Is there a simple proof of the following result? I have been trying to prove the following result:
If $A$ is real symmetric matrix with an eigenvalue $\lambda$ of multiplicity $m$ then $\lambda$ has $m$ linearly independent e.vectors.
 A: Indeed it is as copper.hat pointed out. Pick an eigenvector $v$. Obviously, $A \cdot \text{span}\{v\} = \text{span}\{v\}$. Also, $\langle v, A\cdot \text{span}\{v\}^\bot \rangle = \langle Av, \text{span}\{v\}^\bot \rangle = \langle v, \text{span}\{v\}^\bot \rangle = 0$ thus $A \cdot \text{span}\{v\}^\bot = \text{span}\{v\}^\bot$. Now $v$ is the first element in your orthogonal basis. Repeatedly apply this method to $\text{span}\{v\}^\bot$ to complete the proof.
A: Your statement can fail if the underlying vector space is not an inner product space. For instance, consider
$$A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
over $GF(2)$. We have $\det(\lambda I-A)=(\lambda-1)^2-(-1)^2=\lambda^2-2\lambda+2=\lambda^2$. Therefore the eigenvalue $0$ has multiplicity 2. However, $A$ has only one eigenvector $(1,1)^T$.
For real matrices, you may prove by mathematical induction on the size of the matrix. The 1x1 case is trivial. Suppose $\lambda$ is an eigenvalue of an order-$n$ symmetric matrix $A$ with multiplicity $m$. Let $v_1$ be a corresponding unit eigenvector. Extend $v_1$ to an orthonormal basis $\{v_1,\ldots,v_n\}$ and let $Q=[v_1,\ldots,v_n]$. Then $Q$ is a real orthogonal matrix and $v_1^TAv_j=\lambda v_1^Tv_j=\lambda\delta_{1j}$ for every $j$. Therefore $Q^TAQ=\begin{pmatrix}\lambda\\&B\end{pmatrix}$ for some symmetric matrix $B$, and $\lambda$ is an eigenvalue of $B$ with multiplicity $m-1$. By induction assumption, the dimension of the eigenspace for $\lambda$ in $B$ should be $m-1$. Hence the eigenspaces for $\lambda$ in $Q^TAQ$ and $A$ have dimensions $m$.
