1
$\begingroup$

This seemed like a classic question of diagonalization, however I am a bit confused here since my result does not match my goal.

Let $\left(\begin{matrix} 3 & -1 & 0\\ -1 & 2 & -1 \\ 0 & -1 & 3\end{matrix} \right)$ be a matrix.

  1. Find the quadratic form $q:\Bbb R^3\to \Bbb R$ where q is represented by A.
  2. Find an orthonormal basis of $\Bbb R^3$ where $q$ has a diagonal form (i.e represented by a diagonal matrix).

"Solution"

  1. The quadratic form is the following:

$$q(A)=3x_1^2 -2x_1x_2 +2x_2^2 -2x_2x_3 +3x_3^3$$

  1. Correct me if I'm wrong but I have to diagonalize the matrix and then perform Gram-Schmidt algorithm. Another option was use express the quadratic form in a diagonal form (using Lagrange Method) and then use Gram-Schmidt but I have found out it might not be correct).

The characteristic polynomial of $A$ is: $$P_A=(t-4)(t-3)(t-1)$$ If the matrix is diagonalizable then the algebraic and geometric multiplicity were the same, however calculating the eigenspace for the eigenvalues gives different outcome, for example for $t=4$: $$(A-4I)=\left(\begin{matrix} - 1 & 0 & 0\\ 0 & - 1 & -1 \\ 0 & 0 & 0\end{matrix} \right) $$ Which gives the basis $$span \left\{\left(1,0,0\right),\left( 0,1,1\right) \right\}$$ Which gives geometric multiplicity of 2 for the eigenvalues $t=4$ which has an algebraic multiplicity of 1.

This means that the matrix is not diagonalizable but I think I am wrong somewhere.

Any ideas where I'm wrong?

Thanks,

Alan

$\endgroup$
  • $\begingroup$ Your $A-4I$ is wrong. $\endgroup$ – Emilio Novati Nov 6 '15 at 11:06
  • $\begingroup$ The matrix or the final basis? Or both? thank you. $\endgroup$ – Alan Nov 6 '15 at 11:08
  • $\begingroup$ I have found a mistake indeed, and fixed it, however the basis is still of dimension=2, $span \left\{ (1,0,0),(0,1,1) \right\}$ which is not good for me. $\endgroup$ – Alan Nov 6 '15 at 11:13
1
$\begingroup$

Hint:

Your $A-4I$ is wrong. We have: $$ \left(\begin{matrix} 3 & -1 & 0\\ -1 & 2 & -1 \\ 0 & -1 & 3\end{matrix} \right)- 4\left(\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)= \left(\begin{matrix} 3 & -1 & 0\\ -1 & 2 & -1 \\ 0 & -1 & 3\end{matrix} \right)- \left(\begin{matrix} 4 & 0 & 0\\ 0 & 4 & 0 \\ 0 & 0 & 4\end{matrix}\right) = \left(\begin{matrix} -1 & -1 & 0\\ -1 & -2 & -1 \\ 0 & -1 & -1\end{matrix}\right) $$

Use this and you can find an eigenvector $v_{\lambda=4}=(1,-1,1)^T$:

$$ (a-4I)\vec x=0 \Rightarrow \left(\begin{matrix} -1 & -1 & 0\\ -1 & -2 & -1 \\ 0 & -1 & -1\end{matrix}\right) \left(\begin{matrix}x\\ y \\ z\end{matrix}\right)=\left(\begin{matrix}0\\ 0 \\ 0\end{matrix}\right) \Rightarrow $$ $$ \begin{cases} -x-y=0\\ -x-2y-z=0\\ -y-z=0 \end{cases} \Rightarrow \begin{cases} x=-y\\ y=-z\\ z=t \in \mathbb{R} \end{cases} $$ so the eigenspace is $(t,-t,t)^T$ and an eigenvector is $(1,-1,1)^T$

$\endgroup$
  • $\begingroup$ Somehow I am still not getting it. The final matrix I get has two rows, meaning I have two vectors in the basis, or can I find a basis with one vector even though the final matrix has two rows? I get $\left( \begin{matrix} -1 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{matrix} \right)$ $\endgroup$ – Alan Nov 6 '15 at 11:19
  • 1
    $\begingroup$ As noted in my answer your matrix $A-4I$ is wrong and I cannot see how you find this result. I've added something that I hope can be helpful. In you want a more specific help you have to show the work that gives your result. $\endgroup$ – Emilio Novati Nov 6 '15 at 12:47
  • $\begingroup$ Thank you. I don't know why when I use gauss elimination I get two rows in my canonical matrix. $\endgroup$ – Alan Nov 6 '15 at 13:11
  • $\begingroup$ Do you think my way of solving the question is the right one, giving the fact that I use your method on the other eigenvectors? $\endgroup$ – Alan Nov 6 '15 at 13:15
  • 1
    $\begingroup$ Yes, your method is correct. You can see a diagonalization here: wolframalpha.com/input/… $\endgroup$ – Emilio Novati Nov 6 '15 at 13:48

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.