# Trace of Matrix Product as Scalar Product

1. Aug 2, 2013

### unscientific

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

Let V be the real vector space of all real symmetric n × n matrices and define the scalar product of two matrices A, B by (Tr (A) denotes the trace of A)

Show that this indeed fulfils the requirements on a scalar product.

2. Relevant equations

Conditions for a scalar product:

3. The attempt at a solution

I'm not sure how to show the last part. Which can be summarized as:

<A|B> = 0 if ATA = I and BTB = I

The first 3 parts of my attempt are shown below:

2. Aug 2, 2013

### micromass

Staff Emeritus
For (2), you say $\mathrm{Tr}(AB)^T$. I don't really know what you mean with this. What is the transpose of a number? And why should

$$\mathrm{Tr}(AB)^T = \mathrm{Tr}(B^T A^T)$$

For (3), you should still show that the inner product is $\geq 0$ and that it is $=0$ iff $A=0$.

3. Aug 3, 2013

### Vic Sandler

For rule 1, I think the OP means $Tr ((AB)^T)$, not $(Tr(AB))^T$. The 1st line of the proof is unnecessary. The 2nd line looks good to me. However, you are using the fact that $Tr(A) = Tr(A^T)$. This is easy to prove and you should add that proof in. For rule 2, you still need to prove that <A|A> = 0 implies A = 0. You will need to use the fact that the underlying space only includes symmetric matrices.

Last edited: Aug 3, 2013
4. Aug 5, 2013

### unscientific

Yeah $Tr(A) = Tr(A^T)$ because for any $A_{ij}$ component where i=j, switching their positions don't change anything.

I'm more concerned about the point number 4. Which can be summarized as:

<A|B> = 0 if ATA = I and BTB = I

5. Aug 5, 2013

### D H

Staff Emeritus
Point number 4 says nothing of the kind. Consider A=B=I. Obviously ATA = I, as does BTB. Yet <A,B> is not zero.

Instead think of point #4 as being a definition of what it means for two quantities to be deemed as being "orthogonal" to one another.