In quantum mechanics, we can represent functions as vectors, and unitary operators as matrices, with given orthogonal function system (basis).
I wonder if we could notate binary operators like + ,- ,× ,÷ in matrix notation, or matrix-like candidates.
In addition, can we notate some mathematical actions like assignment in matrix form?
Also if then, how can we notate assigning values to multi-input functions like f(x,y)?
Thanks in advance.
Unfortunately, because I am a high school student yet, I hope that I could understand your answers.
Edit
My final concept was substituting differential equations to matrix(or tensor like) equation, so that I could write an algorithm that computers can solve any kind of differntial equation.
According to @Ricky, it seems that some operations cannot be substituted by simple matrix notation, as matrices can only represent linear operations.
Hope this question is not a kind of XY problem.
Edit #2
It seems that I can generalize @Ricky's answer by concatenating two operand vectors into a matrix, can I?
Edit #3
What I mean by assignment for example:
Assign(f,x)=f(x)
Where Assign is a kind of mathematical action, f a function, f(x) a value of f when its argument is x.
So Assign can be a functional.
If anyone thinks that my question should be splitted into several questions, please don't hesitate to comment about it.
For addition and subtraction, the answer to your question is a straightforward yes. Let's say you want to add together two numbers, $a$ and $b$. We can put them in the vector $\begin{pmatrix}a\\b\end{pmatrix}$.
Then we have: $ a+b = \begin{pmatrix}1 & 1 \end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} $.
Similarly, $ a-b = \begin{pmatrix}1 & -1 \end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} $.
Multiplication is a bit of a more difficult nut to crack. Multiplication is a bilinear operation, while matrices themselves only represent linear operations. (Definition of bilinear here: https://en.wikipedia.org/wiki/Bilinear_map). What this means is that there is no single matrix $M$ such that $M\begin{pmatrix}a\\b\end{pmatrix} = ab$ for all real numbers $a$ and $b$. However, we can arrange things a bit differently and get the answer we want, I think. Write each of the numbers we want to multiply as a $1\times 1$ matrix. Then $\begin{pmatrix}a\end{pmatrix}\begin{pmatrix}b\end{pmatrix} = \begin{pmatrix}ab\end{pmatrix}$.
This works because while each individual matrix is a linear operator on its own, the operation of matrix multiplication itself is a bilinear operation. (You can think of each matrix as a function that takes in a vector and outputs another vector. This function is linear in its single input. Meanwhile, matrix multiplication can be thought of as a function that takes in two matrices and outputs another matrix. This function is bilinear in its two inputs.)
You can also do division if you're allowed to use matrix inversion, but that's kind of cheap in my opinion: $\begin{pmatrix}a\end{pmatrix}\begin{pmatrix}b\end{pmatrix}^{-1} = \begin{pmatrix}\frac ab\end{pmatrix}$.
I don't really know what you mean by "assignment", so I'll just ignore that for now.
So there you go. Hopefully I didn't completely misunderstand your question.