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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?

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    $\begingroup$ 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. $\endgroup$ – amd Jul 19 '16 at 21:50
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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 $$ \dim\DeclareMathOperator{Null}{Null}\Null A=\#\text{ columns of }A-\dim\Col A=6-3=3 $$

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  • $\begingroup$ Since the column space turns out to be good old $\mathbb R^3$, why not just use its standard basis? $\endgroup$ – Henning Makholm Jul 19 '16 at 21:54
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    $\begingroup$ @HenningMakholm Seems pedagogically incorrect, no? Listing the columns corresponding to the pivot columns works in general. $\endgroup$ – Brian Fitzpatrick Jul 19 '16 at 21:55
  • $\begingroup$ What seems pedagogically incorrect to me is to try to hide the (pretty important) fact that the column space is the entire $\mathbb R^n$. $\endgroup$ – Henning Makholm Jul 19 '16 at 21:57
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    $\begingroup$ @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 $\Bbb R^3$. $\endgroup$ – Brian Fitzpatrick Jul 19 '16 at 22:04

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