Quadratic form of the $3\times 3$ symmetric matrix

Question

The quadratic form of the $3\times 3$ matrix is given as follows $$z^TXz=x_1z_1^2+2x_2z_1z_2+2x_3z_1z_3+x_4z_2^2+2x_5z_2z_3+x_6z_3^2$$ If $x_1=0$ then why should $x_2=x_3=0$ if we want $z^TXz\geq 0$? I read it in solution manual of the convex optimization book (by Stephen Boyd).

Answer 1

Because otherwise

$$2x_2z_1z_2+2x_3z_1z_3=2z_1(x_2z_2+x_3z_3)$$ could take arbitrarily large negative values (via $z_1$) that cannot be compensated by the remaining terms.

Answer 2

The associated matrix is

\begin{bmatrix}x_1&x_2&x_3\\x_2&x_4&x_5\\x_3&x_5&x_6 \end{bmatrix}

and if $x_1=0$

\begin{bmatrix}0&x_2&x_3\\x_2&x_4&x_5\\x_3&x_5&x_6 \end{bmatrix}

we have that

$$\begin{vmatrix}0&x_2\\x_2&x_4\end{vmatrix}=-x_2^2$$

which should be always negative an then, by Sylvester criterion, we need $x_2=0$ and then

$$\begin{vmatrix}0&0&x_3\\0&x_4&x_5\\x_3&x_5&x_6 \end{vmatrix}=-x_3^2x_4$$

and by Sylvester criterion, since we need $x_4> 0$, we need also $x_3=0$ and finally we need

• $x_4,x_6> 0$
• $\begin{vmatrix}x_4&x_5\\x_5&x_6\end{vmatrix}=x_4x_6-x_5^2\ge 0\implies$

then to achieve $z^TXz\ge 0$ with $x_1=0$ we need $x_2=x_3=0$, $x_4,x_6>0$ and $x_5^2\le x_4x_6$.

Answer 3

Think that with

$$ X = \left( \begin{array}{ccc} x_1 & x_2 & x_3 \\ x_2 & x_4 & x_5 \\ x_3 & x_5 & x_6 \\ \end{array} \right) $$

in $ z^{\top}X z$ to have $z^{\top}X z \ge 0$ it is enough to have $x_1> 0, x_4 > 0, x_6 > 0, x_2=x_3=x_5=0$

NOTE

If it's leading principal minors are all positive the the necessary and sufficient conditions for positive definiteness are accomplished.