# For a given value of $x$, find the basis

We have the matrix $X = \begin{bmatrix} 2 & 0 & -1 & 1 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ -1 & x & 2 & 1 \end{bmatrix}$

We want to find a basis for the row space, column space and null space of $X$ for values of $x \in \mathbb{R}$. What I did is put the matrix in rref, but I had to do it twice: once for $x=0$, once for $x \neq 0$.

• $x=0 \implies$ $\text{rref}(X) = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

• $x \neq 0 \implies$ $\text{rref}(X) = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

And from there, we can find the row space (the above two rows of the rref if $x=0$ and the above three of the rref is $x \neq 0$). Then the column spaec is column 1 and column 3 of $X$ if $x=0$ and column 1,2,3 of $X$ if $x \neq 0$.

I'm wondering if my ideas here are correct. I separated the problem into two cases, $x=0$ and $x \neq 0$, because to get the rref of $X$ I had to divide by $x$ at one point in a row operation. So I had to obtain the rref twice. If this is correct, is this also the best way of solving this problem, or can we do it in an easier way?

What you did was fine, but you could’ve saved yourself a little work by taking advantage of the structure of $X$. Since all of the other entries in $x$’s column are zero, if $x\ne0$ then its row is linearly independent of the other rows. Similarly, if $x\ne0$ then its column will end up containing a pivot and so will be linearly independent of the other columns. We can conclude from this the bases for the row and column spaces for the case $x\ne0$ can be obtained by extending those for the case $x=0$ by the second row of the rref and the second column of the original matrix, respectively.
So, start by assuming that $x\ne0$ and row-reduce as before, but carry $x$ along unchanged, giving $$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & x & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$ We can see that when $x=0$, we end up with an all-zero row, but it would not be the second row if using conventional row-reduction. However, by using the last method described here to compute the kernel basis, we’d end up with the second row being all-zero anyway, which means that we can read all of the bases for both the $x=0$ and $x\ne0$ cases directly from this matrix. For the case $x=0$, we get kernel basis $((0,-1,0,0)^T, (1,0,1,-1)^T)$, row space basis $((1,0,0,1)^T,(0,0,1,1)^T)$ and column space basis $((2,0,1,-1)^T,(-1,1,0,2)^T)$. For the case $x\ne0$, we set $x=1$ and read kernel basis $((1,0,1,-1)^T)$, row space basis $((1,0,0,1)^T,(0,1,0,0)^T,(0,0,1,1)^T)$ and column space basis $((2,0,1,-1)^T,(0,0,0,1)^T,(-1,1,0,2)^T)$. The respective row and column space bases differ by the inclusion $x$’s row and column as expected, as do the two kernel bases.