# Proof involving complex conjugates and Matrices

1. Sep 19, 2015

### RJLiberator

1. The problem statement, all variables and given/known data

Show that (A+B)*=A*+B*

2. Relevant equations
I think I am missing a property to prove this.

3. The attempt at a solution
This should be easier then I am making it out to be. But I seem to be missing one key property to do this.

A*+B* is just A(ij)*+B(ij)* = Right hand side
Left hand side is just (A+B)(ij)*

What am I missing here to make this concrete?

I've looked around the web and found things such as A(ij)* = bar(A(ji))

2. Sep 19, 2015

### davidbenari

Well you know how each element of the matrix $A+B$ looks like right? Namely its $a_{ij}+b_{ij}$ and you know that the conjugate of a sum of numbers is the sum of the conjugates.

Which makes you get that each element of the new matrix $(A+B)^*$ is $a_{ij}^*+b_{ij}^*$ which should remind you of something important

3. Sep 19, 2015

### Ray Vickson

What is your definition of A*? Is it the element-by-element complex conjugate of A, or is it the element-by-element complex conjugate of the transpose of A? I have seen the same notation A* used in both ways, so you really do need to clear that up.

4. Sep 19, 2015

### lautaaf

You could prove it by definition:
Let $H$ be a Hilbert space and let $A : H \rightarrow H$ be a bounded linear operator. Then, $A^{*}$, the adjoint of $A$, is defined by the following equation: $$<Ax,y>=<x,A^{*}y> for \: all \: x,y \in H$$
Thus, you have to prove that: $$<x,(A+B)^{*}y>=<x,(A^{*}+B^{*})y> for \: all \: x,y \in H$$

5. Sep 19, 2015

### RJLiberator

Is this what we are trying to prove? Essentially?

If I say Left hand side: (Aij+Bij)*=Aij*+Bij*
and so this equals the right hand side Aij* + Bij*

Isn't that what we are just trying to prove? That would be a 2 line proof for it. That simple?

6. Sep 19, 2015

### RJLiberator

Most simple form used here: Element-by-element complex conjugate of A.

7. Sep 19, 2015

### davidbenari

Yes it would look like that. Just think of what happens to each element. It is a simple proof after all.

8. Sep 19, 2015

### RJLiberator

Okay, so
A*+B*=Aij*+Bij* = Right hand side
(A+B)*=Aij*+Bij* = Left Hand side due to the conjugate of a sum of numbers is the sum of the conjugates.

0_o

9. Sep 19, 2015

### davidbenari

Yep

10. Sep 19, 2015

### RJLiberator

11. Sep 19, 2015

### davidbenari

12. Sep 19, 2015

### lautaaf

It`s important to keep in mind that the operator $^{*}$ also involves taking the transpose of the matrix (if you choose the appropriate basis)

13. Sep 19, 2015

### vela

Staff Emeritus
Your confusion probably arises because you're not paying enough attention to details, which is suggested by what you've written. For example, A*+B* isn't equal to Aij* + Bij*. A*+B* is a matrix. Aij* + Bij* is a number. It's not correct to say the two are equal. A clue is that neither i nor j appear on the lefthand side of the equation.

It's probably been established (or assumed) in your class that if z and w are complex numbers, then (z+w)* = z* + w*. It's not "obvious" that the same relationship holds with a completely different type of object. When you write A*+B*, the + and * mean completely different things than the + and * in z* + w* because A and B are completely different mathematical objects than z and w. Recognizing this should help you understand what you need to show.

14. Sep 19, 2015

### RJLiberator

Good post Vela.

What I got from this is that I need to show that the ij'th components from each side are equal.

We actually proved that the conjugate of a sum of numbers is the sum of the conjugates earlier in the assignment, so I will use that to finish off this proof.
I think you are right with where the confusion was, and I now feel very good about this proof.

The right hand side is pretty straight forward when looking at the ij'th components.
The left hand side, I have to first say that I am looking at the ij'th component, and then apply the earlier proof, to show that the left hand side is indeed equal to the right hand side.

And this makes 100% sense to me now .
Kind Regards

15. Sep 19, 2015

### vela

Staff Emeritus
Glad it was helpful. After I posted, I reread your original post and wondered if perhaps you were clear on that and were just being a little sloppy, but I remember being confused the same way when I was first encountered proofs like these.