Eigenvalue proof

Question

How to prove the following relation:

$$ \begin{array}{ccc} Au_i=\lambda_i u_i & \Rightarrow & A=\Sigma\lambda_iu_iu_i^T \end{array} $$

My try:

$$ \left\{ \begin{array}{lc} Au_i=\lambda_i u_i & (1)\\ AU=U \Lambda & (2)\\ U^TAU=U^TU\Lambda=\Lambda & (3)\\ UU^TAUU^T=U\Lambda U^T & (4) \\ A=U\Lambda U^T & (5)\\ A=\Sigma \lambda_i u_iu_i^T & (6) \end{array} \right. $$

Is the mentioned proof correct?

Thanks for any suggestions.

Answer 1

I think you have to state that $\{ u_i \in \mathbb{R}^n | i = 1, \ldots , n \}$ forms an orthonormal basis to justify that $U^TU=I$ and $UU^T=I$.

You might like to state explicitly the definition of $U$ and $\Lambda$ as well.

Name: Orthogonal Diagonalization.

Remark: Since you are able to write $A=U\Lambda U^T$, A is symmetric.