Let $A \in \mathbb{R}^{N \times N}$ have the form $A = (I - 2vv^T)D(I-2vv^T)$ with $D = \text{diag} (\lambda_1, \lambda_2, ..., \lambda_n) \in \mathbb R^{N \times N}, v \in \mathbb{R}^n, v^Tv=1$
Show that $A$ is symmetric and that for $j = 1, ...,N$, the $j$th-column from $I - 2vv^T$ is the respective eigenvector to $\lambda_j$.
Here are my ideas so far. I remember this form $(I - 2vv^T)D(I-2vv^T)$ from matrices that can be diagonalized. This means it's symmetric, right? If I rewrite $(I - 2vv^T)$ as $Q$, then maybe I could do something like $A = QDQ^{T}$, if $Q$ is symmetric. So I guess my first step should be showing that $(I - 2vv^T)=(I - 2vv^T)^T$.
So if I show $Q$ is symmetric, it is proven that $A$ is also symmetric. Am I going in the right direction? How can I show that its eigenvectors are the columns of $Q$? Thanks in advance!
Edit: Here's my proof that $Q$ is symmetric: $Q^T=(I-2vv^T)^T = I^T -2(vv^T)^T =I -2(v^T)^Tv^T =I - 2vv^T=Q$
Edit2: Apparently $Q$ is also orthogonal: $QQ^T=(I - 2vv^T)(I - 2vv^T)=I-2vv^T-2vv^T+4vv^Tvv^T=I$