Positive (semi?) definite problem for 2 x 2 Hermitian matrix.

Question

I have some problem for positive semi-definite and Hermitiian matrix

Problem is following (this problem comes from Linear algebra done wrong by Sergei Treil page 213)

Let $A$ be a $2\times 2$ Hermitian matrix, such that $a_{1,1}\geq0$, $\det A\geq0$. Prove that $A$ is positive semi-definite.

how can i prove this?

My trial is ..

Since $A$ is Hermitian, i can choose orthogonal basis that makes $A$ to diagonal. And, it's diagonal entries are all real. $\det A\geq0$ implies some sign conditions..

I can even think that how can i use the condition $a_{1,1}\geq0$.

Is there any hint for this problem ?

Answer 1

Hint: Start with the definition. The $n\times n$ symmetric matrix $A$ is positive semidefinite if $x^TAx\geq 0$ for all $x\in\mathbb{R}^n$.