Eigenvalue of multiplicity k of a real symmetric matrix has exactly k linearly independent eigenvector If A is an nxn real symmetric matrix then A is diagonalisable. In other words, If A is a symmetric nxn matrix, then there exists an orthogonal matrix $P$ such
that $P_{-1}AP=P_{T}AP=D$, a diagonal matrix. The eigenvalues of A lie on the
main diagonal of D.


Proof of this statement must be considered in two cases.


It is clear that all eigenvalues of a real symmetric matrix are real and if they are all distinct, then eigenvectors $x_i$, i=1,2,..,n  corresponding to $\lambda_i$ i=1,2,..,n are orthogonal. We can obtain an orthonormal set of eigenvectors eigenvectors $u_i$, i=1,2,..,n using these eigenvectors $x_i$, i=1,2,..,n Then we can construct matrix $P$ as $P=[u_1 u_2 ... u_n]$ such that $D=P_{-1}AP$.


I have some troubles in the second case that is not all eigenvalues are simple.

How can I prove that eigenvalue of  multiplicity k of a real symmetric matrix A has exactly k linearly independent eigenvectors, i.e., dimension of solution space of $(A-\lambda I)$ is k?
 A: I'd say you have three ways:


*

*Prove first that a symmetric matrix is diagonalizable. Then see that your desired property is true for diagonal matrices and that a similarity transform preserves both algebraic and geometric multiplicities (eg: see prop 7.5 here). I think this is the most usual way.

*Continuity argument: in the space of symmetric matrices, we can go continuosly from different eigenvalues to repeated eigenvalues; because the starting point had orthogonal eigenvectors, these cannot degenerate in LD eigenvectors. This argument is easy to visualize but hard to formalize.

*Prove it directly. Here's an sketch (essentially from here)
Let ${\bf A}$ be a  $n \times n$ symmetric matrix, let $\lambda_i$ be an eigenvalue with (algebraic) multiplicity $1<m \le n$. Then there exists some eigenvector ${\bf p}_{i1}$ with $|{\bf p}_{i1}|=1$.
Let ${\bf B}=({\bf p}_{i1} \, {\bf C})$ be an orthogonal matrix (orthonormal columns) with ${\bf p}_{i1}$ as first column (it can be constructed by Grand-Schmidt process). Then consider
$$ {\bf B}' {\bf A} {\bf B} = 
\begin{pmatrix} \lambda_i &0 \\
0 & {\bf C}' {\bf A} {\bf C}
\end{pmatrix}$$
By considering the characteristic polynomial, we see that (because the multiplicity of $\lambda_i>1$)$|{\bf C}' {\bf A} {\bf C} -\lambda_i I_{n-1}|=0$
Hence there exists some non null ${\bf q}$ with $({\bf C}' {\bf A} {\bf C} -\lambda_i I_{n-1})){\bf q}=0$. Next, see that ${\bf p}_{i2}={\bf C}{\bf q}$ is eigenvector of ${\bf A}$, and is orthogonal to ${\bf p}_{i1}$. We can repeat the procedure (if $m>2$), by rebuilding ${\bf B}$ with this eigenvector as second column, etc.
