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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 $ad > b^2$ ?

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Do you mean, a $2\times 2$ matrix $$\left(\begin{array}{cc}a&b\\d&c\end{array}\right)\ ?$$ –  Arturo Magidin Apr 19 '12 at 4:30
    
Welcome to math.SE. For future reference, soft-questions are more non-mathematical in nature. I've changed to more appropriate tags. –  Brett Frankel Apr 19 '12 at 4:30
    
@ArturoMagidin: The $c$ and $d$ are interchanged. –  Robb Apr 19 '12 at 4:33
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@Robb: Arturo wrote it like you said it. If you interchange $c$ and $d$, then they're not clockwise. –  joriki Apr 19 '12 at 4:33
    
@ArturoMagidin: Actually you are correct. –  Robb Apr 19 '12 at 4:38

4 Answers 4

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 $a,b,c,d$ ...?

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That all the eigenvalues are positive. Also $a+d > 0$ and $ad-bc 0$. –  Robb Apr 19 '12 at 4:42

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.

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And for negative definite, it would be $a<0$ and $ac < b^2$? –  Robb Apr 19 '12 at 4:44
    
@Robb: No. $ac\lt b^2$ ensures that the two eigenvalues are of opposite sign, so that's the condition for indefiniteness. The condition for negative-definiteness is simply the condition for positive-definiteness of $-A$, which is $a\lt0$ and $ac\gt b^2$. –  joriki Apr 19 '12 at 4:46
    
So then it is just $a<0$ and $ac > b^2$? –  Robb Apr 19 '12 at 4:49
    
@Robb: Yes, I'd already edited that into my comment while you were responding. All this is of course under the assumption that the matrix is symmetric and you label the entries like this: $$\pmatrix{a&b\\b&c}\;.$$ –  joriki Apr 19 '12 at 4:49
    
It should be $a <0$ and $ad > b^2$? In other words, replace the c with a d Edit: Just sa your comment. –  Robb Apr 19 '12 at 4:54

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?

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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 $2 \times 2$.)

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