Finding the basis for null space of the matrix $[0^T, 1^T]$

Question

As part of some work I am doing, I am looking into the basis for the null space of the following matrix $$ [\mathbf{0}^T, \mathbf{1}^T], $$ where the vector of zeros is of length n and the vector of ones is of length m. Since the matrix is already in reduced row echelon form, I multiply it by an arbirtary vector of length n+m and equate to zero to give the following: $$ x_{n+1}+\dots+x_{n+m}=0. $$ Now, $x_{n+1}$ will be my dependent variable, with $x_{n+1}, \dots,x_{n+m}$ my independent variables. This leads to a basis with a matrix $M$ of size $n \times (m-1)$, a vector of $-1$ of length $m-1$ and an identity matrix of size $m-1$. My basis has $m-1$ columns, however this was not what I was anticipating. I carried out the same calculation on Matlab (using rank command) and I was given a basis of size $(n+m) \times (n+m-1)$, which was what I was looking for.

Could anyone give me any pointers as to where I am failing to understand why i've only managed to obtain $m-1$ basis vectors, yet Matlab is able to generate $n+m-1$ of them?

Answer 1

You have $n+m-1$ free variables, since the matrix has rank $1$. You can choose the non-free variable as any one among the last $m$, for instance the very last one; the equation can be written $$ x_{m+n}=-x_{n+1}-\dots-x_{n+m-1} $$ so the basis is $$ \underbrace{\begin{bmatrix} 1\\0\\0\\\vdots\\0\\0\\0\\\vdots\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\1\\0\\\vdots\\0\\0\\0\\\vdots\\0\\0 \end{bmatrix}, \dots, \begin{bmatrix} 0\\0\\0\\\vdots\\1\\0\\0\\\vdots\\0\\0 \end{bmatrix}}_n, \underbrace{ \begin{bmatrix} 0\\0\\0\\\vdots\\0\\1\\0\\\vdots\\0\\-1 \end{bmatrix}, \begin{bmatrix} 0\\0\\0\\\vdots\\0\\0\\1\\\vdots\\0\\-1 \end{bmatrix}, \dots, \begin{bmatrix} 0\\0\\0\\\vdots\\0\\0\\0\\\vdots\\1\\-1 \end{bmatrix}}_{m-1} $$

Answer 2

As you are trying to find the "rank" of this nonzero vector as a $1 \times n+m$ matrix, it has one linearly independent row, so it has rank $1$. thus the dimension of its null space is $n+m-1$.

You have, I believe, identified the $m-1$ solutions to $vx=0$ involving the nonzero elements of $v$; but notice how, for instance, $[0,0,0,1,1][1,0,0,0,0]^\text{T}= 0$; for each of the $n$ zeroes, a similar unit vector is a solution.

Thus there is a basis of size $n+m-1$.

When I ask matlab rank(null([zeros(n) ones(m)])) , for a handful of test values $n,m$, it gives the correct answer $n+m-1$.