Trace of a particular matrix product

Bashyboy
Messages
1,419
Reaction score
5

Homework Statement


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}##

Homework Equations

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?
 
Physics news on Phys.org
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 ;).
 
Intuitively it is obvious. But shouldn't we use the definition to formally show it is valid?
 
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.
 
Perhaps, although I am having double the difficulty with two sums involved.
 
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.

How can this help you argue only two terms survive?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top