# Proving symmetric matrices are diagonalizable using fact eigenvectors must be orthogonal

I'm treating the fact that we know eigenvectors of symmetric matrices corresponding to distinct eigenvalues must be orthogonal as a given and trying to show (real) symmetric matrices are diagonalizable.

Knowing the aforementioned fact, we can conclude that there exists an orthogonal basis of eigenvectors for any symmetric matrix. We know a matrix A is diagonalizable iff there exists a basis of eigenvectors. So therefore, symmetric matrices are diagonalizable. ∎

I'm not sure if I'm making a bit of a logical leap when I can conclude that there exists an orthogonal basis of eigenvectors, would appreciate any criticism/advice.

• The orthogonality of the eigenvectors is not true in general, but in the case of symmetric matrices, the eigenvectors corresponding to distinct eigenvalues are always orthogonal. – Peter Sep 17 '17 at 22:33
• I think you are. You would need a lemma to say "Any symmetric matrix with characteristic polynomial of the form $f(t)=(t-\lambda)^{n},\,\lambda\in\mathbb{R},$ is diagonalizable." Otherwise it might happen that a symmetric matrix has an eigenvalue with multiplicity $>1,$ and only has one (lin. ind.) eigenvector for that eigenvalue. – RideTheWavelet Sep 17 '17 at 22:37
• Knowing that the eigenvectors are orthogonal does not entitle you to conclude that there are enough linearly independent eigenvectors to form a basis. There is some nontrivial work left to do to reach that desired conclusion. – Gerry Myerson Sep 17 '17 at 23:11

• The one-dimensional case is obvious, since a $1\times 1$ matrix multiplication is the same as a scalar one.
• Suppose all $n \times n$ symmetric matrices are diagonalisable. If $A$ is $(n+1) \times (n+1)$ and symmetric, we know it has an eigenvector $v$. Suppose that $u \in v^{\perp}$, that is, the space of $u$ with $\langle u,v\rangle = 0$. Then $$\langle u,Av\rangle = \langle u,\lambda v\rangle = \lambda \langle u,v\rangle = 0.$$ So $A(v^{\perp}) \subseteq v^{\perp}$, and therefore we can define an operator $B:v^{\perp} \to v^{\perp}$ by $Bu=Au$. This is of course also symmetric. But $v^{\perp}$ is $n$-dimensional, so given a basis of $v^{\perp}$, $B$ is represented for this basis by an $n \times n$ symmetric matrix. But this is diagonalisable by the induction hypothesis. Hence the whole operator is, since we can write it as $A = B + \lambda v\langle v, -\rangle$ where $B$ is extended to $v$ by $Bv=0$.