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?

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 $1+2||v||^2,1,\ldots,1.$
