Proving properties of a 2x2 complex positive matrix

Adgorn
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Homework Statement


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##

Homework Equations


N/A

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.
 
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Adgorn said:

Homework Statement


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

Homework Statement


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##

...

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.
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:
Mark44 said:
Is the last part missing "are real"?
That seems to be the case based on what you wrote in your attempt.
Are real and positive yes, forgot to write that for some reason.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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