# Proving properties of a 2x2 complex positive matrix

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1. Jun 28, 2017

1. The problem statement, all variables and given/known data
Prove that a 2x2 complex matrix $A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ is positive if and only if (i) $A=A*$. and (ii) $a, d$ and $\left| A \right| = ad-bc$

2. Relevant equations
N/A

3. The attempt at a solution
I got stuck at the first part. if $A$ is positive then by definition $A=S^*S$ for some matrix $S$ and thus $A^*=(S^*S)^*=S^*S=A$ and so $A=A^*$.
However I cannot prove the opposite, that if $A=A^*$ then A is positive. Since $A=A^*$ A is normal and thus there exists an orthonormal basis of $C^2$ comprised of eigenvectors of A, say, {$v_1.v_2$}. Furthermore since $A$ is self-adjoint all eigenvalues of $A$ are real. If I could prove the eigenvalues of $A$ are also nonnegative the proof will be complete, but I cannot seem to manage to prove that and it might not even be true.

Regarding the second part, assuming I have proven the first part it is easy to show how $a$, $d$ and $ad-bc$ must be real. However I can't seem to figure out how to prove they are nonnegative.
Help would be appriciated.

2. Jun 28, 2017

### Staff: Mentor

Is the last part missing "are real"?
That seems to be the case based on what you wrote in your attempt.

3. Jun 28, 2017

### StoneTemplePython

I would spend some time thinking about the second part. notice that ad-bc is the determinant. If the product of two eigenvalues is negative, and both eigs are real (as they are in Hermitian operator) then what does that tell you about the eigenvalues. What if the product is positive?

Now let's think about the trace for a minute...

you said if Hermitian positive (semi) definite, then $A=S^* S$

I claim that I can get the squared Frobenius norm of $S$, by taking the $\big \Vert S \big \Vert_F^2= trace(A) = trace(S^* S)$? Why is this true? What do the diagonal entries of $A$ look like if they come from $S^* S$?

Last edited: Jun 28, 2017
4. Jun 28, 2017