1$ because that would mean it is not invertible?

Thank you

Answer 1

I will leave it to you to show that

$$(\mathbf{I} + \mathbf{uv^T})\cdot\left(I - \frac{\mathbf{uv^T}}{1+\mathbf{u^Tv}}\right) = I$$

and the same for its reverse. But let's see what happens if $\mathbf{u^Tv} = -1$. Apply the original matrix to the vector $\mathbf{u}$:

$$(\mathbf{I} + \mathbf{uv^T})\cdot\mathbf{u} = \mathbf{u} + \mathbf{uv^Tu} = \mathbf{u} - \mathbf{u} = \mathbf{0}$$

Thus $\mathbf{u}\in \ker(\mathbf{I} + \mathbf{uv^T})$ and the transformation is not invertible.

Answer 2

Hint

Note that $I+uv^T$ is a matrix with $n-1$ eigenvalues equal to $1$ and an eigenvalue equal to $1+v^Tu$ with the corresponding eigenvector $u$. Hence $(I+uv^T)^{-1}$ has $n-1$ eigenvalues equal to $1$ and an eigenvalue equal to $1\over 1+v^Tu$ and must be represented as $$(I+uv^T)^{-1}=I+xy^T$$where $x$ and $y$ are 1-dimensional vectors. You only need to find suitable vectors $x$ and $y$.