Show eigenvalues are real and positive (too easy?)

[ognik]

I'm given that the matrices J_x, J_y, J_z are Hermitian (they are angular momentum components). Show the eigenvalues of J² = J_x² + J_y² + J_z² are real and non-negative.

My proof seems too easy, I'd appreciate improvements to it:
i) The eigenvalues of an Hermitian matrix are real
ii) The square of any real number is $$\geq$$ 0
Therefore the eigenvalues of each of J_x, J_y, J_z are real and their squares are non-negative, therefore the eigenvalues of J² are also real and non-negative.

Its just too easy, is there a more formal approach?

 

[GJA]

Hi ognik,

Everything you've said is correct, but you haven't stated a key fact that really proves the "therefore" portion of your argument.

As you've said, the eigenvalues of $$J_{i}$$, $$i=x,y,z$$ are real, but you haven't mentioned how you know the sum of the square of these numbers is actually an eigenvalue of $$J^2.$$

Also, even if the eigenvalues of individual operators are nonnegative, that does not imply the operator given by their sum has nonnegative eigenvalues. For example, take the $$2\times 2$$ matrices

$$A=
\begin{bmatrix}
1 & 0\\
4 & 1
\end{bmatrix}$$

and

$$B =
\begin{bmatrix}
1 & 4\\
0 & 1
\end{bmatrix}
$$

The eigenvalues of both of these matrices are positive, but the eigenvalues of their sum $$A+B$$ are $$\lambda = -2,6.$$

This example further demonstrates that, in general, the eigenvalues of $$A+B$$ are not simply the sum of the individual eigenvalues of $$A$$ and $$B$$.

All that being said, you really only need to add one statement to your argument to make it complete. Try to think it over and see what you can come up with. Let me know if anything is unclear/not quite right. Good luck!

 

[ognik]

Fair enough, so switching tack slightly,
J_i with i=x,y,z are all Hermitian. The sum of Hermitian matrices is also Hermitian. J=J_x + J_y + J_z, so J is also Hermitian, therefore the eigenvalues of J are real.
Then J² = J.J so the square of real eigenvalues makes the eigenvalues of J² real and non-negative.

Also from (AB)^$$\dagger$$ = B^$$\dagger$$A^$$\dagger$$, ie the product of 2 hermitian matrices is also hermitian, therefore J² is hermitian... Have I got close yet?