Exercise 1221Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Subspace generated by columns of a matrix

#column space#generators#rank

Statement

Let
$A = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 1 & 3 & 2 \end{pmatrix}.$
Find a basis of the column space of $A$ .

Hint

Column-reduce or transpose and row-reduce; the pivot columns of $A$ give a basis.

Show solutionHide solution

Observation. Column 3 = column 1 + column 2, so column 3 is redundant.

Independence of columns 1 and 2. The two columns $(1,0,1)$ and $(2,1,3)$ are not proportional, so they are independent.

Conclusion. A basis of the column space is $((1,0,1),(2,1,3))$ , and the rank is $2$ .

Check. Row reduction confirms rank $2$ : $L_3\leftarrow L_3-L_1-L_2$ gives the zero row.