# Trace of a particular matrix product

1. Apr 10, 2015

### Bashyboy

1. The problem statement, all variables and given/known data
Claim: If $A \in \mathcal{M}_n (\mathbb{C})$ is arbitrary, and $D$ is a matrix with $\beta$ in its $(i-j)$-th entry, and $\overline{\beta}$ in its $(j-i)$-th, where $i \ne j$, and with zeros elsewhere, then $Tr(AD) = a_{ij} \beta + a_{ji} \overline{\beta}$

2. Relevant equations

3. The attempt at a solution

I am having difficulty with sum notation. By definition, the $(l-m)$-th entry of the matrix product $AD$ is

$\displaystyle (AD)_{lm} = \sum_{k=1}^n a_{lk} D_{km}$

And so the trace should be

$\displaystyle Tr(AD) = \sum_{q = 1}^n ((AD)_{lm})_q = \sum_{q=1}^n \sum_{k=1}^n a_{lk} D_{km}$

Given the description of the matrix $D$, it would seem that $(D)_{lm} = 0$ whenever $l \ne i$, $l \ne j$, $m \ne i$, or $m \ne j$. However, I am unsure about this and am having difficulty properly splitting up the sum. Could someone guide me along?

2. Apr 10, 2015

### Zondrina

There are only two elements of that sum which survive. The $a_{ij}D_{ij}$ term and the $a_{ji}D_{ji}$ term. Every other term will be zero.

You and I both know who those survivors are ;).

3. Apr 10, 2015

### Bashyboy

Intuitively it is obvious. But shouldn't we use the definition to formally show it is valid?

4. Apr 10, 2015

### Zondrina

Okay so we know:

$$Tr(AD) = \sum_{i=1} \sum_{j=1} A_{ij} D_{ji}$$

Maybe this notation will be more convenient for this problem.

5. Apr 10, 2015

### Bashyboy

Perhaps, although I am having double the difficulty with two sums involved.

6. Apr 10, 2015

### Zondrina

It might be easier to think about it in more steps. What would the matrix product $AD$ produce on its own? Take the trace of the resulting matrix, which happens to be the sum of the two diagonal entries.