# Find a basis of nullspace(A)

Let $A = \begin{bmatrix}4&-4&2&-6\\2&-2&1&-3\end{bmatrix}$ Find a basis of nullspace$(A)$

I first put $A$ in RREF to get: $\begin{bmatrix}1&-1&1/2&-3/2\\0&0&0&0\end{bmatrix}$

I then found that $x_1 = 1x_2 - \frac{1}{2}x_3 + \frac{3}{2}x_4$

I then got $x_2 \begin{bmatrix}1\\0\\0\\0\end{bmatrix}, x_3 \begin{bmatrix}-1/2\\1\\0\\0\end{bmatrix}, x_4 \begin{bmatrix}3/2\\0\\1\\0\end{bmatrix}$

For some reason my online homework program is saying I am incorrect and that the answer should be:

$\begin{bmatrix}2\\2\\0\\0\end{bmatrix}, \begin{bmatrix}0\\1\\2\\0\end{bmatrix}, \begin{bmatrix}0\\-3\\0\\2\end{bmatrix}$ What am I doing wrong?

• Your basis is not exactly correct: $$\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4} = \pmatrix{x_2-\frac 12x_3+\frac 32x_4 \\ x_2 \\ x_3 \\ x_4} =\ \ ?$$ – user137731 Jul 19 '16 at 20:55
• $x_2 \begin{bmatrix}1\\1\\0\\0\end{bmatrix}, x_3 \begin{bmatrix}-1/2\\1\\0\\0\end{bmatrix}, x_4 \begin{bmatrix}3/2\\0\\1\\0\end{bmatrix}$ – Shammy Jul 19 '16 at 20:58
• Now your first is correct, but the second and third vectors are still wrong. – user137731 Jul 19 '16 at 20:58
• I think it's worth noting that the procedure you've gone through to get your basis is completely standard (even if there are other procedures, e.g. solving for $x_2$ instead of $x_1$), and aside from scaling the vectors by something obvious to clear denominators (i.e. multiplying them by 2), I would expect a reasonable computer program to follow the same strategy as you and produce a basis that is more obviously connected to yours (i.e. differing only up to some scalings) as opposed to linear combinations. – Keenan Kidwell Jul 19 '16 at 21:09
• Dear @Shammy, You might try the linear algebra toolkit: math.odu.edu/~bogacki/cgi-bin/lat.cgi. It's user-friendly and follows standard conventions in e.g. computing nullspaces (to wit, its nullspace calculator, when fed your matrix, produces the basis you've written down). I've used it extensively to check computations when teaching linear algebra and it's never given me anything unexpected. – Keenan Kidwell Jul 19 '16 at 21:12

The first vector you got is missing the $1$ in the second entry, since it represents the $x_2$ free variable (this is equivalent to the first solution vector from the homework program). The $x_3$ vector should have a $-1/2$ in the first entry and a $1$ in the third entry. The $x_4$ vector should have a $3/2$ in the first entry, and a $1$ in the fourth entry. Your only problem is that you need to put a $1$ in the entry of the vector that corresponds to the free variable (e.g. the $x_2$ vector should have a $1$ in the second entry). The basis that was given to you as a homework answer is another basis that also works, however involves a slightly different approach, which may have been causing the confusion.