Intereting Posts

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Why is this 'Proof' by induction not valid?
How to prove $\frac{1}{x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\sqrt{\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}}$

there is a similar thread here Coordinate-free proof of $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$?, but I’m only looking for a simple linear algebra proof.

- Rank of sum of rank-1 matrices
- Induced group homomorphism $\text{SL}_n(\mathbb{Z}) \twoheadrightarrow \text{SL}_n(\mathbb{Z}/m\mathbb{Z})$ surjective?
- bilinear form - proof
- What should be the characteristic polynomial for $A^{-1}$ and adj$A$ if the characteristic polynomial of $A$ be given?
- Short matrix algebra question
- Determining whether a symmetric matrix is positive-definite (algorithm)
- Matrix is conjugate to its own transpose
- Are one-by-one matrices equivalent to scalars?
- matrix representations and polynomials
- Are complex determinants for matrices possible and if so, how can they be interpreted?

Observe that if $A$ and $B$ are $n\times n$ matrices, $A=(a_{ij})$, and $B=(b_{ij})$, then

$$(AB)_{ii} = \sum_{k=1}^n a_{ik}b_{ki},$$

so

$$

\operatorname{Tr}(AB) = \sum_{j=1}^n\sum_{k=1}^n a_{jk}b_{kj}.

$$

Conclude calculating the term $(BA)_{ii}$ and comparing both traces.

The efficient @hjhjhj57: answer

$$\text{Tr}(AB) = \text{Tr}(BA)= \sum a_{ij} b_{ji}$$

Now we can start to understand why if we do a circular permutation of the factors the expression

$$\text{Tr}( A_1 A_2 \ldots A_m)$$

does not change, and what is the expression.

Assume now $A$,$B$ square. Then certainly $\det(AB) = \det(BA)$, using the multiplicative property of the $\det$.

In fact, the matrices $AB$ and $BA$ have the same characteristic polynomial, so in particular the same trace, and the same determinant.

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