Is there a linear transformation that would flip the elements of a matrix around its center?

Question

Is there a linear transformation that would move the elements of a matrix in the following fashion? or some combination of other matrix operations including the transpose etc?

$$ \begin{gather} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \xrightarrow{\;\;T\;\;} \begin{bmatrix} \color{red}{a_{33}} & \color{blue}{a_{32}} & \color{green}{a_{31}} \\ \color{orange}{a_{23}} & a_{22} & \color{orange}{a_{21}} \\ \color{green}{a_{13}} & \color{blue}{a_{12}} & \color{red}{a_{11}} \end{bmatrix} \end{gather} $$

Answer 1

Fix a field $\mathbb{K}$ and $n\in\mathbb{Z}_{>0}$. Then, $[n]:=\{1,2,\ldots,n\}$ and $\text{Mat}_{n\times n}(\mathbb{K})$ is the vector space of $n$-by-$n$ matrices over $\mathbb{K}$. Let $T:\text{Mat}_{n\times n}(\mathbb{K})\to\text{Mat}_{n\times n}(\mathbb{K})$ be the flipping map sending an element $\big[a_{i,j}\big]_{i\in[n],j\in[n]}$ of $\text{Mat}_{n\times n}(\mathbb{K})$ to $\big[a_{n+1-i,n+1-j}\big]_{i\in[n],j\in[n]}$. It is actually clear that $T$ is a $\mathbb{K}$-linear map. However, I believe that the OP wants more.

If $P$ denotes the $n$-by-$n$ permutation matrix $$\newcommand\iddots{\mathinner{ \kern1mu\raise1pt{.} \kern2mu\raise4pt{.} \kern2mu\raise7pt{\Rule{0pt}{7pt}{0pt}.} \kern1mu }}\begin{bmatrix}0&0&\dots&0&0&1\\ 0&0&\dots&0&1&0\\ 0&0&\dots&1&0&0\\ \vdots&\vdots&\iddots&\vdots&\vdots&\vdots\\0&1&\dots&0&0&0\\ 1&0&\dots&0&0&0\end{bmatrix}\,,$$ then $T$ is the same as conjugation by $P$. That is, $$T(A)=PAP^{-1}$$ for all $A\in\text{Mat}_{n\times n}(\mathbb{K})$. (Since $P^{-1}=P$, we can also write $T(A)=PAP$ for all $A\in\text{Mat}_{n\times n}(\mathbb{K})$.) In particular, when $n:=3$, we have $$ \begin{bmatrix} 0& 0 & 1 \\ 0 &1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\, \begin{bmatrix} \color{red}{a_{11}} & \color{blue}{a_{12}} & \color{green}{a_{13}} \\ \color{orange}{a_{21}} & \color{grey}{a_{22}} & \color{magenta}{a_{23}} \\ \color{teal}{a_{31}} & \color{cyan}{a_{32}} & \color{brown}{a_{33}} \end{bmatrix}\, \begin{bmatrix} 0& 0 & 1 \\ 0 &1 & 0 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \color{brown}{a_{33}} & \color{cyan}{a_{32}} & \color{teal}{a_{31}} \\ \color{magenta}{a_{23}} & \color{grey}{a_{22}} & \color{orange}{a_{21}} \\ \color{green}{a_{13}} & \color{blue}{a_{12}} & \color{red}{a_{11}} \end{bmatrix}\,. $$

Answer 2

Of course:

Note that $e_ie_j^T$ where $i,j \in \{1, \ldots, 3\}$ and $e_i$ is the standard unit basis in $\mathbb{R}^3$ is a basis for $\mathbb{R}^{3 \times 3}$.

The linear map is:

$$L\left( e_ie_j^T \right)=e_{4-i}e_{4-j}^T$$

For example:

$$L\left( e_1e_1^T \right)=e_{3}e_{3}^T$$

We can permute the entries of a matrix and it is linear, we just have to describe the image of each basis element.