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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 $\dim (\ker(\lambda I-A))=k$?

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    Here is one possible way to go about. You should first show that there are only real eigenvalues. Then you know $A$ to be trigonalizable. Then use the Gram-Schmidt process to orthonormalize this basis. Because of how this process is done, the orthonormal basis you end up with is stil a basis of trigonalization. But being an orthonormal basis, the matrix has to be symmetrical. As a result, your matrix is both upper triangular and symmetrical thus is diagonal. At first I wanted to look at symmetrical nilpotent matrices, but this is quicker, albeit a little furthee away from your request.2012-05-03

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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 $\lambda \left=\left=\left+\left=\left+\left=\lambda\left+\left$ Which means $y=0$.

If you believe that there is a basis of generalised eigenvectors you are done.