Question2. Show that the sets consisting of vectors of the following form are subspaces of $\mathbf{R}^{3}$ or $\mathbf{R}^{4}$ . (c) $(a, 2 a,-a)$

Studdy Solution

STEP 1

1. We are given a set of vectors in the form $(a, 2a, -a)$ .
2. We need to determine if this set is a subspace of $\mathbf{R}^3$ .
3. A subset of a vector space is a subspace if it satisfies three conditions: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication.

STEP 2

1. Verify that the zero vector is in the set.
2. Check if the set is closed under vector addition.
3. Check if the set is closed under scalar multiplication.

STEP 3

Verify that the zero vector is in the set:
The zero vector in $\mathbf{R}^3$ is $(0, 0, 0)$ . For the vector $(a, 2a, -a)$ to be the zero vector, set $a = 0$ :
$(0, 2 \times 0, -0) = (0, 0, 0)$
Since the zero vector is in the set, this condition is satisfied.

STEP 4

Check if the set is closed under vector addition:
Take two arbitrary vectors from the set, $(a_1, 2a_1, -a_1)$ and $(a_2, 2a_2, -a_2)$ . Add them together:
$(a_1, 2a_1, -a_1) + (a_2, 2a_2, -a_2) = (a_1 + a_2, 2a_1 + 2a_2, -a_1 - a_2)$
This simplifies to:
$(a_1 + a_2, 2(a_1 + a_2), -(a_1 + a_2))$
Let $a = a_1 + a_2$ . The resulting vector $(a, 2a, -a)$ is in the set, so the set is closed under vector addition.

STEP 5

Check if the set is closed under scalar multiplication:
Take an arbitrary vector from the set, $(a, 2a, -a)$ , and a scalar $c$ . Multiply the vector by the scalar:
$c(a, 2a, -a) = (ca, 2ca, -ca)$
Let $a' = ca$ . The resulting vector $(a', 2a', -a')$ is in the set, so the set is closed under scalar multiplication.
Since all three conditions are satisfied, the set of vectors $(a, 2a, -a)$ is a subspace of $\mathbf{R}^3$ .

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