Which property should the vectors $u$ and $v$ satisfy in order for $A$ to be singular 
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 $λ_2 = 1-v^Tu$.


b) Which property should the vectors $u$ and $v$ satisfy
in order for $A$ to be singular (not invertible)?

My attempts:
a) $\det(A-\lambda I) = \det(I-uv^T - \lambda I) = \det(I(1-uv^T-\lambda) = \det(I) × \det(1-uv^T-\lambda) = 0$.
Since $\det(I) \not= 0$,then $ \det(1-uv^T-\lambda) = 0 \tag{*}$
So for $\lambda_1 =1$, we replace it in $(*)$, we get $-\det(uv^T) = 0$. So $\lambda_1$ satisfies this equation.
For $\lambda_2 = 1-v^Tu$, we get $\det(1-uv^T-1+v^Tu) = \det(-uv^T+v^Tu)$. But how to show that this is zero?
b) 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 $v^Tu = 1$. Is this correct?
 A: In your answer to part (a), it isn't clear why the two equalities marked in red below hold:
$$
\det(A-\lambda I) = \det(I-uv^T - \lambda I) \color{red}{=} \det(I(1-uv^T-\lambda) \color{red}{=} \det(I) × \det(1-uv^T-\lambda) = 0.
$$
As $uv^T$ is not a scalar, the expression $1-uv^T-\lambda$ does not make sense.
There is actually no need to use any determinant. Consider the matrix $B=uv^T$ first. When $n\ge2$, 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 $1-v^Tu$.
Your answer to part (b) is correct, but you are overthinking the problem. A square matrix is singular if and only if it has a zero eigenvalue. There is no need to use any determinant.
