Positive-definite, Symmetric Matrix Problem

Question

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.

Answer 1

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 $\lambda_i > 0$.

Answer 2

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 $A$.