Why is $Av_j=\lambda_jv_j$? I don't quite understand the solution give to this exercise, so I would like some clarification on that:
Let $v_1,...,v_p \in \mathbb{R}^p$ be orthonormal vectors, and for some $$-1< \lambda_p<\lambda_{p-1}<...<\lambda_2<\lambda_1=1$$
let $$A=\lambda_1v_1v_1^T+...+\lambda_pv_pv_p^T$$
What are the eigenvalues and eigenvectors of $A$?
The solution the book gives is:
For any $j=1,...,p$ we can clearly see that $$Av_j=\lambda_jv_j$$
by the orthonormality. so, the eigenvalues of $A$ are $\lambda_1,...,\lambda_p$ and the eigenvectors are $v_1,...,v_p$.
The thing I don't see clearly is why $$Av_j=\lambda_jv_j$$
Now, if I wouldn't know the spectral decomposition theorem I would be very confused as to why $A$ is equal to $\lambda_1v_1v_1^T+...+\lambda_pv_pv_p^T$, but I don't actually understand why the above is true.
 A: You have the condition that all those $v_i$ are orthonormal. That's nothing else than
$v_i^Tv_j = \delta_{ij}$ the Kronecker-delta ($0$ if $i\neq j$ and $1$ if $i=j$).
So if you calculate the multiplication we get:
$$Av_j = \left(\sum_{i=1}^p \lambda_i v_i v_i^T\right) v_j = \sum_{i=1}^p \lambda_i v_i (v_i^T v_j) =\sum_{i=1}^p \lambda_i v_i \delta_{ij} = \lambda_jv_j + \sum_{i\neq j} \lambda_i \cdot 0 = \lambda_jv_j$$
The multiplication $v_iv_i^T$ is nothing else than a matrix which is symmetric of rank $1$ with $v_i$ horizontally and vertically. E.g. $v_i = \left(\begin{array} &1&5&2\end{array}\right)^T$ will give $v_iv_i^T = \left(\begin{array}& 1 & 5 & 2 \\ 5 & 25 & 10 \\ 2 & 10 & 4 \end{array}\right)$
A: By definition
$$A = \sum_{i=1}^p (\lambda_i v_i v_i^\top).$$
Then:
$$Av_j = \sum_{i=1}^p (\lambda_i v_i v_i^\top)v_j = \sum_{i=1}^p (\lambda_i v_i) (v_i^\top v_j).$$
Since the $v_i$ are orthonormal, then $v_i^\top v_j = 0 ~\forall i \neq j$ and  $v_i^\top v_j = 0 \iff i = j.$
Hence:
$$Av_j = \lambda_j v_j,$$
and hence $\lambda_j$ and $v_j$ are an eigenvalue and an eigenvector of $A$.
