Sufficient condition for a matrix to be positive definite Is a sufficient condition for a $2\times 2$ matrix
$$\left(\begin{array}{cc}a&b\\b&d\end{array}\right)$$ to be positive definite that $a >0$ and $ad > b^2$ ?
 A: Hint: A sufficient condition for a $2\times 2$ matrix to be positive definite is
$$\det A > 0 \qquad \text{and} \qquad \operatorname{Tr} A > 0.$$
For you to find out:


*

*This is true due to the fact that ...?

*This implies for $a,b,c,d$ ...?
A: The answer (irrespective of how $c$ and $d$ are positioned) is no. The conditions are independent of $d$, and the positive definiteness of the matrix can't be independent of one of the entries. Fix any values for the remaining entries that satisfy your conditions, then make $d$ arbitrarily large and negative. Then a vector $x$ with $1$s in the right places will lead to a value $x^\top Ax\approx d\lt0$.
Note, however, that some people define positive-definitness only for Hermitian matrices. Judging from your $b^2$, I presume that you're dealing with real matrices; in that case "Hermitian" would mean "symmetric". Your labelling of the entries makes no sense for symmetric matrices, but if you adjust it to that case by replacing $d$ by $b$, then indeed the two conditions you state are the conditions that the two principal minors of the matrix are positive, which is a necessary and sufficient condition for the matrix to be positive-definite.
[Edit:] That last paragraph depended on your original labelling of the entries. Now that you've switched to the usual labelling in Latin reading order, $b$ and $d$ are no longer related by symmetry. My impression is that your labelling may be confused precisely because you're mixing up the general case and the symmetric case.
A: Yes. By Sylvester's criterion, a symmetric matrix is positive definite iff all leading pricipal minors are positive.
(This holds for any size, not just $2 \times 2$.)
A: Slightly high-brow route: if a symmetric matrix $\mathbf A$ is positive definite, then the matrix $\mathbf X\mathbf A\mathbf X^\top$ is positive definite as well (Sylvester). Now, consider the decomposition
$$\begin{pmatrix}a&b\\b&d\end{pmatrix}=\begin{pmatrix}1&0\\\frac{b}{a}&1\end{pmatrix}\cdot\begin{pmatrix}a&0\\0&d-\frac{b^2}{a}\end{pmatrix}\cdot\begin{pmatrix}1&0\\\frac{b}{a}&1\end{pmatrix}^\top$$
How does one check if a diagonal matrix is positive definite?
