I have a matrix:
$$ A = \begin{bmatrix} 4&1 \\ 1&5 \end{bmatrix} $$
I want to determine whether $A$ is positive definite.
Therefore, I used the definition, $A$ is positive definite iff:
$$ x^*Ax > 0, \forall x \in C^2, x \neq 0 $$
$$ x = [a,b]^t , a,b \in C, a.b \neq 0\\ x^*Ax = [a,b]^tA[a,b] = 4|a|^2 + \bar{b}a + \bar{a}b + 5|b|^2 $$
Now, I know that $z + \bar{z} = 2\operatorname{Re}z \Rightarrow \bar{b}a + \bar{a}b = 2\operatorname{Re}(\bar{b}a)$
But still, I don't know whether I get a positive number for $x^*Ax$, I probably miss something, what am I missing?
Thank you.
You can use the fact that $|a-b|^2 \geq 0$ and $|a+b|^2 \geq 0$, so after expanding, you get the famous inequality $$ - (|a|^2 + |b|^2) \leq 2 \Re(a \bar{b}) \leq |a|^2 + |b|^2. $$ Plugging this into your result gives $x^*Ax \geq 3 |a|^2 + 4 |b|^2 \geq 0$.
Another option is to check positive-definiteness by computing the eigenvalues and checking that they are both positive, or by using the characterization with the minors.