Trace of a matrix elements of a product of operators

Question

The trace of a matrix is defined to be the sum of its diagonal matrix elements

$$ Tr(\Omega) = \sum_{i} \Omega_{ii} $$

Show that $Tr(\Omega \Lambda \theta) = Tr(\Lambda \theta \Omega) = Tr(\theta \Omega \Lambda).$ You may recall that $\Omega, \Lambda,$ and $\theta$ are linear operators.

When seeing the solution of this problem that's what I did not understand: $$Tr(\Omega \Lambda \theta) = \sum_{i}(\Omega \Lambda \theta)_{ii} = \sum_{i}\sum_{j}\sum_{k}\Omega_{ij}\Lambda_{jk}\theta_{ki} = \sum_{j}\sum_{k}\sum_{i}\Lambda_{jk}\theta_{ki}\Omega_{ij} = $$ $$ \sum_{j}(\Lambda\theta\Omega)_{jj} = Tr(\Lambda\theta\Omega)$$

May someone explain how these passages from an equality to the other happened? I have no clue.

Thank you in advance!

score 1 · Answer 1 · answered Jan 08 '20 at 21:32

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I think you can better understand the problem if you first make the case of two operators. If $A, B$ are linear operators and define $C = AB, \tilde{C} = BA$ then

\begin{equation*} Tr(C) = \sum_{i=1}^{n}c_{ii} = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}b_{ji} = \sum_{j=1}^{n}\sum_{i=1}^{n}b_{ji}a_{ij} = \sum_{j=1}^{n}\tilde{c}_{jj} = Tr(\tilde{C}) \end{equation*} the above implies that $Tr(AB) = Tr(C) = Tr(\tilde{C}) = Tr(BA)$.

Now define $A = \Omega$ and $B = \Lambda \theta$.

answered Jan 08 '20 at 21:32

Eduardo Maza

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I'm sorry if it's a silly question, I am just starting with linear algebra, what is this symbol in the top of your C? – Victor Lins Jan 08 '20 at 21:40
Further, why do you need to exchange the order of the summations? – Victor Lins Jan 08 '20 at 21:41
The symbol above the C is a tilde. In latex it comes out as \ tilde {} – Eduardo Maza Jan 08 '20 at 21:43
I don't understand your second question well – Eduardo Maza Jan 08 '20 at 21:46
I mean, what is the meaning of the tilde up to the C? I asked the second question because I did not understand why you exchanged the order of the summations when you passed from $a_{ij}b_{ji}$ to $b_{ji}a_{ij}$ – Victor Lins Jan 08 '20 at 21:56
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If you prefer, replace $\tilde C$ with the letter $D$. The point is that $\tilde C$ is another variable-name. – Ben Grossmann Jan 08 '20 at 22:06
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@Victor You might find this proof a bit more understandable – Ben Grossmann Jan 08 '20 at 22:07
@Victor You might also find this post helpful. – Ben Grossmann Jan 08 '20 at 22:11
Thank you guys, I will see all the material carefully. – Victor Lins Jan 08 '20 at 22:13

Trace of a matrix elements of a product of operators

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