Positive-definite, Symmetric Matrix Problem

determinantlinear algebramatricessymmetric matricestrace

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.

Best Answer

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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