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How can we show that a Hermitian $2\times 2$ matrix $M$ is positive definite if and only if $$M=\begin{pmatrix}a&b\\ \overline{b} &d\end{pmatrix}$$ for $a,d$ positive real numbers and $ad-b\overline{b}>0$?

I can show that any Hermitian $2\times 2$ matrix has the form $M=\left(\begin{smallmatrix}a&b\\ \overline{b} &d\end{smallmatrix}\right)$ for $a,d\in\mathbb{R}$ and then it is clear that $a,d>0$ if $M$ is positive definite. But where does the condition $ad-b\overline{b}>0$ comes from?

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Note: the condition $d>0$ is redundant.

A $2\times 2$ Hermitian matrix $M$ is diagonalizable with real spectrum $\{\lambda_1,\lambda_2\}$.Then it is easily seen to be positive definite if and only if $\lambda_1$ and $\lambda_2$ are both positive.

Now with $M$ written like yours, $M$ is positive definite if and only if $$ a>0\qquad\mbox{and}\qquad \det M=ad-b\overline{b}>0. $$

Necessary: we have $a=(e_1,Me_1)>0$ where $e_1=(1,0)$, and $\det M=\lambda_1\lambda_2>0$.

Sufficient: Let $\lambda_1$ be the maximal eigenvalue and let $(f_1,f_2)$ be an orthonormal diagonalization basis for $M$. Writing $x=x_1f_1+x_2f_2$, we have $(x,Mx)=\lambda_1|x_1|^2+\lambda_2|x_2|^2\leq \lambda_1 \|x\|_2$. Applying this to $x=e_1$ yields $0<a=(e_1,Me_1)\leq \lambda_1\|x\|^2$, hence $\lambda_1>0$. Now since $\det M=\lambda_1\lambda_2>0$, it follows that $\lambda_2>0$ also.

Generalization: An $n\times n$ Hermitian matrix $M$ is positive definite if and only the principal minors satisfy $$ \det M_k=\det (M_{i,j})_{1\leq i,j\leq k}>0 \quad\forall k=1,\ldots,n. $$

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If an $n\times n$ matrix is positive definite, it has $n$ positive eigenvalues and hence its determinant is positive. So, in your case, if $M$ is positive definite, we must have $ad-|b|^2>0$. Conversely, if $a,\,d,\,ad-|b|^2>0$, then either both eigenvalues of $M$ are positive or both of them are negative. But $M$ cannot be negative definite because $a>0$. Hence $M$ is positive definite.

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