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I have a question that I've been working on for a bit now, and it says, "Let $A\in M_{2\times2}(\mathbb{R})$ be a symmetric matrix. We say that A is positive definite if all of the eigenvalues of $A$ (which are necessarily real) are positive. Show that $A$ is symmetric and positive definite if and only if $Tr(A)>0$ and det$(A)>0$." I know how to do the forwards proof, but I'm kind of stuck with the backwards proof. I know that if $$det(A)=ad-bc>0$$ and Tr$(A)>0$, then $a$ and $d$ must both be positive, and $ad>bc$, but I'm stuck at this point. Help would be appreciated. Thanks in advance.

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  • $\begingroup$ $A$ is symmetric means $b=c$. Also, you need to look at the characteristic equation if you are going to say anything about the eigenvalues. $\endgroup$ – saulspatz Jul 29 '18 at 17:29
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Guide:

  • determinant of $A$ is equal to the product of the eigenvalues. Hence you have $\lambda_1 \cdot \lambda_2 >0$.

  • trace of $A$ is equal to the sum of the eigenvalues. Hence you have $\lambda_1 + \lambda_2 > 0$.

  • Use those two information to show that $\lambda_i > 0$.

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    $\begingroup$ It's important $A$ be symmetric, so that the eigenvalues are real. Otherwise, you could take $\lambda_1 = \epsilon + i$ and $\lambda_2 = \epsilon - i$ as solutions to your inequalities. A real matrix with these eigenvalues is $\epsilon I + [ [ 0, -1], [1, 0] ]$. The statement of the problem is a little weird: I think it should be if $A$ is real and symmetric then [A is positive definite iff tr(A) > 0 and det(A) > 0]. $\endgroup$ – Lorenzo Najt Jul 29 '18 at 17:37
  • $\begingroup$ good point! thanks for the input. $\endgroup$ – Siong Thye Goh Jul 29 '18 at 17:39
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    $\begingroup$ I thought that the determinant of a symmetric matrix was $ad-b^2$. $\endgroup$ – Mr. Frothingslosh Jul 29 '18 at 18:14
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    $\begingroup$ It is, it is also equal to the product of the eigenvalues, Let $A=UDU^T$ be the EVD where $U$ is orthogonal. If you take the determinant, you can show that $\det(A)=\det(D)$. $\endgroup$ – Siong Thye Goh Jul 29 '18 at 18:16
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Siong Thye Goh's answer is excellent, but here is another way to attack the problem (although both ways are similar).

For $2\times 2$ matrices one can show easily that the characteristic polynomial for $A$, call it $p_A$, is given by $$p_A(t)=t^2-\operatorname{tr}(A)t+\det(A)$$ Then the roots of $p_A$ are $$t=\frac{\operatorname{tr}(A)\pm\sqrt{\operatorname{tr}(A)^2-4\det(A)}}{2}$$ So suppose $\operatorname{tr}(A)=a+d$ and $\det(A)=ad-bc=ad-b^2$ are positive, then what kind of roots can $p_A(t)$ have (i.e. what sign will they have)? Then recall that the roots of $p_A$ are precisely the eigenvalues of $A$.

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  • $\begingroup$ This was the way I was trying to approach it actually $\endgroup$ – Mr. Frothingslosh Jul 29 '18 at 19:21
  • $\begingroup$ Well you just have to show those roots are positive, and that comes down to showing that $$\operatorname{tr}(A)-\sqrt{\operatorname{tr}(A)^2-4\det(A)}>0$$ $\endgroup$ – Dave Jul 29 '18 at 19:43
  • $\begingroup$ Makes sense actually. Thanks a lot $\endgroup$ – Mr. Frothingslosh Jul 29 '18 at 19:47

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