An other way of proving that : $(^tA A)_{i,j} = \langle a_i, a_j \rangle$

Question

Thanks' to a recent question I know the following :

If $A \in M_{n,k}(\mathbb{R})$ with columns $(a_1, ..., a_k)$ then $(^tA A)_{i,j} = \langle a_i, a_j \rangle$.

It's very easy to prove since you just need to multiply the matrices. I am wondering if there is less magical way of proving this. That's to say the proof shouldn't use matrix multiplication but only the property of the transpose, supplementary spaces... so that this result doesn't seem magical.

I thought about using the fact that : $\operatorname{Ker}(A)^{\perp} = \operatorname{Im}(^tA)$ but it doesn't seems to help.

Thank you !

Answer 1

For any matrix $M$, we have $M_{i,j} = \langle e_i,Me_j\rangle$. Therefore, $$ (A^t A)_{i,j} = \langle e_i,A^t Ae_j\rangle = \langle A e_i,Ae_j\rangle = \langle a_i,a_j\rangle.$$