## Linear Algebra and Its Applications, Exercise 2.1.2

Exercise 2.1.2. Of the following subsets of $\mathbf{R}^3$, which are subspaces and which are not?

(a) the set of vectors $(0, b_2, b_3)$ with the first component $b_1 = 0$

(b) the set of vectors $(1, b_2, b_3)$ with the first component $b_1 = 1$

(c) the set of vectors $(0, b_2, b_3)$ and $(b_1, 0, b_3)$ for which $b_1b_2 = 0$

(d) the single vector $(0, 0, 0)$

(e) all linear combinations of the vectors $(1, 1, 0)$ and $(2, 0, 1)$

(f) all vectors $(b_1, b_2, b_3)$ for which $b_3 - b_2 + 3b_1 = 0$

Answer: (a) This set is a subspace: It is closed under vector addition, since for two members of the set $u = (0, u_2, u_3)$ and $v = (0, v_2, v_3)$ the sum $(0, u_2+v_2, u_3+v_3)$ is also in the set. It is also closed under scalar multiplication, since for any vector $v = (0, v_2, v_3)$ the product $cv = (0, cv_2, cv_3)$ is also in the set.

(b) This set is not a subspace: It is not closed under scalar multiplication, since for a vector $v = (1, v_2, v_3)$ the product $0 \cdot v = (0, 0, 0)$ is not in the set. It is also not closed under vector addition, since for the vectors $u = (1, u_2, u_3)$ and $v = (1, v_2, v_3)$ the sum $(2, u_2+v_2, u_3+v_3)$ is not in the set.

(c) This set is not a subspace: It is not closed under vector addition, since the vectors $(1, 0, 0)$ and $(0, 1, 0)$ are in the set but their sum $(1, 1, 0)$ is not.

(d) This set is a subspace: It is closed under vector addition, since the sum of $(0, 0, 0)$ and $(0, 0, 0)$ is $(0, 0, 0)$. It is also closed under scalar multiplication, since for any scalar $c$ the product $c \cdot (0, 0, 0) = (0, 0, 0)$.

(e) This set is a subspace: It is closed under scalar multiplication, since for any vector $u = a \cdot (1, 1, 0) + b \cdot (2, 0, 1)$ the product $cu = ca \cdot (1, 1, 0) + cb \cdot (2, 0, 1)$ is also a linear combination of $(1, 1, 0)$ and $(2, 0, 1)$. It is also closed under vector addition, since for any two vectors $u = a \cdot (1, 1, 0) + b \cdot (2, 0, 1)$ and $v = c \cdot (1, 1, 0) + d \cdot (2, 0, 1)$ their sum $u + v = (a+c) \cdot (1, 1, 0) + (b+d) \cdot (2, 0, 1)$ is also a linear combination of $(1, 1, 0)$ and $(2, 0, 1)$.

(f) This is a subspace: It is closed under scalar multiplication, since given a vector $u = (u_1, u_2, u_3)$ in the set, for the scalar product $cu = (cu_1, cu_2, cu_3)$ we have $cu_3 - cu_2 +3cu_1 = c(u_3 - u_2 + 3u_1) = c \cdot 0 = 0$

It is also closed under vector addition, since given two vectors $u = (u_1, u_2, u_3)$ and $u = (u_1, u_2, u_3)$ in the set, for their sum $u + v = (u_1 + v_1, u_2 + v_2, u_3 + v_3)$ we have $(u_3+v_3) - (u_2+v_2) + 3 \cdot (u_1+v_1) = (u_3 - u_2 + 3u_1) + (v_3 - v_2 + 3v_1) = 0 + 0 = 0$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

This entry was posted in linear algebra. Bookmark the permalink.