Consider the matrix $A$ given by $A=I-alpha vv^{T}$ we want to show that there are two distinct eigenvalues $lambda_{1},lambda_{2}$ to be found with their corresponding eigenvectors $x_{lambda_{1}}$ and $x_{lambda_{2}}$.

**My attempt** : By definition, we have that $Ax=lambda x$ thus :

$$

(I-alpha vv^{T})x=x-xalpha vv^{T}=x-(alpha v^{T}x)v

$$

One can easily notice that $v$ is nothing but a scalar multiple of $x$ that is to say $x=beta v$ and thus we have that for $x=v$ we get :

$$

Av=(1-alpha v^{T}v)v

$$

Thus, an eigenvalue of $A$ is $lambda_{1}=1-alpha v^{T}v$. I am unable to find the second eigenvalue nor the corresponding eigenvectors. I would truly appreciate help as I am lost in the process.