QuestionSubspaces
1. Show that the sets consisting of vectors of the following form are subspaces of $\mathbf{R}^{2}$ by showing that they are closed under addition and under scalar multiplication. (a) $(a, 3 a)$

Studdy Solution

STEP 1

1. We are given a set of vectors of the form $(a, 3a)$ where $a$ is a real number.
2. We need to show that this set is a subspace of $\mathbf{R}^2$ .
3. To prove a set is a subspace, we need to verify that it is closed under vector addition and scalar multiplication, and that it contains the zero vector.

STEP 2

1. Verify the set contains the zero vector.
2. Verify closure under vector addition.
3. Verify closure under scalar multiplication.

STEP 3

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

STEP 4

Verify closure under vector addition:
Consider two arbitrary vectors $(a_1, 3a_1)$ and $(a_2, 3a_2)$ from the set. Their sum is:
$(a_1, 3a_1) + (a_2, 3a_2) = (a_1 + a_2, 3a_1 + 3a_2) = (a_1 + a_2, 3(a_1 + a_2))$
Since $(a_1 + a_2, 3(a_1 + a_2))$ is of the form $(a, 3a)$ , the set is closed under addition.

STEP 5

Verify closure under scalar multiplication:
Consider an arbitrary vector $(a, 3a)$ from the set and a scalar $c$ . The scalar multiplication is:
$c \cdot (a, 3a) = (ca, 3ca)$
Since $(ca, 3ca)$ is of the form $(a', 3a')$ , the set is closed under scalar multiplication.
The set consisting of vectors of the form $(a, 3a)$ is a subspace of $\mathbf{R}^2$ .

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