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Chapter 4: Problem 32

Determine whether the set \(W\) is a subspace of \(R^{3}\) with the standard operations. Justify your answer. \(W=\left\\{\left(x_{1}, x_{2}, 4\right): x_{1} \text { and } x_{2} \text { are real numbers }\right\\}\)

Short Answer

Expert verified
No, the set \(W\) defined above is not a subspace of \(R^{3}\) because it does not contain the zero vector of \(R^{3}\), it is not closed under vector addition, and it is not closed under scalar multiplication.

Step by step solution

01

Check for zero vector

The zero vector in \(R^{3}\) is (0,0,0). We can check if this is in set \(W\) by plugging 0 into the variables \(x_1\) and \(x_2\). Doing so, we find that the result is (0,0,4), not the zero vector. Hence, \(W\) does not contain the zero vector from \(R^{3}\).
02

Check for closure under addition

Even though we already concluded that \(W\) is not a subspace basis of Step 1, for understanding sake, we proceed to check the other two properties as well. For \(W\) to be closed under addition, any two vectors from \(W\) when added, should also fall inside \(W\). Let's take 2 arbitrary vectors from \(W\), say \((x_1, x_2, 4)\) and \((y_1, y_2, 4)\). Their sum is \((x_1 + y_1, x_2 + y_2, 4 + 4) = (x_1 + y_1, x_2 + y_2, 8)\), which is not in \(W\) since third component is not 4.
03

Check for closure under scalar multiplication

Similarly for closure under scalar multiplication, any vector from \(W\) when multiplied by a scalar, should also fall inside \(W\). Consider an arbitrary vector from \(W\) say \((x_1, x_2, 4)\) and a scalar \(k\). The result of the multiplication is \((kx_1, kx_2, 4k)\). This is not a member of \(W\) except for the special case where \(k\) = 0, because for any other scalar, the third component will not be equal to 4, and so the resulting vector cannot be in \(W\).

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