Discuss the eigenvalues and eigenvectors of $A=I+2vv^T$.. I need some help with this question: Let $v\in\mathbb R^n$. Discuss the eigenvalues and eigenvectors of $$A=I+2vv^T$$. Can anyone help me?

If $v=0$, then the matrix $A$ is $I_n$ and the problem is trivial.
Now assume $v\neq 0$.
For every $x$ in the orthogonal of $v$ (ie $v^Tx=0$), which is an $n-1$ dimensional subspace, we have
$$
Ax=x+2vv^Tx=x.
$$
So $1$ is an eigenvalue with mutilplicity at least $n-1$.
Then for $x=v$, we have
$$
Av=v+2vv^Tv=(1+2\|v\|^2)v.
$$
So $1+2\|v\|^2$ is an eigenvalue distinct from $1$. 
Therefore
$$
\sigma(A)=\{1,1+2\|v\|^2\}.
$$
The eigenspace of $1$ is the $n-1$ dimensional subspace $v^\perp=\{x\;;\;v^Tx=0\}$.
The eigenspace of $1+2\|v\|^2$ is the $1$ dimensional subspace $\mathbb{R}v$.
