Lecture 19 · Week 5· 12:54

Subspaces of Rⁿ

Column space, null space.

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In this lecture

Today: subspaces
Subspaces
Sets that are not subspaces
Spans are subspaces
The column space
The null space
Computing the null space
What the two spaces say

Materials

Check yourself

Try it before you look

Which of these subsets of $R^{3}$ are subspaces? $H_{1} = {(x, y, z) : x - y + 2 z = 0}$ , $H_{2} = {(x, y, z) : x - y + 2 z = 1}$ , $H_{3} = {(x, y, z) : z = x^{2}}$ . Give a reason for each.
Let $A = [103001 2 - 1]$ . Find vectors that span $Nul A$ , and decide whether $Col A = R^{2}$ .
True or false: if $A$ is a $3 \times 5$ matrix, then $Nul A \neq = {0}$ .

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 4.1 of the course book. Open it at page 1 of the chapter PDF; any theorem the question names is stated there, with an example worked the same way.

2Answer

$H_{1}$ is a subspace: it is $Nul [1 - 1 2]$ (Theorem 4.1.12). $H_{2}$ is not: $0 - 0 + 2 \cdot 0 = 0 \neq = 1$ , so $0 \in / H_{2}$ . $H_{3}$ is not: it contains $0$ and $(1, 0, 1)$ , but $2 (1, 0, 1) = (2, 0, 2)$ has $z = 2 \neq = 4 = x^{2}$ , so it is not closed under scalar multiplication.
$A$ is already in reduced row echelon form, with pivots in columns $1$ and $3$ and free variables $x_{2} = s$ , $x_{4} = t$ . Then $x_{1} = - 3 s - 2 t$ and $x_{3} = t$ , so $x = s (- 3, 1, 0, 0) + t (- 2, 0, 1, 1)$ and $Nul A = Span {(- 3, 1, 0, 0), (- 2, 0, 1, 1)}$ (Proposition 4.1.15). Both rows have a pivot, so $Col A = R^{2}$ (Theorem 4.1.16(a)).
True. $A x = 0$ has $5$ unknowns and $3$ equations, so it has a nontrivial solution (Theorem 2.5.9), and that solution lies in $Nul A$ .