Stuck on a proof about rotating a matrix

Question

Given a matrix $A \in M^{n×n}(F)$, let $A^{\rho}$ denote the matrix obtained from $A$ by ‘rotating’ it $90^{\circ}$ clockwise. For example, $$\begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}^\rho =\begin{bmatrix} 3&1\\ 4&2 \end{bmatrix}.$$ Find (with proof) an expression for $A^\rho$ in terms of $A$.

Through analyzing $2\times 2, 3\times 3, 4\times 4$, and $5\times 5$ cases, it seems that the rotation is equivalent to transposing A and then swapping the first column with the $n^{th}$ column, the second column with the $(n-1)^{th}$ column, and so on. Performing these column operations is equivalent to right multiplication by a permutation matrix.

So I've concluded that $A^\rho$ is equivalent to $A^TP$ where $P$ is the the "mirror image" of $I_n$.

For example in the $5 \times 5$ case, $P= \begin{bmatrix} 0&0&0&0&1\\ 0&0&0&1&0\\ 0&0&1&0&0\\ 0&1&0&0&0\\ 1&0&0&0&0 \end{bmatrix}.$

I'm struggling with how to prove this in the general case, or how to express $P$ for any $n$. Any help is appreciated!

Answer 1

A good way to prove this is to take $A_{ij}$ and track where it should end up after a rotation, and where it does end up after your expression.

To get you started: $A_{i,j} = A^{\rho}_{j , n-i} $

Now show that $A_{i,j} = (A^TP)_{j, n-i}$. This will show that they are equivalent matrices, since all their entries are the same. If you found a good expression, it should work. As for defining $P$, your current definition should be good enough.

Answer 2

Anthony Ter's suggestion of verifying $(A^{\rm T} P)_{j,\ n-i+1}$ actually equals $A_{i,j}$ is key (I'm assuming indexes go from $1$ to $n$), but in order to do so, it's helpful to rely on a way of defining $P$ entrywise. Consider the Kronecker delta: $$\delta_{i,j} :=\begin{cases} 1,& \text{if }\ i=j;\\ 0,& \text{if }\ i\neq j.\end{cases}$$ Then you can define $I_n$ entrywise as $(I_n)_{i,j} :=\delta_{i,j}$; and similarly, $(P)_{i,j} :=\delta_{i,n-j+1}$. Then, you can compute $(A^{\rm T} P)_{j,\ n-i+1}$ directly by using the matrix-product entrywise definition and the matrix transpose property: $(A^{\rm T})_{i,j} := A_{j,i}$.

\begin{align} (A^\rho)_{j,\ n-i+1} &=(A^{\rm T} P)_{j,\ n-i+1} =\sum_{k=1}^n (A^{\rm T})_{j,k} \ P_{k,\ n-i+1}\\ &=\sum_{k=1}^n A_{k,j} \ \delta_{k,\ n-(n-i+1)+1} =\sum_{k=1}^n A_{k,j} \ \delta_{k,i}\\ &=A_{1,j}\ \delta_{1,i} +A_{2,j}\ \delta_{2,i} +A_{3,j}\ \delta_{3,i} +\ldots +A_{n,j}\ \delta_{n,i}.\end{align} Note that inside the summation, $k$ is variable, whereas $i$ and $j$ are fixed. For $\delta_{k,i}$ not to be zero (i.e., one) then $k$ shall equal $i$. Thus, $(A^\rho)_{j,\ n-i+1} =A_{i,j} \ \delta_{i,i} =A_{i,j}$, as desired.