Find a basis for the space of all $2$ by $3$ matrices whose rows and columns sum to zero.

Question

Find a basis for the space of all $2$ by $3$ matrices whose rows and columns sum to zero.

I'm having trouble doing these kinds of questions methodically. I can 'guess' at the answer but I'm not sure of a clear method to get the result. Can anyone offer a step by step method or at least express clearly the process of reasoning to get the result?

Answer 1

This is a basis $$\begin{pmatrix} 1&0&-1\\-1&0&1 \end{pmatrix}$$ $$\begin{pmatrix} 0&1&-1\\0&-1&1 \end{pmatrix}$$Proof: The requested matrices are of form:$$\begin{pmatrix} a&b&c\\d&e&f \end{pmatrix}$$where$$a+d=0\\b+e=0\\c+f=0$$therefore $$a=-d\\b=-e\\c=-f$$and the matrix gets as following$$\begin{pmatrix} a&b&c\\-a&-b&-c \end{pmatrix}$$and $$a+b+c=0$$which gives us the most general form of such matrices:$$\begin{pmatrix} a&b&-a-b\\-a&-b&a+b \end{pmatrix}$$. Since we have two degree of freedom we one basis can be obtained using $a=0,b=1$ and one other using $a=1,b=0$

Answer 2

Let's consider $m$-by-$n$ matrices in place of $2$-by-$3$ matrices. Take $A=(a_{ij})$ to be a general matrix Then the entries in the last column will depend on the first $n-1$ columns. The top row will be $$\pmatrix{a_{11}&a_{12}&\cdots&a_{1\,n-1}&-a_{11}-a_{12}-\cdots-a_{1\,n-1}}$$ etc. Likewise the last row will depend on the top $m-1$ rows. The upshot is that $A$ is determined by $(m-1)(n-1)$ parameters $a_{ij}$ for $1\le i\le m-1$ and $1\le j\le n-1$. Looking at the general form of the final matrix will enable you to write down a basis.