Is the matrix $A$ positive (negative) (semi-) definite? 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: 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.
A: 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?
A: 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.
A: 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.
