A matrix is diagonalizable iff it has a set of eigen vectors which are linearly independent. Now, why is this satisfied in case of symmetric matrix ?

  • 6
    $\begingroup$ See Proof 2: maecourses.ucsd.edu/~mdeolive/mae280a/lecture11.pdf $\endgroup$ – Amzoti Sep 3 '13 at 1:14
  • $\begingroup$ IIRC, this is proved in Sheldon Axler's Down with Determinants, a very nice article. I would link to it, but I'm on my phone. $\endgroup$ – user641 Sep 3 '13 at 5:19

Here's some intuition (but not a rigorous proof).

If $A$ is hermitian (with entries in $\mathbb C$), you can easily show that the eigenvalues of $A$ are real and that eigenvectors corresponding to distinct eigenvalues are orthogonal.

Typically, all the eigenvalues of $A$ are distinct. (It is in some sense a huge coincidence if two eigenvalues turn out to be equal.) So, typically $A$ has an orthonormal basis of eigenvectors.

Even if $A$ has some repeated eigenvalues, perturbing $A$ slightly will probably cause the eigenvalues to become distinct, in which case there is an orthonormal basis of eigenvectors. By thinking of $A$ as a limit of slight perturbations of $A$, each of which has an ON basis of eigenvectors, it seems plausible that $A$ also has an ON basis of eigenvectors.

  • 2
    $\begingroup$ I really love this idea, however it seems very tricky to turn it into a readable proof: 1) the traditional perturbation technique i.e. $+\epsilon I$ doesn't work here and it seems rather murky to have any simple perturbation path that can keep all eigenvalues distinct; 2) eigenvectors being orthogonal isn't a topological concept and what would be a nice way to associate it continuously with entries of the matrices? $\endgroup$ – Vim Sep 17 '17 at 2:19

Suppose the ground field is $\mathbb C$. It is immediate then that every square matrix can be triangulated. Now, symmetry certainly implies normality ($A$ is normal if $AA^t=A^tA$ in the real case, and $AA^*=A^*A$ in the complex case). Since normality is preserved by similarity, it follows that if $A$ is symmetric, then the triangular matrix $A$ is similar to is normal. But obviously (compute!) the only normal triangular matrix is diagonal, so in fact $A$ is diagonalizable.

So it turns out that the criterion you mentioned for diagonalizability is not the most useful in this case. The one that is useful here is: A matrix is diagonalizable iff it is similar to a diagonal matrix.

Of course, the result shows that every normal matrix is diagonalizable. Of course, symmetric matrices are much more special than just being normal, and indeed the argument above does not prove the stronger result that symmetric matrices are orthogonaly diagonalizable.

Comment: To triangulate the matrix, use induction of the order of the matrix. For $1\times 1$ it's trivial. For $n\times n$, first find any arbitrary eigenvector $v_1$ (one such must exist). Thinking of the matrix as a linear transformation on a vector space $V$ of dimension $n$, write $V$ as $V=V_1\oplus W$, where $V_1$ is the subspace spanned by $v_1$. Then $W$ is $n-1$-dimensional, apply the induction hypothesis to $A|_{W}$ to obtain a base $v_2,\ldots, v_n$ in which $A|_W$ is triangular. It now follows that in the base $v_1,\ldots, v_n$ $A$ is triangular.

  • $\begingroup$ "symmetry is preserved by similarity" - what does this mean? It's certainly not true that if $A$ is symmetric then any matrix similar to $A$ is symmetric. $\endgroup$ – Ted Sep 3 '13 at 5:11
  • $\begingroup$ @Ted thank you for this! I meant to use normality rather than symmetry. I made the corrections. Thanks! $\endgroup$ – Ittay Weiss Sep 3 '13 at 5:29
  • 1
    $\begingroup$ @IttayWeiss Could you (or anyone else) comment on why it's immediate that every square matrix can be triangulated? I'm not sure that part is as obvious to me as it should be. $\endgroup$ – littleO Sep 3 '13 at 5:46
  • $\begingroup$ @littleO I added the proof of that as a comment in the answer text. $\endgroup$ – Ittay Weiss Sep 3 '13 at 5:54
  • 1
    $\begingroup$ @IttayWeiss thanks! $\endgroup$ – littleO Sep 3 '13 at 5:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.