Which of the following are subspaces of R?W1 =y : X,y,z are real and x- 3y+ 4z =0%3Dx2W2 =z2X,y,z are real-y?.X-yW3=y+z:X,y are real%3DX-4z|

Answered: Which of the following are subspaces of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Which of the following are subspaces of R?
W1 =
y : X,y,z are real and x- 3y+ 4z =0
%3D
x2
W2 =
z2
X,y,z are real
-y?.
X-y
W3=
y+z
:X,y are real
%3D
X-4z
|

Expert Solution

Step 1

PART - A-

Concept Used:

$W$ is said to be subspace of the vector space V if

$W \neq \emptyset$

$u, v \in W \Rightarrow u + v \in W u \in W, k - s c a l a r \Rightarrow k u \in W$

Given:

$W_{1} = \{[\begin{matrix} x \\ y \\ z \end{matrix}] : x, y, z \in ℝ | x - 3 y + 4 z = 0\} . . . . . . . . . . . . . . . (1)$

We want to check is it subspace of $ℝ^{3}$

Solution:

Let $u = (x, y, z), v = (x', y', z')$ be any vectors in $W_{1}$ Then from (1) we get

$x - 3 y + 4 z = 0 . . . . . . . . . . . . . . . . (2) x' - 3 y' + 4 z' = 0 . . . . . . . . . . . . . . . . . (3)$

Now $u + v = (x + x', y + y', z + z')$

we need to show that $(x + x') - 3 (y + y') + 4 (z + z') = 0$

Now

$\begin{array}{rcl} (x + x') - 3 (y + y') + 4 (z + z') & = & x + x' - 3 y - 3 y' + 4 z + 4 z' \\ = & (x - 3 y + 4 z) + (x' - 3 y' + 4 z') \\ = & 0 + 0 . . . . . \{f r o m 1 a n d 2\} \\ (x + x') - 3 (y + y') + 4 (z + z') & = & 0 \end{array}$

Therefore $u + v \in W_{1}$

Also for any scalar k to show that $k u \in W_{1}$ that is $(k x, k y, k z) \in W_{1}$ so

We have to show that $k x - 3 (k y) + 4 (k z) = 0$

But

$\begin{array}{rcl} k x - 3 (k y) + 4 (k z) & = & k (x - 3 y + 4 z) \\ k x - 3 (k y) + 4 (k z) & = & 0 \end{array}$

Thus $k u \in W_{1}$

Therefore both conditions are satisfied

Hence $W_{1}$ is subspace of $ℝ^{3}$

Step 2

PART - B-

Concept Used:

$W$ is said to be subspace of the vector space V if

$W \neq \emptyset$

$u, v \in W \Rightarrow u + v \in W u \in W, k - s c a l a r \Rightarrow k u \in W$

Given:

$W_{2} = \{[\begin{matrix} x^{2} \\ z^{2} \\ - y^{2} \end{matrix}] : x, y, z \in ℝ\} . . . . . . . . . . . . . . . (1)$

We want to check is it subspace of $ℝ^{3}$

Solution:

Let $u = (x, y, z), v = (x', y', z')$ be any vectors in $W_{2}$ Then from (1) we get

Now consider

$\begin{array}{rcl} u + v & = & (x, y, z) + (x', y', z') \\ = & [\begin{matrix} x^{2} \\ z^{2} \\ - y^{2} \end{matrix}] + [\begin{matrix} x'^{2} \\ z'^{2} \\ - y'^{2} \end{matrix}] \\ = & [\begin{matrix} x^{2} + x'^{2} \\ z^{2} + z'^{2} \\ - (y^{2} + y'^{2}) \end{matrix}] \end{array}$

Therefore $u + v \notin W_{2}$

Therefore first conditions are not satisfied

Hence $W_{2}$ is not subspace of $ℝ^{3}$

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