Prove that if $\mathbf A$ is an $n\times m$ matrix, then $\text{tr}(\mathbf A \mathbf A')=\text{tr}(\mathbf A' \mathbf A) $

Question

If $\mathbf A$ is an $n\times m$ matrix, then $\text{tr}(\mathbf A \mathbf A')=\text{tr}(\mathbf A' \mathbf A) \text{ where } \mathbf A'\text{ is transpose of }\mathbf A\text{ and tr}(\mathbf A )\text{ is trace of }\mathbf A.$

Answer 1

Let $A=(a_{ij})$ and $AA'=(c_{ij})$ then

$$c_{ij}=\sum_{k=1}^na_{ik}a_{jk}$$ $$\operatorname{tr}(AA')=\sum_{i=1}^n c_{ii}=\sum_{i=1}^n\sum_{k=1}^n a^2_{ik}=\operatorname{tr}(A'A)$$

Answer 2

$$\text{tr}(AA') = \sum_{i=1}^{i=n}(AA')_{ii}=\sum_{i=1}^{i=n}\sum_{j=1}^{j=m}A_{ij}A'_{ji}=\sum_{i=1}^{i=n}\sum_{j=1}^{j=m}A_{ij}^2$$

Similarly, $$\text{tr}(A'A) = \sum_{i=1}^{i=n}\sum_{j=1}^{j=m}A_{ij}^2$$

Answer 3

Generally, $\operatorname{tr} (AB) = \sum_k [AB]_{kk} = \sum_k \sum_i [A]_{ki}[B]_{ik} = \sum_i \sum_k [A]_{ki}[B]_{ik} = \sum_i \sum_k [B]_{ik} [A]_{ki} = \operatorname{tr} (BA)$.