Let $A$ be an $n \times n$ symmetric matrix and let $\{(\lambda_i, u_i)\}^n$ be its eigenpairs. Assume all eigenvectors are normalized. Prove that $$A = \sum_{i=1}^n \lambda_i u_i u_i^T$$
I am stuck on this problem. I think I am missing an important proof or definition. First, I set $u_i u_i^T = 1$ since symmetric matrices have an orthonormal basis of eigenvectors, which is orthogonal by definition. So now:
$$ A = \sum_{i=1}^n \lambda_i $$
I also know that I can define a matrix $D$ which is a diagonal matrix with the eigenvalues of A in the entries on the main diagonal, $A = P D P^{-1}$ which has a trace tr(PDP$^-$$^1$)=tr(DP$^-$$^1$P)=tr(D)=$\sum_{i=i}^n \lambda_i$. Unfortunately it is here that I am stuck since I do not believe that I can apply the trace operation to only one side of the equation. Any assistance is most appreciated.
Best Answer
$u_i^Tu_i$ is indeed $1$, but $u_iu_i^T$ is an $n\times n$-matrix.
A hint for the proof: $u_i$ is an orthonormal basis, and a matrix is determined by how it acts on basis vectors. What happens if you multiply $\sum\limits_i \lambda_i u_i u_i^T$ and $u_j$ for some $j$?