# Find an orthonormal basis for $\Bbb R^3$ where $q$ has a diagonal form

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

• Your $A-4I$ is wrong. – Emilio Novati Nov 6 '15 at 11:06
• The matrix or the final basis? Or both? thank you. – Alan Nov 6 '15 at 11:08
• 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. – Alan Nov 6 '15 at 11:13

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 $(1,-1,1)^T$

• 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 $\left( \begin{matrix} -1 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \end{matrix} \right)$ – Alan Nov 6 '15 at 11:19
• 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. – Emilio Novati Nov 6 '15 at 12:47
• Thank you. I don't know why when I use gauss elimination I get two rows in my canonical matrix. – Alan Nov 6 '15 at 13:11
• 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? – Alan Nov 6 '15 at 13:15
• Yes, your method is correct. You can see a diagonalization here: wolframalpha.com/input/… – Emilio Novati Nov 6 '15 at 13:48