Math / Algebra

QuestionDetermine if the subset of $\mathbb{R}^{3}$ consisting of vectors of the form $\left[\begin{array}{l}a \\ b \\ c\end{array}\right]$ , where $a \geq 0, b \geq 0$ , and $c \geq 0$ is a subspace. Select true or false for each statement.
True $\square$ 1. The set contains the zero vector
True 2. This set is closed under vector addition $\square$ 3. This set is a subspace
False $\square$ 4. This set is closed under scalar multiplications

Studdy Solution

STEP 1

What is this asking? We need to figure out if a set of vectors with non-negative components forms a subspace of $\mathbb{R}^3$ .
A subspace has to satisfy three conditions: contain the zero vector, be closed under addition, and be closed under scalar multiplication. Watch out! Remember, closed under an operation means that applying that operation to any elements in the set always produces another element in the set.
If we can find even one counterexample, the set is not closed.

STEP 2

1. Zero Vector Check
2. Closure under Addition Check
3. Closure under Scalar Multiplication Check
4. Subspace Determination

STEP 3

Does the set contain the zero vector?
The zero vector in $\mathbb{R}^3$ is $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ .

STEP 4

Since $\textbf{0} \geq 0$ , all the components of the zero vector are non-negative.
So, yes, the set does contain the zero vector!

STEP 5

Let's pick two vectors in the set.
Let's say $\textbf{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $\textbf{v} = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$ .
Both vectors have non-negative components.

STEP 6

Now, let's add them: $\textbf{u} + \textbf{v} = \begin{bmatrix} 1+4 \\ 2+5 \\ 3+6 \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \\ 9 \end{bmatrix}$ .
The resulting vector also has non-negative components.
This makes it look like it's closed under addition, but one example isn't enough!
We need to think generally.
Since adding any two non-negative numbers always results in a non-negative number, this set is closed under addition.

STEP 7

Let's take a vector from our set, say $\textbf{w} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , and a scalar.
Let's choose a negative scalar, like $k = -2$ .

STEP 8

Now, let's multiply: $k\textbf{w} = -2 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} -2 \\ -4 \\ -6 \end{bmatrix}$ .
Uh oh!
The resulting vector has negative components, so it's not in our set.

STEP 9

This single counterexample shows that the set is not closed under scalar multiplication.

STEP 10

To be a subspace, the set must satisfy all three conditions.
It contains the zero vector and is closed under addition, but it's not closed under scalar multiplication.

STEP 11

Therefore, this set is not a subspace of $\mathbb{R}^3$ .

STEP 12

1. True: The set contains the zero vector.
2. True: The set is closed under vector addition.
3. False: The set is not a subspace.
4. False: The set is not closed under scalar multiplication.

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Studdy Solution

STEP 1

STEP 2

1. Zero Vector Check
2. Closure under Addition Check
3. Closure under Scalar Multiplication Check
4. Subspace Determination

STEP 3

Does the set contain the zero vector?
The zero vector in $\mathbb{R}^3$ is $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ .

STEP 4

Since $\textbf{0} \geq 0$ , all the components of the zero vector are non-negative.
So, yes, the set does contain the zero vector!

STEP 5

Let's pick two vectors in the set.
Let's say $\textbf{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $\textbf{v} = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$ .
Both vectors have non-negative components.

STEP 6

STEP 7

Let's take a vector from our set, say $\textbf{w} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , and a scalar.
Let's choose a negative scalar, like $k = -2$ .

STEP 8

Now, let's multiply: $k\textbf{w} = -2 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} -2 \\ -4 \\ -6 \end{bmatrix}$ .
Uh oh!
The resulting vector has negative components, so it's not in our set.

STEP 9

This single counterexample shows that the set is not closed under scalar multiplication.

STEP 10

To be a subspace, the set must satisfy all three conditions.
It contains the zero vector and is closed under addition, but it's not closed under scalar multiplication.

STEP 11

Therefore, this set is not a subspace of $\mathbb{R}^3$ .

STEP 12

1. True: The set contains the zero vector.
2. True: The set is closed under vector addition.
3. False: The set is not a subspace.
4. False: The set is not closed under scalar multiplication.

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Studdy Solution

STEP 1

STEP 2

1. Zero Vector Check2. Closure under Addition Check3. Closure under Scalar Multiplication Check4. Subspace Determination

STEP 3

Does the set contain the **zero vector**?The zero vector in R3\mathbb{R}^3R3 is [000]\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}⎣⎡​000​⎦⎤​.

STEP 4

Since 0≥0\textbf{0} \geq 00≥0, all the components of the zero vector are non-negative.So, yes, the set *does* contain the zero vector!

STEP 5

Let's pick two vectors in the set.Let's say u=[123]\textbf{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}u=⎣⎡​123​⎦⎤​ and v=[456]\textbf{v} = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}v=⎣⎡​456​⎦⎤​.Both vectors have non-negative components.

STEP 6

STEP 7

Let's take a vector from our set, say w=[123]\textbf{w} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}w=⎣⎡​123​⎦⎤​, and a scalar.Let's choose a **negative** scalar, like k=−2k = -2k=−2.

STEP 8

STEP 9

This single **counterexample** shows that the set is *not* closed under scalar multiplication.

STEP 10

To be a subspace, the set must satisfy *all three* conditions.It contains the zero vector and is closed under addition, but it's *not* closed under scalar multiplication.

STEP 11

Therefore, this set is *not* a subspace of R3\mathbb{R}^3R3.

STEP 12

1. True: The set contains the zero vector.2. True: The set is closed under vector addition.3. False: The set is *not* a subspace.4. False: The set is *not* closed under scalar multiplication.

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1. Zero Vector Check
2. Closure under Addition Check
3. Closure under Scalar Multiplication Check
4. Subspace Determination

Does the set contain the zero vector?
The zero vector in $\mathbb{R}^3$ is $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ .

Since $\textbf{0} \geq 0$ , all the components of the zero vector are non-negative.
So, yes, the set does contain the zero vector!

Let's pick two vectors in the set.
Let's say $\textbf{u} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $\textbf{v} = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$ .
Both vectors have non-negative components.

Let's take a vector from our set, say $\textbf{w} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ , and a scalar.
Let's choose a negative scalar, like $k = -2$ .

This single counterexample shows that the set is not closed under scalar multiplication.

To be a subspace, the set must satisfy all three conditions.
It contains the zero vector and is closed under addition, but it's not closed under scalar multiplication.

Therefore, this set is not a subspace of $\mathbb{R}^3$ .

1. True: The set contains the zero vector.
2. True: The set is closed under vector addition.
3. False: The set is not a subspace.
4. False: The set is not closed under scalar multiplication.