Given, $$A = \begin{bmatrix}
2 &-1 & -1\\
-1&2 & -1\\
-1& -1& 2
\end{bmatrix}.$$
I want to see if the matrix $A$ positive (negative) (semi-) definite.
Define the quadratic form as $Q(x)=x'Ax$.
Let $x \in \mathbb{R}^{3}$, with $x \neq 0$.
So, $Q(x)=x'Ax = \begin{bmatrix}
x_{1} &x_{2} &x_{3}
\end{bmatrix} \begin{bmatrix}
2 &-1 & -1\\
-1&2 & -1\\
-1& -1& 2
\end{bmatrix} \begin{bmatrix}
x_{1}\\x_{2}
\\x_{3}
\end{bmatrix}$.
After multiplying out the matrices I am left with $$Q(x) = 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-x_{1}x_{2} – x_{1}x_{3}-x_{2}x_{3}).$$
Not sure what I can do with this result. Any suggestions on how to proceed would be appreciated.
Best Answer
A simple way is to calculate all principle minors of $A$. If they are all positive, then $A$ is positive definite.
For example, $|A|_1=2>0$
$$ |A|_2=\left|\begin{array}{}{\quad2 \quad-1\\ -1\quad 2} \end{array}\right|=3>0 $$ Then calculate $|A|_3=|A|$.
If $|A|_i\geqslant0,1\leqslant i\leqslant n$, then $A$ is semi-positive definite.
If $|A|_i<0$ for $i$ is odd and $|A|_i>0$ for $i$ is even, then $A$ is negative definite.
If $|A|_i\leqslant 0$ for $i$ is odd and $|A|_i\geqslant 0$ for $i$ is even, then $A$ is semi-negative definite.