In full-disclosure I asked this question on Quora anonymously, because I thought the answer was going to be embarrassingly self-evident, but given the single, tangential answer (albeit very interesting), I want to maximize my chances by posting the question here with a follow-up immediately related secondary question.

  1. Before convolving a filter with an image, or a kernel with a layer in convolutional neural networks, the filter (or kernel) is flipped in its rows and columns. I am looking for the name of this flipped matrix.

The matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}\color{red}{\blacksquare}&\color{blue}{\blacksquare}\\\color{green}{\blacksquare}&\color{aqua}{\blacksquare}\end{bmatrix}$ with flipped columns and rows would be $\begin{bmatrix}d&c\\b&a\end{bmatrix}=\begin{bmatrix}\color{aqua}{\blacksquare}&\color{green}{\blacksquare}\\\color{blue}{\blacksquare}&\color{red}{\blacksquare}\end{bmatrix}.$

To be clear (given prior experience - I know it is by and large completely unnecessary), I am not asking for the transpose:


  1. Immediately after getting this flipped filter or kernel the convolution consists of a sum of all the entries of a Hadamard product, which really is sort of a "dot product of matrices". What is the name of this matrix operation in general:

$$\text{elementwise}\sum\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\circ \begin{bmatrix}z&w\\v &y\end{bmatrix}\right)=\text{elementwise}\sum\begin{bmatrix}az&bw\\cv&dy\end{bmatrix}=az+bw+cv+dy$$


Thanks to @Omnomnomnom the answer to the second question is the Frobenius inner product:

For $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $\begin{bmatrix}z&w\\v &y\end{bmatrix}$, the Frobenius inner product, $\langle A,B \rangle_\mathbf F$ takes two matrices and returns a number.

$$\small\sum_{\text{el.wise}}\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\circ \begin{bmatrix}z&w\\v &y\end{bmatrix}\right) = tr\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}^\top\cdot \begin{bmatrix}z&w\\v &y\end{bmatrix}\right)=tr\left(\begin{bmatrix}a&c\\b&d\end{bmatrix}\cdot \begin{bmatrix}z&w\\v &y\end{bmatrix}\right)=\tiny az+cv+bw+dy$$

and thanks to @Hurkyl

$$\small\sum_{\text{el.wise}}\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\circ \begin{bmatrix}z&w\\v &y\end{bmatrix}\right)=\small\sum_{\text{el.wise}}\left(\begin{bmatrix}az&bw\\cv&dy\end{bmatrix}\right)=\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}az&bw\\cv&dy\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\tiny az+cv+bw+dy $$

  • $\begingroup$ I don't think that there is one for the first. I use the second and generally just refer to it as the element-sum of the matrix (or in your case the element-sum of the convolution). $\endgroup$ – Stella Biderman Mar 26 '17 at 13:15
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    $\begingroup$ For that second operation: taking the "dot-product" of matrices is sometimes called the "Frobeinius" or "Hilbert-Schmidt" inner-product of matrices. Notably, if $A$ and $B$ are matrices, then $tr(A^TB)$ is exactly this dot-product. $\endgroup$ – Omnomnomnom Mar 26 '17 at 15:02
  • $\begingroup$ It doesn't really give a name for it, but if you use lexical ordering for the vectorization you could build the first as the composition of a reflection and a transpose where the reflection in question is the "opposite diagonal matrix identity". Maybe "reflected transpose". $\endgroup$ – mathreadler Mar 26 '17 at 16:27
  • $\begingroup$ Aside: the sum of the elements of a matrix can be given by $z^T A z$, where $z$ is the all ones vector. $\endgroup$ – Hurkyl Mar 26 '17 at 17:12

I do not know of any name for the operation so I will try and make an argument for why it could be called "reflected transpose".

Consider the $3\times 3$ matrix :

$${\bf A} = \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$$

using lexical vectorization, we get:

$$\mathrm{vec}({\bf A}) = \left[\begin{array}{ccccccccc} 1&4&7&2&5&8&3&6&9 \end{array}\right]^T$$

We can now define this "flipped identity" reflection in the vectorization:

$${\bf R} = \left[\begin{array}{ccccccccc}0&0&0&0&0&0&0&0&1\\0&0&0&0&0&0&0&1&0\\0&0&0&0&0&0&1&0&0\\0&0&0&0&0&1&0&0&0\\0&0&0&0&1&0&0&0&0\\0&0&0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0&0&0\\0&1&0&0&0&0&0&0&0\\1&0&0&0&0&0&0&0&0\end{array}\right]$$

The matrix you seek is then:

$$(\textrm{vec}^{-1}({\bf R}\textrm{vec}({\bf A})))^T = \left[\begin{array}{ccc}9&6&3\\8&5&2\\7&4&1\end{array}\right]$$

Which is the transpose of the special reflection $\bf R$ of the vectorization.

In some sense a reflected transpose.

We can of course do all operations inside the vectorization with $\bf T$ transpose:

$${\bf T} = \left[\begin{array}{ccccccccc}1&0&0&0&0&0&0&0&0\\0&0&0&1&0&0&0&0&0\\0&0&0&0&0&0&1&0&0\\0&1&0&0&0&0&0&0&0\\0&0&0&0&1&0&0&0&0\\0&0&0&0&0&0&0&1&0\\0&0&1&0&0&0&0&0&0\\0&0&0&0&0&1&0&0&0\\0&0&0&0&0&0&0&0&1\end{array}\right]$$

And we get:

$$\textrm{vec}^{-1}({\bf TR}\textrm{vec}({\bf A})) = \left[\begin{array}{ccc}9&6&3\\8&5&2\\7&4&1\end{array}\right]$$

  • $\begingroup$ Easier to stick to $3 \times 3$ matrices that describe "flip the rows" and "flip the columns" than to unfold into a 9-dimensional space. But either way this doesn't give a name to the operation, just a formula. $\endgroup$ – Hurkyl Mar 26 '17 at 17:07
  • $\begingroup$ Yes easier, but does not add as much room for understanding. I was just about to add that operation commutes with the transpose, which we could not show with matrices in the smaller space. $\endgroup$ – mathreadler Mar 26 '17 at 17:13
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    $\begingroup$ $(R_3AR_3)^\top = R_3 A^\top R_3$ follows by matrix algebra from $R_3^\top = R_3$. $\endgroup$ – Hurkyl Mar 26 '17 at 17:14
  • $\begingroup$ Yes. But it's good to learn as it's a very powerful tool in other contexts. $\endgroup$ – mathreadler Mar 26 '17 at 17:28

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