# Showing a matrix is symmetric and its eigenvectors are in the columns of $(I - 2vv^T)$

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: $$QQ^T=(I - 2vv^T)(I - 2vv^T)=I-2vv^T-2vv^T+4vv^Tvv^T=I$$

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What is $Q^T * Q$? Calculating that might help. – Cocopuffs Jun 16 '12 at 16:37
@Cocopuffs, $Q^TQ$ got me the identity matrix. What does this mean? – Clash Jun 16 '12 at 16:45

This means that $AQ = QDQ^TQ = QD$

By the definition of matrix multiplication, $A$ times the $j$-th column vector of $Q$ gives the $j$-th column vector of $QD$, which can be seen to be $\lambda_j$ times the $j$-th column vector of $Q$.

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thanks for the fast reply cocopuffs! How can I prove that $A$ is symmetric? Is it correct to say that because it's diagonalizable, that it's also symmetric? – Clash Jun 16 '12 at 18:16
Not symmetric matrices can also be diagonalizable... All you need is that $A^T = (QDQ)^T = Q^TD^TQ^T = QDQ = A$, since $Q$ and $D$ are symmetric. – Cocopuffs Jun 16 '12 at 18:17
awesome, thanks! – Clash Jun 16 '12 at 18:32

Since $$QQ^T=Q^TQ=Q^2=I-4vv^T+4v(v^Tv)v^T=I-4vv^T+4vv^T=I,$$ i.e. $Q \in O(N)$, then $u_i^Tu_j=0$ for $i\ne j$ and $u_i^Tu_i=1$, where $u_i$ is the $i$-th column from Q (and consequentely $u_i^T$ is the $i$-th row from $Q$).

It follows that $Qu_i=e_i$ for every $i=1,\ldots, N$, with $e_1,\ldots, e_N$ the canonical basis of $\mathbb{R}^N$. Hence $$Au_i=QDQu_i=QDe_i=Q(\lambda_ie_i)=\lambda_iQe_i=\lambda_iu_i \quad \forall\ i.$$

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Thanks for the reply, but what's $O$? – Clash Jun 16 '12 at 18:33
$O(N)=\{A \in \mathbb{R}^{N\times N}:\ \ A^TA=I=AA^T\}$ – Mercy King Jun 16 '12 at 19:26
Oh it's $O$ from $o$rthogonal? – Clash Jun 16 '12 at 20:18