Math / Algebra

Question5. Consider the sets of vectors of the following form. Determine whether the sets are subspaces of an appropriate vector space. If a set is a subspace give a basis and its dimension. (a) $a(1,2,3)$

Studdy Solution

STEP 1

1. We are working in the vector space $\mathbb{R}^3$ .
2. The set of vectors is of the form $a(1,2,3)$ , where $a$ is a scalar.
3. We need to determine if this set is a subspace of $\mathbb{R}^3$ .
4. If it is a subspace, we need to find a basis and its dimension.

STEP 2

1. Verify if the set is a subspace of $\mathbb{R}^3$ .
2. Determine a basis for the subspace if it is a subspace.
3. Calculate the dimension of the subspace.

STEP 3

To determine if the set is a subspace, check if it satisfies the three subspace criteria: - Contains the zero vector. - Closed under vector addition. - Closed under scalar multiplication.

STEP 4

Check if the zero vector is in the set. Set $a = 0$ in $a(1,2,3)$ to get the zero vector $(0,0,0)$ . Thus, the set contains the zero vector.

STEP 5

Check closure under addition. Take two arbitrary vectors $a(1,2,3)$ and $b(1,2,3)$ from the set. Their sum is $(a+b)(1,2,3)$ , which is also in the set. Thus, it is closed under addition.

STEP 6

Check closure under scalar multiplication. Take an arbitrary vector $a(1,2,3)$ and a scalar $c$ . The product is $ca(1,2,3)$ , which is in the set. Thus, it is closed under scalar multiplication.

STEP 7

Since the set satisfies all subspace criteria, it is a subspace. A basis for this subspace is $\{(1,2,3)\}$ .

STEP 8

The dimension of the subspace is the number of vectors in the basis. Since the basis is $\{(1,2,3)\}$ , the dimension is 1.
The set is a subspace of $\mathbb{R}^3$ . A basis is $\{(1,2,3)\}$ and its dimension is 1.

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Question5. Consider the sets of vectors of the following form. Determine whether the sets are subspaces of an appropriate vector space. If a set is a subspace give a basis and its dimension. (a) $a(1,2,3)$

Studdy Solution

STEP 1

1. We are working in the vector space $\mathbb{R}^3$ .
2. The set of vectors is of the form $a(1,2,3)$ , where $a$ is a scalar.
3. We need to determine if this set is a subspace of $\mathbb{R}^3$ .
4. If it is a subspace, we need to find a basis and its dimension.

STEP 2

1. Verify if the set is a subspace of $\mathbb{R}^3$ .
2. Determine a basis for the subspace if it is a subspace.
3. Calculate the dimension of the subspace.

STEP 3

To determine if the set is a subspace, check if it satisfies the three subspace criteria: - Contains the zero vector. - Closed under vector addition. - Closed under scalar multiplication.

STEP 4

Check if the zero vector is in the set. Set $a = 0$ in $a(1,2,3)$ to get the zero vector $(0,0,0)$ . Thus, the set contains the zero vector.

STEP 5

Check closure under addition. Take two arbitrary vectors $a(1,2,3)$ and $b(1,2,3)$ from the set. Their sum is $(a+b)(1,2,3)$ , which is also in the set. Thus, it is closed under addition.

STEP 6

Check closure under scalar multiplication. Take an arbitrary vector $a(1,2,3)$ and a scalar $c$ . The product is $ca(1,2,3)$ , which is in the set. Thus, it is closed under scalar multiplication.

STEP 7

Since the set satisfies all subspace criteria, it is a subspace. A basis for this subspace is $\{(1,2,3)\}$ .

STEP 8

The dimension of the subspace is the number of vectors in the basis. Since the basis is $\{(1,2,3)\}$ , the dimension is 1.
The set is a subspace of $\mathbb{R}^3$ . A basis is $\{(1,2,3)\}$ and its dimension is 1.

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Question5. Consider the sets of vectors of the following form. Determine whether the sets are subspaces of an appropriate vector space. If a set is a subspace give a basis and its dimension. (a) a(1,2,3)a(1,2,3)a(1,2,3)

Studdy Solution

STEP 1

1. We are working in the vector space R3\mathbb{R}^3R3.2. The set of vectors is of the form a(1,2,3)a(1,2,3)a(1,2,3), where aaa is a scalar.3. We need to determine if this set is a subspace of R3\mathbb{R}^3R3.4. If it is a subspace, we need to find a basis and its dimension.

STEP 2

1. Verify if the set is a subspace of R3\mathbb{R}^3R3.2. Determine a basis for the subspace if it is a subspace.3. Calculate the dimension of the subspace.

STEP 3

To determine if the set is a subspace, check if it satisfies the three subspace criteria: - Contains the zero vector. - Closed under vector addition. - Closed under scalar multiplication.

STEP 4

Check if the zero vector is in the set. Set a=0a = 0a=0 in a(1,2,3)a(1,2,3)a(1,2,3) to get the zero vector (0,0,0)(0,0,0)(0,0,0). Thus, the set contains the zero vector.

STEP 5

Check closure under addition. Take two arbitrary vectors a(1,2,3)a(1,2,3)a(1,2,3) and b(1,2,3)b(1,2,3)b(1,2,3) from the set. Their sum is (a+b)(1,2,3)(a+b)(1,2,3)(a+b)(1,2,3), which is also in the set. Thus, it is closed under addition.

STEP 6

Check closure under scalar multiplication. Take an arbitrary vector a(1,2,3)a(1,2,3)a(1,2,3) and a scalar ccc. The product is ca(1,2,3)ca(1,2,3)ca(1,2,3), which is in the set. Thus, it is closed under scalar multiplication.

STEP 7

Since the set satisfies all subspace criteria, it is a subspace. A basis for this subspace is {(1,2,3)}\{(1,2,3)\}{(1,2,3)}.

STEP 8

The dimension of the subspace is the number of vectors in the basis. Since the basis is {(1,2,3)}\{(1,2,3)\}{(1,2,3)}, the dimension is 1. The set is a subspace of R3\mathbb{R}^3R3. A basis is {(1,2,3)}\{(1,2,3)\}{(1,2,3)} and its dimension is 1.

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Question5. Consider the sets of vectors of the following form. Determine whether the sets are subspaces of an appropriate vector space. If a set is a subspace give a basis and its dimension. (a) $a(1,2,3)$

1. We are working in the vector space $\mathbb{R}^3$ .
2. The set of vectors is of the form $a(1,2,3)$ , where $a$ is a scalar.
3. We need to determine if this set is a subspace of $\mathbb{R}^3$ .
4. If it is a subspace, we need to find a basis and its dimension.

1. Verify if the set is a subspace of $\mathbb{R}^3$ .
2. Determine a basis for the subspace if it is a subspace.
3. Calculate the dimension of the subspace.

Check if the zero vector is in the set. Set $a = 0$ in $a(1,2,3)$ to get the zero vector $(0,0,0)$ . Thus, the set contains the zero vector.

Check closure under addition. Take two arbitrary vectors $a(1,2,3)$ and $b(1,2,3)$ from the set. Their sum is $(a+b)(1,2,3)$ , which is also in the set. Thus, it is closed under addition.

Check closure under scalar multiplication. Take an arbitrary vector $a(1,2,3)$ and a scalar $c$ . The product is $ca(1,2,3)$ , which is in the set. Thus, it is closed under scalar multiplication.

Since the set satisfies all subspace criteria, it is a subspace. A basis for this subspace is $\{(1,2,3)\}$ .

The dimension of the subspace is the number of vectors in the basis. Since the basis is $\{(1,2,3)\}$ , the dimension is 1.
The set is a subspace of $\mathbb{R}^3$ . A basis is $\{(1,2,3)\}$ and its dimension is 1.