Question

1)Let V= R3. Show that W is a subspace of V where W = {(a, b, 0): a, b ϵ R}, i.e. W is the xy plane consisting of those vectors whose third component is 0.

2)Let V= R3. Show that W is not a subspace of V where W = {(a, b, c): a, b, c ϵ Q}, i.e. W consists of those vectors whose components are rational numbers.

3)Determine whether the vectors v1 = (2, -1, 3), v2 = (4, 1, 2) and v3 = (8, -1, 8) span R3.

4)Use system of linear equations form and row echelon form to show that the vectors (2, -1, 4), (3, 6, 2) and (2, 10, -4) are linearly independent.

Answer 1

1) Let $V = R3$ . Show that $W$ is a subspace of $V$ where $W = \{(a, b, 0) : a, b \in R\}$ , i.e. $W$ is the xy plane consisting of those vectors whose third component is 0.

2) Let $V = R3$ . Show that $W$ is not a subspace of $V$ where $W = \{(a, b, c) : a, b, c \in Q\}$ , i.e. $W$ consists of those vectors whose components are rational numbers.

3) Determine whether the vectors $v1 = (2, -1, 3)$ , $v2 = (4, 1, 2)$ and $v3 = (8, -1, 8)$ span R3.

4) Use system of linear equations form and row echelon form to show that the vectors $(2, -1, 4)$ , $(3, 6, 2)$ and $(2, 10, -4)$ are linearly independent.

Solution.

1) $W$ is a subspace of $\pmb{R}^3$ .

For proving this fact it is necessary to show that for any $\nu_{1} = (x1, y1, 0)$ and $\nu_{2} = (x2, y2, 0)$ belonging to $W$ and any $a1, a2 \in R$ their linear combination $a1^{*}\nu_{1} + a2^{*}\nu_{2} \in W$ . We have $a1^{*}(x1, y1, 0) + a2^{*}(x2, y2, 0) = (a1^{*}x1 + a2^{*}x2, a1^{*}y1 + a2^{*}y2, 0) \in W$ . Q.E.D.

2) The set $W = \{(a, b, c,): a, b, c, \in Q\}$ is not a subspace over the set real numbers as, for example,

$$\sqrt {2} (a, b, c) = (\sqrt {2} a, \sqrt {2} b, \sqrt {2} c) \notin W,$$

as $\sqrt{2} a, \sqrt{2} b, \sqrt{2} c \notin Q$ .

3) Vectors $\nu_{1}, \nu_{2}, \nu_{3}$ form a spanning set of $R^3$ if any vector $\nu \in R^3$ may be represented as a linear combination of $\nu_{1}, \nu_{2}, \nu_{3}$ . This can be done always if a determinant formed by the vectors $\nu_{1}, \nu_{2}, \nu_{3}$ is not equal to zero. So, consider

$$\det = \left| \begin{array}{c c c} 2 & 4 & 8 \\ - 1 & 1 & - 1 \\ 3 & 2 & 8 \end{array} \right|$$

and using Leibniz formula, for example, we have $\det = 16 - 12 - 16 - 24 + 4 + 32 = 0$ . The given vectors do not form a spanning set of $R^3$ .

4) Consider a system

$$x (2, - 1, 4) + y (3, 6, 2) + z (2, 1 0, - 4) = 0$$

or

$$\begin{array}{l} 2 x + 3 y + 2 z = 0 \\ - x + 6 y + 1 0 z = 0. \\ 4 x + 2 y - 4 z = 0 \\ \end{array}$$

It has only trivial solution if its determinant equals zero. So, forming and computing determinant we have

$$\det = \left| \begin{array}{ccc} 2 & 3 & 2 \\ -1 & 6 & 10 \\ 4 & 2 & -4 \end{array} \right| = -48 + 120 - 4 - 48 - 40 - 12 = -32 \neq 0.$$

Hence the given vectors are linear independent.

Do the same using row echelon form. With the help of Gaussian elimination we get

$$\left| \begin{array}{cccccc} 2 & 3 & 2 & 0 & -1 & 6 & 10 \\ -1 & 6 & 10 & 0 & \rightarrow & 2 & 3 & 2 \\ 4 & 2 & -4 & 0 & \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \rightarrow \left| \begin{array}{cccccc} -1 & 6 & 10 & 0 & -1 & 6 & 10 \\ 0 & 15 & 22 & 0 & \rightarrow & 0 & 15 & 22 \\ 0 & 26 & 36 & 0 & \end{array} \right| \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \Rightarrow x = y = z = 0$$

and vectors are linear independent.