Question

Verify that W is a subspce of V. Assume that V has the standard operations.

W={(x1,x2,x3,0)} where x1,x2,x3 are real numbers.
V=R4

Answer 1

Answer on Question #80925 – Math – Linear Algebra

Question

Verify that $W$ is a subspace of $V$. Assume that $V$ has the standard operations.

$W = \{(x_{1},x_{2},x_{3},0)\}$, where $x_{1},x_{2},x_{3}$ are real numbers.

$V = \mathbb{R}^4$.

Solution

$W$ is a subspace, if the following three conditions are satisfied:

1) $W$ is non-empty (the zero vector is in $W$).

2) $W$ is closed under addition: if $\vec{u}$ and $\vec{w}$ are in $W$, then $\vec{u} + \vec{w}$ is in $W$.

3) $W$ is closed under scalar multiplication: if $\vec{u}$ is in $W$, and $c$ is a scalar, then $c\vec{u} \in W$.

1) $W$ is non-empty because it contains the zero vector $(0,0,0,0)$.

2) Let $\vec{u} = (u_{1}, u_{2}, u_{3}, 0)$ and $\vec{w} = (w_{1}, w_{2}, w_{3}, 0)$ be two vectors in $W$. Show that $W$ is closed under addition

$$\begin{array}{l} \vec{u} + \vec{w} = (u_{1}, u_{2}, u_{3}, 0) + (w_{1}, w_{2}, w_{3}, 0) = (u_{1} + w_{1}, u_{2} + w_{2}, u_{3} + w_{3}, 0) = \\ = (x_{1}, x_{2}, x_{3}, 0) \end{array}$$

where $x_{1} = u_{1} + w_{1}, x_{2} = u_{2} + w_{2}$ and $x_{3} = u_{3} + w_{3}$ are real numbers.

Hence, $\vec{u} + \vec{w}$ is in $W$.

3) Let $\vec{u} = (u_{1}, u_{2}, u_{3}, 0)$ be a vector in $W$, and let $c$ be any real number. Show that $W$ is closed under scalar multiplication

$$\begin{array}{l} \overline{c \vec{u}} = c (u_{1}, u_{2}, u_{3}, 0) = (c u_{1}, c u_{2}, c u_{3}, 0) = (x_{1}, x_{2}, x_{3}, 0) \end{array}$$

where $x_{1} = c u_{1}, x_{2} = c u_{2}$ and $x_{3} = c u_{3}$ are real numbers.

Hence, $c\vec{u}$ is in $W$.

Finally, because all three conditions are satisfied, we can conclude that $W$ is a subspace of $\mathbb{R}^4$.

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