completing squares in positive definite matrices

Question

n the following question, I had a solution by using determinant test but my instructor wants me to solve this problem by using energy test.

$$B= \begin{bmatrix} -1 & \alpha & -1 \\ \alpha & -4 & \alpha \\ -1 & \alpha &-1 \\ \end{bmatrix}$$

Determine, without computing eigenvalues, the interval of $\alpha$ for which they are

1. positive definite

2. negative definite

3. positive semidefinite

4. negative semidefinite

5. indefinite.

First I calculated $x^T Ax$ as following:

$$ x^T Ax = \begin{bmatrix} x_{1} & x_{2} & x_{3} \\ \end{bmatrix} \cdot B \cdot \begin{bmatrix} x_{1} \\x_{2} \\x_{3} \\\end{bmatrix} $$

Then I get the following;

$-x_{1}^2+2\alpha x_{1}x_{2}-2x_{1}x_{3}-4x_{2}^2+2\alpha x_{2}x_{3}-x_{3}^2$

After this point, I could not complete squares and could not make an evaluation about the definiteness of the matrix. Could you please help me about completing squares?

Answer 1

We have \begin{aligned} \pmatrix{-1&a&-1\\ a&-4&a\\ -1&a&-1} &=-\pmatrix{1&-a&1\\ -a&4&-a\\ 1&-a&1}\\ &=-\pmatrix{1\\ -a\\ 1}\pmatrix{1&-a&1} -(4-a^2)\pmatrix{0\\ 1\\ 0}\pmatrix{0&1&0} \end{aligned} and hence $x^TBx=-(x_1-ax_2+x_3)^2 - (4-a^2)x_2^2$. Therefore $B$ is never positive semidefinite, and it is negative semidefinite (resp. negative definite) if and only if $4-a^2$ is nonnegative (resp. negative).

Answer 2

One approach to completing the squares here is to deal with each cross-term individually. For instance, we can absorb the $x_1x_2$ term into a square as follows: $$ -x_{1}^2+ \underline{2\alpha x_{1}x_{2}}-2x_{1}x_{3}-4x_{2}^2+2\alpha x_{2}x_{3}-x_{3}^2 =\\ -(x_{1}^2 - 2\alpha x_{1}x_{2})-2x_{1}x_{3}-4x_{2}^2+2\alpha x_{2}x_{3}-x_{3}^2 =\\ -(x_{1}^2 - 2\alpha x_{1}x_{2} + \alpha^2 x_2^2) + \alpha^2x_2^2-2x_{1}x_{3}-4x_{2}^2+2\alpha x_{2}x_{3}-x_{3}^2 =\\ -(x_1 - \alpha x_2)^2 + \alpha^2x_2^2-2x_{1}x_{3}-4x_{2}^2+2\alpha x_{2}x_{3}-x_{3}^2 =\\ -(x_1 - \alpha x_2)^2 - 2x_{1}x_{3}+(\alpha^2-4)x_{2}^2+2\alpha x_{2}x_{3}-x_{3}^2 $$ Once you have moved every cross-term into a square in this fashion, you will have a sum of squares.