Why Does Tr(AB) Equal Tr(BA) for Matrices?

[Bipolarity]

Homework Statement

Prove the following statement, where A and B are nxn matrices.

$$Tr(AB) = Tr(BA)$$

Homework Equations

The Attempt at a Solution

Using some manipulations, I arrived at
$$\sum^{n}_{p=1}\sum^{n}_{k=1}A_{pk}B_{kp} = \sum^{n}_{p=1}\sum^{n}_{k=1}B_{pk}A_{kp}$$

If I can prove the above, I am done, but I have never worked with double sigma notations before, so any advice folks?

BiP

 

[Dick]

Homework Statement

Prove the following statement, where A and B are nxn matrices.

$$Tr(AB) = Tr(BA)$$

Homework Equations

The Attempt at a Solution

Using some manipulations, I arrived at
$$\sum^{n}_{p=1}\sum^{n}_{k=1}A_{pk}B_{kp} = \sum^{n}_{p=1}\sum^{n}_{k=1}B_{pk}A_{kp}$$

If I can prove the above, I am done, but I have never worked with double sigma notations before, so any advice folks?

BiP

k and p are what are called 'dummy indices'. You can change k to any other letter, like q and you still have a good formula for Tr. Just interchange k and p.

 

[Bipolarity]

k and p are what are called 'dummy indices'. You can change k to any other letter, like q and you still have a good formula for Tr. Just interchange k and p.

Ah, ingenious! Thanks!

BiP