Trace of Matrix Product as Scalar Product

[unscientific]

Homework Statement

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.

Homework Equations

Conditions for a scalar product:

The Attempt at a Solution

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

<A|B> = 0 if A^TA = I and B^TB = I

The first 3 parts of my attempt are shown below:

 

[micromass]

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 $A=0$.

 

[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.

 

[unscientific]

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.

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 A^TA = I and B^TB = I

 

[D H]

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

<A|B> = 0 if A^TA = I and B^TB = I

Point number 4 says nothing of the kind. Consider A=B=I. Obviously A^TA = I, as does B^TB. 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.