2
$\begingroup$

Consider the following quadratic form over $\mathbb{R}^3$:

$$q = x_1^2+4x_1x_2-2x_1x_3+8x_2^2+2x_3^2-8x_2x_3$$

It's fairly easy to arrive at the diagonal form of $q$ - by using Lagrange's method (repeated complete the square), we get

$$q = (x_1+2x_2-x_3)^2+(2x_2-x_3)^2$$

So, the "canonical" diagonal form (with only $1$s, $-1$s and $0$s in the diagonal) of $q$ is:

$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

Q. Is there an orthonormal basis in which the representing matrix of $q$ is canonically diagonal?

Intuitively, is seems that no such basis exists, since we've arrived at the diagonal form with a non-orthonormal basis, but I couldn't find a formal proof for that.

Any ideas?

$\endgroup$
  • $\begingroup$ I see you've been trying to edit my post. It would be better to ask questions in the comments. $\endgroup$ – Yly Sep 27 '18 at 16:42
  • $\begingroup$ I've revised the question itself due to a calculation error in the original version (see that now $q$ is congruent to $\text{diag}(1,1,0)$), and wanted to update the answer accordingly. Also I just wanted to add a clarification about congruence vs diagonalization by eigenvectors $\endgroup$ – matan129 Sep 27 '18 at 16:46
  • 1
    $\begingroup$ I've made a modification to my answer (changed a $=$ to a $\leq$) which handles your modification to the question. $\endgroup$ – Yly Sep 27 '18 at 16:49
  • $\begingroup$ As for diagonalization vs. congruence, you should make your remarks either in the question itself, or here in the comments. $\endgroup$ – Yly Sep 27 '18 at 16:50
1
$\begingroup$

There can't be an orthonormal basis for which the quadratic form has canonical diagonal form, because if there were such a basis $\{v_1,v_2,v_3\}$, then for any unit vector $u = \sum_i \lambda_i v_i$ we would have $$q(u) = \left(\sum_i \lambda_i v_i\right)^T A \left(\sum_i \lambda_i v_i\right) = \begin{bmatrix}\lambda_1 & \lambda_2 & \lambda_3\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}\lambda_1\\ \lambda_2 \\ \lambda_3\end{bmatrix} \leq \sum_i \lambda_i^2 = 1$$

But this is false, because from your definition of $q$, we see that e.g. $q(0,1,0)=8$.

By writing $A$ as a symmetric matrix you can find an orthonormal basis that diagnonalizes it (because a symmetric matrix always can always be diagonalized by a rotation matrix), but the diagonal entries will then be the eigenvalues of $A$, not $1$.

$\endgroup$

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.