How to prove real symmetric matrix can be diagonalized via my approach? My idea of proving every real symmetric matrix can be diagonalized is that, first prove two eigenvectors with different eigenvalues must be orthogonal, then I failed to prove that all the eigenvectors span the whole vector space.
To be specific, my question is, if $A$ is a real symmetric $n\times n$ matrix, let $p(t)=\det(tI-A)$ be the characteristic polynomial of $A$, and $\lambda$ be some eigenvalue of $A$, and $\lambda$ is a root of $p(t)$ of order $k$, then how to prove $\dim (\ker(\lambda I-A))=k$?
 A: There are obviously many ways to prove your statement. Some of the comments suggest to use the following:
Let $v$ be an eigenvector to the eigenvalue $\lambda$. Set $U=\mathbb R v$ and write $\mathbb R^n=U\bot U^\bot$. Then $A(U)\subseteq U$ and $A(U^\bot)\subseteq U^\bot$ (you use symmetry here). Therefore we can restrict $A$ to $U^\bot$, get a symmetric matrix and proceed by induction. This is somewhat the standard proof of the spectral theorem.
Another way - I think a way closer to what you asked - would be the following lemma:
If $x\in Ker(\lambda I-A)^k$, then $x\in Ker(\lambda I-A)$. 
For simplicity I will proof the case $k=2$. Let $x\in Ker(\lambda I-A)^2$ and $y=(\lambda I-A)x$. We want to show $y=0$. We have
$$\lambda x=Ax+y$$
and
$$\lambda y=Ay$$
It follows
$$\lambda \left<y,x\right >=\left<y,Ax+y\right >=\left<y,Ax\right >+\left<y,y\right >=\left<Ay,x\right >+\left<y,y\right >=\lambda\left<y,x\right >+\left<y,y\right >$$
Which means $y=0$. 
If you believe that there is a basis of generalised eigenvectors you are done.
