# Scalar product square matrix hermitian adjoint proof

1. Nov 23, 2012

### bossman007

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

If M is a square matrix, prove:

(A, MB) = (adj(M)A, B)

where (A, MB) denotes the scalar product of the matrices and adj() is the adjoint (hermitian adjoint, transpose of complex conjugate, M-dagger, whatever you want to call it!)

2. Relevant equations

adj(M)=M(transpose of the complex conjugate)

scalar product : A dot B = (A, B) = adj(A)B

I'm supposed to first write (A, MB) = (MB,A)^(complex conjugate) and apply the definitions of the scalar product above and adj(M)...and am warned to take complex conjugates carefully

3. The attempt at a solution

My attempted work is this:...i didnt follow the hint above because i just applied the inner product definition as in the picture and got both sides to be equivalent

[PLAIN]http://postimage.org/image/v3127r3w5/ [Broken][/PLAIN]

Last edited by a moderator: May 6, 2017
2. Nov 23, 2012

### HallsofIvy

Staff Emeritus
Are you given that these matrices are all 3 by 3? Because all you prove this for.

3. Nov 23, 2012

### bossman007

It doesnt say, I listed everything about the problem. I'm still not sure what to do

4. Nov 23, 2012

### Robert1986

Have you learned that $(a,Mb)=(Mb)^\top a$ and that $(Mb)^\top = b^\top M^\top$?

Of course the formula is the same for conjugate transpose as well (ie you can replace the T by *).

5. Nov 23, 2012

### bossman007

No I haven't learned that yet. What I have to do has something to do with the hint in OP but I can't make sense of

6. Nov 23, 2012

### bossman007

this is as far as i got using the hint...

[url=http://postimage.org/][PLAIN]http://s18.postimage.org/oz17lwl4p/photo_1.jpg[/url] free photo hosting[/PLAIN]

M is supposed to be a square matrix. The hint says after I applied the first step and the definitions of the scalar product and M(adjoint), I need to take complex conjugates? Where did I go wrong, or what do I have to do next in this proof? Many thanks

7. Nov 23, 2012

### Robert1986

Wait a second, here. I think you actually do know those formulas I gave you before, but the notation you are using doesn't seem to be what the standard notation is. Usually T means transpose, * means Hermitian adjoint (or conjugate transpose, both mean the same) and a bar means complex conjugate. Also, I believe that your formula is incorrect for $(A,B)$. It should be $B^*A$ (in "standard notation") or $B^\top A$ (in your notation.)

Whose notation is this, btw? Is it yours? Your prof's? Your book's?

8. Nov 23, 2012

### bossman007

its my books notation as seen here.

The difference in notation arises from the fact my book uses column matrix representation for vectors instead of row matrix representation , you think?

9. Nov 23, 2012

### bossman007

I made a mistake in my work picture above...quick correction here. Still do not know what to do next

[url=http://postimage.org/][PLAIN]http://s9.postimage.org/6luvfiam7/photo_2.jpg[/url] photo sharing websites[/PLAIN]

10. Nov 23, 2012

### Robert1986

So, using your notation, $M^\dagger A,B) = (M^\dagger A)^\dagger B = A^\dagger MB$. Now do you see what to do? Of course, this didn't really use the hint, but it looks like it works.

Again, just out of curiosity, what is the title of this book? Is it a Mathematical Physics book? I am interested because I have never seen the scalar product defined that way in a math book (that's not to say it hasn't happened, I just haven't seen it.)

11. Nov 23, 2012

### bossman007

Yea it's a "tutorial" math methods in physics book written by a professor for us, it's broken into 2 parts (semesters) and I also have Mary boas math methods book but couldn't find any help in there for this problem. Many thanks for the help !!! So what you just posted is basically the proof I did in the very first picture, right ? Would that very first picture I posted consist of the whole proof, like what you just hinted at, which seems correct ?