# Determining a basis for Col($A$) and a dimension for the null space of $A$

Let $$A = \begin{bmatrix}1&-1&1&0&-2&1\\1&-1&1&1&0&0\\-1&1&-1&2&5&-1\end{bmatrix}$$

a) Determine a basis for Col($$A$$)

b) What is the dimension of the null space of $$A$$?

I just want to make sure I am not doing this incorrectly.

part a) I put $$A$$ in rref form to get: $$rref(A) = \begin{bmatrix}1&-1&1&0&-2&1\\0&0&0&1&2&-1\\0&0&0&0&-1&0\end{bmatrix}$$ so I got the basis for Col(A) = $$\begin{bmatrix}1\\1\\-1\end{bmatrix}, \begin{bmatrix}0\\1\\2\end{bmatrix}$$

Then for part b)

I know that nullity($$A$$) = # of non-pivot columns (or free vars) so I got that dim(null(A)) = 4? Since i have 4 free variables?

• You haven’t quite finished row-reducing $A$, but even at this stage you should be able to tell that its column space is three-dimensional, not two-dimensional. Knowing that should make it really easy to find a basis for it.
– amd
Jul 19 '16 at 21:50

Your matrix can be reduced with the following steps

• add $-1$ times row 1 to row 2
• add $1$ times row 1 to row 3
• add $-2$ times row 2 to row 3
• scale row 3 by $-1$
• add $2$ times row 3 to row 1
• add $-2$ times row 3 to row 2

This gives $$\DeclareMathOperator{rref}{rref}\rref A= \left[\begin{array}{rrrrrr} 1 & -1 & 1 & 0 & 0 & -3 \\ 0 & 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 0 & 1 & -2 \end{array}\right]$$ Columns one, four, and five are the pivot columns. Hence $$\DeclareMathOperator{Col}{Col}\Col A=\DeclareMathOperator{Span}{Span}\Span\left\{ \left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right], \left[\begin{array}{r} 0 \\ 1 \\ 2 \end{array}\right], \left[\begin{array}{r} -2 \\ 0 \\ 5 \end{array}\right] \right\}$$ To get a feel for what $\Col A$ looks like, you can put these basis vectors into the rows of a matrix $$C= \left[\begin{array}{rrr} 1 & 1 & -1 \\ 0 & 1 & 2 \\ -2 & 0 & 5 \end{array}\right]$$ The rows of $\rref C$ also form a basis for $\Col A$. Here, $$\rref C= \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$ Hence $$\Col A=\Span\left\{ \left[\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right], \left[\begin{array}{r} 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 0 \\ 0 \\ 1 \end{array}\right] \right\}=\Bbb R^3$$

So $\dim\Col A=3$ and the rank-nullity theorem implies that $$\dim\DeclareMathOperator{Null}{Null}\Null A=\#\text{ columns of }A-\dim\Col A=6-3=3$$

• Since the column space turns out to be good old $\mathbb R^3$, why not just use its standard basis? Jul 19 '16 at 21:54
• @HenningMakholm Seems pedagogically incorrect, no? Listing the columns corresponding to the pivot columns works in general. Jul 19 '16 at 21:55
• What seems pedagogically incorrect to me is to try to hide the (pretty important) fact that the column space is the entire $\mathbb R^n$. Jul 19 '16 at 21:57
• @HenningMakholm One could equally say that just using the standard basis would hide the fact that the columns corresponding to the pivot columns give a basis for the column space in the general case. Either way I've updated the answer to make it clear that the column space of this matrix is indeed all of $\Bbb R^3$. Jul 19 '16 at 22:04