Is there a standard way of expressing matrix row and column operations algebraically?

Question

I've looked hard and failed to find any convincingly standard algebraic notation for such. I have a specific problem in mind. It is the transformation of this matrix:

$\mathrm{W} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.25 & 0 \\ 0 & 0.75 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $

into this one:

$\mathrm{A} = \begin{bmatrix} 1 & 0 \\ -0.25 & 0.25 \\ -0.75 & 0.75 \\ 0 &-1 \\ \end{bmatrix} $

It's rather easy to see how to do that column by column, basically:

$a_{i,j} = w_{i,j} - w_{i,j+1}$

and I can write code do that but what I'm look for is a nice algebraic nomenclature for the operation, ideally in a standard matrix notation such that I might write:

$\mathrm{A} = f(\mathrm{W})$

and I could write the function f algebraically somehow. I can invent a notation easily enough, for example:

$\mathrm{A} = \mathrm{W}>>1 - \mathrm{W}<<1$

where:

$>>$ is a right shift operator which shifts each column one to right losing the right most column and reducing the number of columns by 1.

$<<$ is a left shift operation which shifts each column one to the left losing the first column and reducing the number of columns by 1.

But I made that up. And of course we can work on transposes and use row operators if they exist instead. By quest here is for the most standard, widely understood and used algebraic notation if it exists, and ideally a reference I could cite that documents it. I've failed to find any such thing.

Answer 1

Thanks to rikhavshah's hint I think I've nailed something I'm satisfied with, as follows:

$\mathrm{A} = \mathrm{W}\mathrm{R} - \mathrm{W}\mathrm{L}$

where:

$\mathrm{W}$ is an $n \times m$ matrix

$\mathrm{A}$ is an $n \times m-1$ matrix being the differences between adjacent columns in $\mathrm{W}$

$\mathrm{R} = \begin{bmatrix} \mathrm{I}_{m-1} \\ 0_{1,m-1} \end{bmatrix} $ is an $m \times m-1$ right truncating matrix

$\mathrm{L} = \begin{bmatrix} 0_{1,m-1}\\ \mathrm{I}_{m-1} \end{bmatrix} $ is an $m \times m-1$ left truncating matrix

and so a worked example for the specific matrices in the question:

$ \begin{align*} \mathrm{A} &= \mathrm{W} \mathrm{R} - \mathrm{W} \mathrm{L} \\&= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.25 & 0 \\ 0 & 0.75 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.25 & 0 \\ 0 & 0.75 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \\&= \begin{bmatrix} 1 & 0 \\ 0 & 0.25 \\ 0 & 0.75 \\ 0 & 0 \\ \end{bmatrix} - \begin{bmatrix} 0 & 0 \\ 0.25 & 0 \\ 0.75 & 0 \\ 0 & 1 \\ \end{bmatrix} \\&= \begin{bmatrix} 1 & 0 \\ -0.25 & 0.25 \\ -0.75 & 0.75 \\ 0 & -1 \\ \end{bmatrix} \end{align*}$

which uses only standard notation from:

• https://en.wikipedia.org/wiki/Identity_matrix
• https://en.wikipedia.org/wiki/Zero_matrix
• https://en.wikipedia.org/wiki/Block_matrix

which I find standard enough with references available for the naive reader.

If there is a better option I'm all ears but this one seems the best so far!

UPDATE:

This can simplify further to:

$\mathrm{A} = \mathrm{W}\boldsymbol\Delta$

Where: $\boldsymbol\Delta = \begin{bmatrix} \mathrm{I}_{m-1} \\ 0_{1,m-1} \end{bmatrix} - \begin{bmatrix} 0_{1,m-1} \\ \mathrm{I}_{m-1} \end{bmatrix}$

Which is easily the most elegant solution to date. Defining on generic multiplier for the conversion. We can think of $\boldsymbol\Delta$ as a differencing matrix methinks which creates for any matrix a new on which has one fewer columns and each column is the difference between two adjacent columns in the original.

Answer 2

One potential answer is to use colon notation $$ f(W) = W(:,\text{1:end-1}) - W(:,\text{2:end}) $$ It is well-known within the numeric linear algebra community; see for example Golub & Van Loan (2013).

Answer 3

This is simply the linear transformation $$A=WP$$ where $$ P = \begin{bmatrix} \,\,\,1 & \,\,\,0 \\ -1 & \,\,\,1 \\ \,\,\,0 & -1 \\ \end{bmatrix} $$

Update

Let's consider the case $m=4$, and the permutation matrix with ones on the first sub-diagonal $$ B = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix}\in{\mathbb R}^{m\times m} $$

Then $B^m=I,\,$ and the product $$ W(I-B) = \big[\,A\,\,c\,\big] $$ yields the desired $A$ matrix augmented by an extra column $(c)$ which contains the difference of the last & first columns of $W$. This is the circulant extension of your differencing algorithm.