# Prove that $v_1,v_2,…,v_k$ are eigenvectors if $A=\lambda_1v_1v_1^{T}+\lambda_2v_2v_2^{T}+…+\lambda_kv_kv_k^{T}$

Given are vectors $v_1,v_2,...,v_k \in \mathbb{R}^{n}$ with $k<n$, all orthogonal to the standard inner product on $\mathbb{R}^{n}$. Now take $\lambda_1,\lambda_2,...,\lambda_k \in \mathbb{R}$ and suppose

$A=\lambda_1v_1v_1^{T}+\lambda_2v_2v_2^{T}+...+\lambda_kv_kv_k^{T} \in \mathbb{R}^{n\times n}$

Now I'm supposed to prove that $v_1,v_2,...,v_k$ are eigenvectors of $A$, but I don't know how to go about it. I already tried proving it based on the properties of symmetric matrices, but that didn't seem to get me anywhere.

• Isn't $x x^T$ just the definition of $<x, x> = \| x \|^2$? Should it instead say $A = \sum_{j = 1}^{k} \lambda_j v_{j}^{T} v_j$? – AJY Jan 18 '17 at 12:33
• @AJY No, $x^Tx$ is equal to $\langle x,x\rangle$, while $xx^T$ is a $n\times n$ matrix. – 5xum Jan 18 '17 at 12:34
• Okay, so I had it mixed up. In that case, we should still have $\sum_{j = 1}^k \lambda_j v_j (v_{j}^{T} x) = \sum_{j = 1}^k \lambda_j v_j \left< v_j, x \right>$. – AJY Jan 18 '17 at 12:37

For all $i=1,2,\ldots k$: $$Av_i=\left(\sum_{j=1}^k\lambda_jv_jv_j^T\right)v_i=\sum_{j=1}^k\lambda_jv_j(\underbrace{v_j^Tv_i}_{=\langle v_j,v_i \rangle})\underbrace{=}_{\{v_j\}\text{ orthog. system}}\lambda_iv_i|| v_i||^2=\left(\lambda_i|| v_i||^2\right)v_i.$$

• Is $||v_i||^{2}$ always $1$? I don't exactly understand why. – Simon Jan 18 '17 at 12:55
• Sorry, I read orthonormal instead of orthogonal. I've edited the answer. – Fernando Revilla Jan 18 '17 at 13:00

Hint

Notice that if $(v_i)$ is orthonormal then

$$v_iv_i^Tv_j=v_i(v_i^Tv_j)=(v_i^Tv_j)v_i=\delta_{i,j} v_i$$

Hint:

What is $Av_1$?

Remember, $(A+B)v = Av + Bv$...