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Chapter 4: Problem 10

\(W\) is not a subspace of the vector space. Verify this by giving a specific example that violates the test for a vector subspace (Theorem 4.5). \(W\) is the set of all vectors in \(R^{2}\) whose components are integers.

Short Answer

Expert verified
The set \(W\) is not a subspace of the vector space because it does not satisfy all conditions required for a subspace. Specifically, it fails the test for closure under scalar multiplication.

Step by step solution

01

Define the Subspace

The subspace in question is set \(W\), which comprises all vectors in \(R^{2}\) whose components are integers. Now, we will put this to the test to determine if it is a true subspace or not.
02

Check for Zero Vector

The first condition to check is whether vector \(0\) is in the set \(W\). Indeed, the vector (0,0) is an integer and thus, is a part of the set. This means the first condition is satisfied.
03

Checking for Closure Under Addition

Next, we test whether \(W\) is closed under vector addition. This means that if we add any two vectors in \(W\), the result should also be in \(W\). If we take two arbitrary integer vectors, say (1,2) and (3,4), and add them, we get (4,6), which is also an integer vector. Hence, \(W\) is also closed under addition.
04

Checking for Closure Under Scalar Multiplication

The last condition is that \(W\) must be closed under scalar multiplication. Let's take the vector (1,2) from \(W\) and multiply it by 0.5, which is not an integer. The resulting vector is (0.5,1), which is not in \(W\) because its components are not integers. Therefore, \(W\) is not closed under scalar multiplication.

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Most popular questions from this chapter

Determine which functions are solutions of the linear differential equation. \(y^{\prime \prime}+y=0\) (a) \(e^{x}\) (b) \(\sin x\) (c) \(\cos x\) (d) \(\sin x-\cos x\)

Find the coordinate matrix of \(\mathbf{x}\) in \(R^{n}\) relative to the basis \(B\) $$B=\left\\{\left(\frac{3}{2}, 4,1\right),\left(\frac{3}{4}, \frac{5}{2}, 0\right),\left(1, \frac{1}{2}, 2\right)\right\\}, \mathbf{x}=\left(3,-\frac{1}{2}, 8\right)$$

Determine which functions are solutions of the linear differential equation. \(y^{\prime \prime}-y^{\prime}-2 y=0\) (a) \(y=x e^{2 x}\) (b) \(y=2 e^{2 x}\) (c) \(y=2 e^{-2 x}\) (d) \(y=x e^{-x}\)

Use a graphing utility with matrix capabilitics to (a) find the transition matrix from \(B\) to \(B^{\prime},\) (b) find the transition matrix from \(B^{\prime}\) to \(B\), (c) verify that the two transition matrices are inverses of one another, and (d) find [x] \(_{B}\) when provided with \([\mathbf{x}]_{B^{*}}\) $$\begin{array}{l} B=\\{(4,2,-4),(6,-5,-6),(2,-1,8)\\}, \\\ B^{\prime}=\\{(1,0,4),(4,2,8),(2,5,-2)\\} \\\ {[\mathbf{x}]_{B^{\prime}}=\left[\begin{array}{r} 1 \\ -1 \\ 2 \end{array}\right]} \end{array}$$

Use a graphing utility with matrix capabilitics to (a) find the transition matrix from \(B\) to \(B^{\prime},\) (b) find the transition matrix from \(B^{\prime}\) to \(B\), (c) verify that the two transition matrices are inverses of one another, and (d) find [x] \(_{B}\) when provided with \([\mathbf{x}]_{B^{*}}\) $$\begin{array}{l} B=\\{(1,3,4),(2,-5,2),(-4,2,-6)\\} \\\ B^{\prime}=\\{(1,2,-2),(4,1,-4),(-2,5,8)\\} \\\ {[\mathbf{x}]_{B^{\prime}}=\left[\begin{array}{r} -1 \\ 0 \\ 2 \end{array}\right]} \end{array}$$
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