How do you prove that a symmetric matrix is always diagonalizable even though there are eigenvalues whose multiplicity $k\ge2$?

Question

I want to make sure that the symmetric matrix is always diagonalizable. So I'm studying proof 2 here. But I cannot understand why the following formula holds. $$0=v^*(A-\lambda_iI)^2v=v^*(A-\lambda_iI)(A-\lambda_iI)\ne0$$ Where is $v$ in the second equation? Is there any other easy-to-understand proof of the diagonalization of the symmetric matrix?

Answer 1

Yes, the second equation should end with v. Then, for $M=A-\lambda I$, which is symmetric, $$\vec{0} = M^2v,\text{ so }\\0=v^*MMv = v^*M^*Mv = (Mv)^*(Mv) = |Mv|^2 \neq 0 $$

Another explanation is here:

http://www.quandt.com/papers/basicmatrixtheorems.pdf

Answer 2

The $v$ has disappeared in a typo, and that line should've been $$ v^*(A-\lambda_iI)(A-\lambda_iI)v = \|(A-\lambda_iI)v\|^2 $$ The right-hand side is gotten precisely because $A$ is symmetric.

We see this is gives a contradiction because we assumed that $v$ was a generalized eigenvector of order $2$, and thus the right-hand side is non-zero.

Answer 3

The following proof does not require introduction of generalized eigenvectors and demonstrates that all eigenvectors of a normal matrix have order (rank) 1.

Assume that $n\times n$ normal matrix $A$ has an eigenvector $v$ of order (rank) $m>1$ with associated eigenvalue $\lambda$.

In terms of the matrix $$ M=A-\lambda I $$ this means $$ M^m\nu=0; \forall \{k: 1\le k<m\}: M^k\nu\ne0. $$

Define $$ l=\left\lceil\frac{m}{2}\right\rceil.\tag{1} $$ Observe that for odd $m$: $$ M^{2l}\nu=M^{m+1}\nu=MM^m\nu=0. $$ The same equality $M^{2l}\nu=0$ is valid for even $m$, so that one can write irrespective of parity: $$ 0=\nu^*M^{2l}\nu=\nu^*M^{l}M^l\nu=\|M^l\nu\|^2\ne0,\tag{2} $$ obtaining a contradiction. In (2) we used the facts that an integer power of a normal matrix is normal and that $l$ defined in (1) is strictly less than $m$ for $m>1$.