## 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.

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