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Are there any particular types of matrices for which: $tr(AB)=tr(A)tr(B)$.

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    $\begingroup$ $1\times 1$ matrices? $\endgroup$ Mar 15, 2019 at 6:15
  • $\begingroup$ Thanks, are there any other classes? $\endgroup$
    – Mike
    Mar 16, 2019 at 3:00
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    $\begingroup$ Yes, such as the set of all upper triangular matrices with zero diagonals. $\endgroup$
    – user1551
    Mar 16, 2019 at 3:11

1 Answer 1

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$$tr(AB)=tr(A)tr(B)$$ $$\Sigma_{j,i} a_{ji} b_{ij}=\Sigma_{j,i} a_{jj} b_{ii}$$

$$\Sigma_{j,i} a_{ji} b_{ij} - a_{jj} b_{ii}=0$$

From here I am trying to find a class of matrices. To start with, let us assume that we are dealing with diagonal matrices only.

$$\Sigma_{j,i} a_{ji} b_{ij} \delta_{ij} - a_{jj} b_{ii}=0$$

For 2x2 matrices:

$$a_{11} b_{11} + a_{22} b_{22} - (a_{11} + a_{22})(b_{11} + b_{22})=0$$ $$-a_{11} b_{22} - a_{22} b_{11} =0$$ $$-a_{11} b_{22} = a_{22} b_{11}$$

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