Proofs for non symmetric positive definite matrix

Question

I've been trying to find out the exact condition for the 2 by 2 non symmetric matrix $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ to be positive semidefinite.

A related post was this one: criterion for positive definite $2\times 2$ matrix? but I think it's wrong. A useful discussion could be found Do positive semidefinite matrices have to be symmetric?

In Why is positive (semi-)definite only defined for symmetric matrices?? Randy E metioned that the necessary and sufficient condition was for the symmetric part of the matrix

$$A+A^T$$

to be also positive semidefinite. However, I'm not sure where that reference came from or how to prove it.

For the symmetric case, the condition was such that $a\geq 0$ and $\det(A)\geq 0$

But for the non symmetric case, $\det(A)\geq 0$ appeared to be holding, but there were sources saying $a\geq 0$ was replaced with $a\geq 0$, $d\geq 0$ and some other saying it was replaced with $tr(A)=a+d\geq 0$.

What's the condition for the $2$ by $2$ non symmetric matrix to be positive semi definite? Could you provide some references(papers, posts, etc.) and proofs, please?

Answer 1

Let $A$ be the 2x2 matrix in your post. By algebra, $$ v^{\intercal}Av=av_{1}^{2}+(b+c)v_{1}v_{2}+dv_{2}^{2}=\frac{1}{2}v^{\intercal}\left(A+A^{\intercal}\right)v. $$ That is why it is sufficient to consider $A+A^{\intercal}$.

Answer 2

The matrix $A+A^T$ is always symmetric, what Randy E was saying is that $$\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}a & (b+c)/2\\ (b+c)/2 & d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=ax^2+(b+c)xy+dy^2=(*),$$ and so since a matrix is positive semidefinite if and only if $(*)> 0$ for all $x,y\in\mathbb{R}$ with $(x,y)\neq (0,0)$, it follows $A$ is positive semidefinite if and only if $(A+A^T)/2$, and therefore $A+A^T$, is (the reason for this last point is that if $c\geq 0$, then $B$ is positive semidefinite if and only if $cB$ is, which is straightforward to prove). Therefore to figure out whether $A$ is positive semidefinite, one needs only check whether the symmetric matrix $A+A^T$ is.

For an answer in terms of the coefficients of $A$, an answer in your first linked post mentions Sylvester's criterion. Using this with the matrix $A+A^T$ (since $A$ is a real matrix, we need a symmetric matrix to apply Sylvester's criterion), we see that $A+A^T$, and therefore $A$, is positive semidefinite if and only if $$a>0$$ and $$4ad-(b+c)^2>0.$$