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.


A simple way is to calculate all principle minors $A$ and 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.

  • $\begingroup$ Much neater and less complicated. And the same logic applies for negative (semi-) definite as well? $\endgroup$ – OGC Sep 3 '15 at 6:11
  • $\begingroup$ So here I found $|A_{3}|=0$, so $A$ is positive semi-definite. $\endgroup$ – OGC Sep 3 '15 at 6:26
  • $\begingroup$ Yes, you are right. You can see it in another post that quadratic form could be $0$ for nonzero $x$. $\endgroup$ – Math Wizard Sep 3 '15 at 6:27
  • $\begingroup$ Thanks again for this approach! Very convenient. $\endgroup$ – OGC Sep 3 '15 at 6:28

To say about positive (negative) (semi-) definite, you need to find eigenvalues of A. Then, 1) If all eigenvalues are positive, A is positive definite 2) If all eigenvalues are non-negative, A is positive semi-definite 3) If all eigenvalues are negative, A is negative definite 4) If all eigenvalues are non-positive, A is negative semi-definite 3) If some eigenvalues are positive and some are negative, A is neither positive nor negative definite

Eigenvalues of a matrix can be found by solving $det(\lambda I -A)=0$. For your example, this results in: $\lambda(\lambda-3)^2 =0$ which means that eigenvalues are 0, 3, 3. So we are in the second case and A is positive semi-definite.


If you want to proceed with this solution, you should complete the square. It is important that you "complete one variable completely every time". We write $$\begin{aligned} x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3&=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}x_2^2+\frac{3}{4}x_3^2-\frac{3}{2}x_2x_3\\ &=\Bigl(x_1-\frac{1}{2}x_2-\frac{1}{2}x_3\Bigr)^2+\frac{3}{4}\bigl(x_2-x_3\bigr)^2. \end{aligned} $$ Can you conclude from here?

  • $\begingroup$ So then the matrix is positive definite? $\endgroup$ – OGC Sep 3 '15 at 5:56
  • 1
    $\begingroup$ No, it is positive semidefinite. From the calculation above, you find that $Q(x)\geq 0$ for all $x$. The question is: "Does there exist $x\neq 0$ such that $Q(x)=0$ or not?" In this case it does. Take $x_3$ arbitrary, $x_2=x3$ (to make the last parenthesis zero) and $x_1=x_2$ (to make the first parenthesis zero). We conclude that $Q$ is only positive semidefinite. $\endgroup$ – mickep Sep 3 '15 at 6:00
  • $\begingroup$ I see. Thanks a lot! $\endgroup$ – OGC Sep 3 '15 at 6:01
  • $\begingroup$ where did the factor $2$ go on the LHS? $\endgroup$ – OGC Sep 3 '15 at 6:05
  • $\begingroup$ I just skipped the factor 2 since it only multiplies everything and does not change the character of the quadratic form. I should have mentioned this. $\endgroup$ – mickep Sep 3 '15 at 6:07

Find $A$'s eigenvalues first. Once you know them, you know everything you need about $A$.

More explicitly, you can start by calculating $A$'s characteristic polynomial. A straightforward calculation shows that its roots are $0$ and $3$. These are $A$'s eigenvalues, and hence, with respect to an appropriate orthonormal basis, $A$ becomes$$\left(\begin{array}{ccc}0&0&0\\0&3&0\\0&0&3\end{array}\right).$$This means that $A$ is positive semi-definite.

  • $\begingroup$ Could you please elaborate? What steps do I need follow here? Is there a theorem that needs to be applied? $\endgroup$ – OGC Sep 3 '15 at 5:48
  • $\begingroup$ @OGC I added some details to my answer. $\endgroup$ – Amitai Yuval Sep 3 '15 at 5:55

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