Multiplication of diagonal matrices with identity

Question

What would the result of this multiplication be, given that $A$ is an $m \times n$ rectangular diagonal matrix and $I$ is the identity matrix.

$$A^TIA = \cdots$$

Answer 1

The result will be $A^TA$ (because $IA = A$, of course, even for rectangular matrices).

And $A^TA$ will be a diagonal matrix with the diagonal entries being the square of the diagonal entries of $A$ (and some zeroes, if $m < n$).

Proof
Let $A = [a_{ij}]_{m \times n}$. Then $A^TIA = A^TA = B = [b_{ij}]_{n \times n}$ (say).

$$\begin{align} b_{ij} & = \sum_{k = 1}^m a^T_{ik}a_{kj}\\ & = \sum_{k = 1}^m a_{ki}a_{kj} \end{align}$$

As $a_{ki} = 0$ if $i \ne k$, and $a_{kj} = 0$ if $j \ne k$, we have $a_{ki}a_{kj} = 0$ if except when $i = k = j$, in which case $b_{ij} = b_{ii} = a_{ii}^2$.

When $m < n$, obviously $k$ cannot be equal to $i$ for $i \ge n$, so $b_{ii} = 0$ for all $i \ge n$.

Thus, $A^TA$ is a diagonal matrix with the $(i, i)$-element being the square of the $(i,i)$-element of $A$ when $i \le m$, and the remaining $n - m$ diagonal entries being $0$ (if $m < n$).