# linear algebra – Eigenvalues and Eigenvectors of \$A=I-alpha vv^{T}\$

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.