1. Determine which of the following subsets of are sub-spaces. If so, prove it, if not, specify which conditions of a sub-space are violated.
a. Vectors of the form:
Answer:
This is a subspace.
- Closure under addition: If and , then , which is still in the set.
- Closure under scalar multiplication: If and , then , which is still in the set.
- Contains the zero vector: is in the set.
Thus, this is a subspace.
b. Vectors of the form:
Answer:
This is not a subspace.
- Does not contain the zero vector: The zero vector is not in the set because the third component is always .
- Closure under addition: If and , then , which is not in the set.
- Closure under scalar multiplication: If and , then , which is not in the set unless .
Thus, this is not a subspace.
c. Vectors of the form: where
Answer:
This is a subspace.
- Closure under addition: If and satisfy and , then satisfies .
- Closure under scalar multiplication: If satisfies and , then satisfies .
- Contains the zero vector: satisfies .
Thus, this is a subspace.
d. Vectors of the form: where
Answer:
This is not a subspace.
- Does not contain the zero vector: The zero vector satisfies , so it is in the set. However, the set fails other conditions.
- Closure under addition: Let and . Both satisfy , but does not satisfy .
- Closure under scalar multiplication: Let and . Then does not satisfy .
Thus, this is not a subspace.
e. Vectors of the form: where
Answer:
This is not a subspace.
- Closure under scalar multiplication: Let and . Then , which is not in the set because .
Thus, this is not a subspace.
f. Vectors of the form: where and
Answer:
This is a subspace.
- Closure under addition: If and satisfy , , , and , then satisfies and .
- Closure under scalar multiplication: If satisfies and , and , then satisfies and .
- Contains the zero vector: satisfies and .
Thus, this is a subspace.
2. Determine if the shaded regions of define a subspace.
Answer:
The images are not provided, so I cannot analyze the shaded regions. Please provide the images or describe the shaded regions.
3. Find the null space of .
a.
Answer:
The null space of is the set of all vectors such that .
We solve the system:
From the first equation, . Substituting into the second equation:
Thus, the null space is:
b.
Answer:
The null space of is the set of all vectors such that .
We solve the system:
From the second equation, . Substituting into the first equation:
Substituting into the third equation:
Thus, the null space is:
4. Check whether or not is in and is in range where with:
Answer:
-
Check if :
Compute :
Since , .
- Check if :
We need to find such that .
Solve the system:
(1 - 2x_2) + 3x_2 - x_3 = 5 \implies 1 + x_2 - x_3 = 5 \implies x_2 - x_3 = 4.
Let $x_2 = t$, then $x_3 = t - 4$ and $x_1 = 1 - 2t$. Thus, a solution exists:\mathbf{x} = \begin{bmatrix}1 - 2t \ t \ t - 4\end{bmatrix}.
Therefore, $\mathbf{v} \in \operatorname{range}(T)$. --- ### 5. Classify the subspaces of $\mathbb{R}^{2}$. **Answer:** The subspaces of $\mathbb{R}^{2}$ are: 1. The zero subspace: $\{\mathbf{0}\}$. 2. Lines through the origin: Any set of the form $\{k\mathbf{v} \mid k \in \mathbb{R}\}$, where $\mathbf{v} \neq \mathbf{0}$ is a fixed vector. 3. The entire space $\mathbb{R}^{2}$. These are the only subspaces of $\mathbb{R}^{2}$.