Given A is $m\times n$ and B is a $n \times m$ matrix, can we say trace(AB) = trace(BA)? It worked for a few examples with $m = 2$ and $n = 3$.
Trace(AB) = Trace(BA) for rectangular matrices
0
$\begingroup$
linear-algebra
matrices
trace
random-matrices
-
0It's generally true. Just write out the definition of the matrix product and the trace. See also [this answer](https://math.stackexchange.com/questions/252272/trace-is-invariant-under-cyclic-permutation-even-with-rectangular-matrices-invol?rq=1). – 2017-02-20
1 Answers
1
Yes, It is true. Trace(AB) = Trace(BA) for rectangular matrices. You can rewrite the expression used for calculating trace to prove this.
tr(AB) = $\sum_{i=1}^{m}(AB)_{ii} = \sum_{i=1}^{m}\sum_{j=1}^{n}A_{ij}B_{ji} = \sum_{j=1}^{n}\sum_{i=1}^{m}B_{ji}A_{ij} = \sum_{j=1}^{n}(BA)_{jj} $= tr(BA)
For more references, you can see the answers in this link : Read more here..
-
0While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - [From Review](/review/low-quality-posts/765233) – 2017-02-20
-
0Thanks.I'll make sure of this in my future answers.. :) – 2017-02-20
-
0you are welcome. You will do well. – 2017-02-20