Lecture 20 · Week 5· 16:50

Basis and dimension in Rⁿ

Bases of subspaces, dimension, Steinitz exchange idea.

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In this lecture

Today: describing a subspace
Bases
Coordinates
Shrink or grow
The key theorem
The exchange idea
Dimension
Bases of Col A and Nul A
When the count is right
Summary

Materials

Check yourself

Try it before you look

Let $A = [122413]$ . Find a basis of $Col A$ and a basis of $Nul A$ , and check that the dimensions add up to $3$ .
Is $((1, 1, 0), (0, 1, 1), (1, 0, 1))$ a basis of $R^{3}$ ? Is it a basis of $Z_{2}^{3}$ ? (Use Example 2.6.11.)
True or false: if $H$ is a subspace of $R^{4}$ and $dim H = 4$ , then $H = R^{4}$ . Justify your answer.

Work it out on paper first. Then open the hint if you are stuck, and the answer to check.

1Hint

Everything you need is in Section 4.2 of the course book. Open it at page 7 of the chapter PDF; any theorem the question names is stated there, with an example worked the same way.

2Answer

$R_{2} \to R_{2} - 2 R_{1}$ gives row $2$ $= (0, 0, 1)$ ; then $R_{1} \to R_{1} - R_{2}$ gives row $1$ $= (1, 2, 0)$ . So the reduced row echelon form is $[102001]$ , with pivots in columns $1$ and $3$ . By Theorem 4.2.14(a), the pivot columns of $A$ , $(1, 2)$ and $(1, 3)$ , form a basis of $Col A$ , so $dim Col A = 2$ (here $Col A = R^{2}$ ). The free variable is $x_{2} = t$ , and the rows give $x_{1} = - 2 t$ , $x_{3} = 0$ , so $x = t (- 2, 1, 0)$ and $((- 2, 1, 0))$ is a basis of $Nul A$ (Theorem 4.2.14(b)). Check: $A (- 2, 1, 0) = (- 2 + 2 + 0, - 4 + 4 + 0) = (0, 0)$ , and $2 + 1 = 3$ , the number of columns.
Over $R$ , yes: the three vectors are independent (Example 2.6.11), there are $3$ of them and $dim R^{3} = 3$ , so they form a basis by Theorem 4.2.16(a). Over $Z_{2}$ , no: there $(1, 1, 0) + (0, 1, 1) + (1, 0, 1) = (0, 0, 0)$ , so the list is dependent, and a basis must be independent.
True. Apply Theorem 4.2.16
with $R^{4}$ in the role of $H$ and our $H$ in the role of $U$ : $H \subseteq R^{4}$ and $dim H = 4 = dim R^{4}$ (Example 4.2.13), so $H = R^{4}$ .