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?
 A: $\,\exists\;\lambda\in\Bbb R\,$  s.t.:
$$\lambda u=Au=\left(I+2vv^t\right)u=u+2vv^t(u)\iff 2vv^t(u)=(\lambda-1)u\iff$$
$$\iff(2vv^t-(\lambda-1)I)u=0\iff \det(2vv^t-(\lambda -1)I)=0\ldots$$
A: The matrix $vv^T$ is real symmetric, so it's diagonalizable, then there's an invertible matrix $P$ and a diagonal matrix $D$ such that
$$vv^T=PDP^{-1}.$$
Since the rank of $vv^T$ is $0$ or $1$ then $D=diag(||v||^2,0,\ldots,0).$
Now, we have:
$$A=I+2vv^T=P(I+2D)P^{-1}$$
so $A$ is diagonalizable in the same basis of eigenvectors than $vv^T$ and has the eigenvalues $1+2||v||^2,1,\ldots,1.$
A: if $A=A^T$, then


*

*$\lambda(2A)=2\lambda(A)$ 

*$\lambda(I+A)=1+\lambda(A)$

*if also $A=vv^T$, $\lambda(A)=\{
||v||^2,0,\ldots,0\}$
A: 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 $\mathbb{R}v$.
