Find the eigenspaces corresponding respectively to each eigenvector

Let $u$ and $v$ be $2$ non-zero (column) vectors in $mathbb R^n$. Let $A = I- uv^T$ where $I$ is the $n times n$ identity matrix.

a) Prove that $A$ has $2$ eigenvalues $λ_1 = 1$ and $λ_2 = 1-v^Tu$.
Which property should the vectors $u$ and $v$ satisfy
in order for $A$ to be singular?

b) Find the eigenspaces (the set of eigenvectors) $E_1$ and $E_2$ corresponding respectively to $λ_1 = 1$ and $λ_2 = 1 – v^Tu$.

My attempt:

a) Consider the matrix $B=uv^T$ first. When $nge2$, since the rank of $B$ is at most one, $B$ always has a zero eigenvalue and it has at most one nonzero eigenvalue. If $x$ is an eigenvector of $B$ corresponding to a nonzero eigenvalue $gamma$, then $(v^Tx)u=uv^Tx=gamma x$. Therefore $u$ is necessarily a scalar multiple of $x$, meaning that we can take $x=u$. Conversely, $u$ is indeed an eigenvector of $B$ corresponding to the eigenvalue $gamma=v^Tu$. Therefore the eigenvalues of $B$ are $0$ and $v^Tu$. In turn, the eigenvalues of $I-uv^T=I-B$ are $1$ and $1-v^Tu$.

for $A$ to be singular, its $det = 0$,so $det(I- uv^T)=0$, but $det(I- uv^T)= 1 – v^Tu$, hence $1-v^Tu = 0$, which implies that $v^Tu = 1$

b) I know I should solve $(I – uv^T)x=lambda x$. But how to solve this for my two $lambda$‘s. Any help please? Thank you