Undergrad Doubt about proof on self-adjoint operators.

[Rodrigo Schmidt]

So the statement which the proof's about is: For every linear transformation $A$(between finite dimension spaces), the product $A^*A$ is self-adjoint. So, the proof is:
$(A^*A)^*=A^*A^{**}=A^*A$
What i don't understand is why $(A^*A)^*=A^*A^{**}$. Isn't that true only if $A$ and $A^*$commute?

 

[PeroK]

So the statement which the proof's about is: For every linear transformation $A$(between finite dimension spaces), the product $AA^*$ is self-adjoint. So, the proof is:
$(AA^*)^*=A^*A^{**}=A^*A$
What I don't understand is $(AA^*)^*=A^*A^{**}$. Shouldn't $(AA^*)^*$be equal to $A^*A^{**} = A^*A$? Wouldn't this mean that $AA^*$ is self-adjoint only if $A$ and $A^*$ commute?

Your post is very confused, because you have posted the same thing twice as both the correct and wrong versions. What you meant to say, no doubt was:

The proof is given as:

$(AA^*)^* = A^{**}A^* = AA^*$

But, you think it should be:

$(AA^*)^* = A^*A^{**} = A^*A$

But, you are wrong, because:

$(AB)^* = B^*A^*$ and not $A^*B^*$

 

[Rodrigo Schmidt]

AB)∗=B∗A∗(AB)^* = B^*A^* and not A∗B∗

Oh, thank you! I totally forgot that.

 

[mathwonk]

think through the proof of these more basic facts.