What is $\det{(I-\alpha vv^T)}$? Let $v \in \mathbb{R^n}$ and $\alpha \in \mathbb{R}$
(a) What is the determinant of the matrix $M = I - \alpha vv^T$, where $I$ is the $n \times n$ identity matrix?
(b) For what values $\alpha$ is $M$ nonsingular?
For (a) I used a theorem: $\det(I + uv^T) = 1 + u^Tv$.
So,  $\det(I-\alpha vv^T)= 1 -(\alpha v)^Tv = 1 - v^T\alpha^Tv$
We get for $\alpha$ orthogonal, $-\|v\|_{2}^2 \leq v^T\alpha^Tv \leq \|v\|_{2}^2$
Thus, $1-\|v\|_{2}^2 \leq \det{(I-\alpha vv^T)} \leq 1+ \|v\|_{2}^2$.
I don't know how I can answer (b).

For the first part, you can have
$\det(I-\alpha vv^T)= 1 - v^T\alpha^Tv= 1-\alpha v^Tv=1-\alpha||v||^2$
Then put it equal to $0$ and solve it for $\alpha$, to find values of $\alpha$ for which the matrix is singular.
