Consider the matrix $A$ given by $A=I-\alpha vv^{T}$ with $v\neq0$ and $v\in\mathbb{R}^{n}$ and $\alpha\neq0$. 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 : $$ (I-\alpha vv^{T})x=x-\alpha vv^{T}x=x-(\alpha v^{T}x)v $$ 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.
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