Proof that $(A+B)^T=A^T+B^T$ (homework question)

Question

Homework question:

Proof that $(A+B)^T=A^T+B^T$

Let A and B be $m \times n$ matrices. Prove that $(A+B)^T=A^T+B^T$ by comparing the ij-th entries of the matrices on each side of this equation. (Let $A=(a_{ij})$ and $B=(b_{ij}$).)

I am not sure how to do this proof, I know how to prove it by substituting ij-th entries with arbitrary numbers but I do not know how to do it by 'comparing the ij-th entries'

Answer 1

Hint:

$ij^{th}$ entry of $ A^t=a_{ji}$

Answer 2

Let $C=A+B$ then $c_{ij}=a_{ij}+b_{ij}$. So $c_{ji}=a_{ji}+b_{ji}$ for all $i,j$. Hence $C'=A'+B'$ implying $(A+B)'=A'+B'$

Answer 3

So, if $A=(a_{ij})$, $B=(b_{ij})$, and $C=A+B=(c_{ij})$ with $c_{ij}=a_{ij}+b_{ij}$ one has

$(A+B)^T=$

$=[(a_{ij})+(b_{ij})]^T=[(a_{ij}+b_{ij})]^T=(c_{ij})^T=$

$ =(c_{ji})=(a_{ji}+b_{ji})=(a_{ji})+(b_{ji})=(a_{ij})^T+(b_{ij})^T=$

$=A^T+B^T $