0

In other words why is symmetric matrix always diagonalizable? could someone explain intuitively?

Vineel Kumar Veludandi
  • 1,035
  • 1
  • 8
  • 16

1 Answers1

2

Every matrix has an eigenvalue, provided we permit complex eigenvalues. So take an eigenvector $x_1$ of $A$ with eigenvalue $\lambda_1$.

Now suppose $y$ is orthogonal to $x_1$. Then

$$\langle x_1,Ay \rangle = \langle Ax_1,y \rangle = \lambda_1 \langle x_1,y \rangle = 0.$$

That is, if $y$ is orthogonal to $x_1$ then $Ay$ is orthogonal to $x_1$. Put differently, the orthogonal complement of $x_1$, call it $W_1$, is an invariant subspace of $A$. This means we can define a restricted function $A : W_1 \to W_1$. Because this function is a linear map from $W_1$ into itself, it will also have an eigenvalue and eigenvector $x_2 \in W_1$. $x_2$ must then be an eigenvector of $A : \mathbb{C}^n \to \mathbb{C}^n$.

Now do the same thing with $W_1$ to get an invariant subspace $W_2 \subset W_1$, giving an eigenvector $x_3 \in W_2$. Repeating this procedure $n-1$ times gives us $n$ orthogonal (and thus linearly independent) eigenvectors.

Ian
  • 101,645
  • Sounds a little bit like turning round and round though it is much simpler: The characteristic polynomial factors completely over the complex numbers. That means that the eigenvectors provide a basis... – C-star-W-star Jul 19 '14 at 05:00
  • ...but wasn't the question rather that symmetric matrices over the reals do not require the assumption of the existence of eigenvalues? – C-star-W-star Jul 19 '14 at 05:05
  • If $\lambda_1$ is not real, then neither is $x_1$, and talking about orthogonal complement is a bit iffy. Actually the orthogonal complement would then have codimension two, but that's besides my point: You need to include the standard proof for $\lambda_1$ to be real as well for this to be complete argument. – Jyrki Lahtonen Jul 19 '14 at 05:30
  • @Freeze_S: the eigenvectors don't necessarily provide a basis. You do get $n$ eigenvalues algebraically, but even with symmetric matrices these may come with multiplicity, and it is not trivial that the algebraic multiplicity and geometric multiplicity coincide. – Ian Jul 19 '14 at 13:17
  • @JyrkiLahtonen You're right. I think this exposes the intuition however. It also solves the problem in $\mathbb{C}^n$ by itself. – Ian Jul 19 '14 at 13:17
  • @JyrkiLahtonen In fact, I think the eigenvalues and eigenvectors being real is actually an unrelated and simple problem. Consequently I've rephrased my proof in terms of $\mathbb{C}^n$; one can then go back and show that in fact each eigenvalue and eigenvector is real with the usual treatment. – Ian Jul 19 '14 at 14:01
  • Well, I guess it does work. In the complex case you need the matrix to be conjugate to its transpose for that's what you need to do to a matrix when moving it to the other side in the bracket notation. Real & symmetric will, of course, do nicely. – Jyrki Lahtonen Jul 19 '14 at 15:47