Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that $\operatorname{trace}(ABC) = \operatorname{trace}(BCA) = \operatorname{trace}(CAB)$ if $A,B,C$ matrices have the same size.

share|cite|improve this question
up vote 6 down vote accepted

Is it already known that $\operatorname{Tr}(XY) = \operatorname{Tr}(YX)$ when $X$ and $Y$ are square matrices of the same size?

If it is, then simply set $X= AB$ and $Y = C$. It will give you $\operatorname{Tr}(ABC) = \operatorname{Tr}(CAB)$. You can get $\operatorname{Tr}(ABC) = \operatorname{Tr}(BCA)$ in a similar fashion.

share|cite|improve this answer


$$tr(ABC)=\sum_i (ABC)_{ii}=(ABC=A(BC))=\sum_i\sum_j A_{ij}(BC)_{ji}= \sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki};$$

now you can exchange the order of the matrices to arrive at the thesis as each of the $A_{ij}$, $B_{jk}$ and $C_{ki}$ are scalars (considering matrices over $\mathbb R$, for example). We arrive at

$$tr(ABC)=\sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki}=\sum_i\sum_j\sum_kB_{jk}C_{ki}A_{ij}=(BCA=(BC)A)=tr(BCA), $$

and so on.

share|cite|improve this answer
here I used the associativity of the product of matrices. – Avitus Nov 10 '13 at 11:20
got it thanks :) – Wanderer Nov 10 '13 at 11:57
you are welcome! Feel free to upvote it, if it was helpful. – Avitus Nov 10 '13 at 12:15

Hint: use the definition of trace.

$$\text{Tr}(ABC)=\sum_i\sum_j\sum_k A_{ij}B_{jk}C_{ki}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.