4.1 Subspaces

1. Determine which of the following subsets of ${\mathbf{R}}^3$ are sub-spaces. If so, prove it, if not, specify which conditions of a sub-space are violated.

a. Vectors of the form: $\left[\begin{array}{l}a\\ b\\ 0\end{array}\right]$

Answer:
This is a subspace.

• Closure under addition: If $\mathbf{u}=\left[\begin{array}{c}{a}_1\\ b_1\\ 0\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{c}{a}_2\\ b_2\\ 0\end{array}\right]$, then $\mathbf{u}+\mathbf{v}=\left[\begin{array}{c}{a}_1+a_2\\ b_1+b_2\\ 0\end{array}\right]$, which is still in the set.
• Closure under scalar multiplication: If $\mathbf{u}=\left[\begin{array}{c}a\\ b\\ 0\end{array}\right]$ and $c\in \mathbb{R}$, then $c\mathbf{u}=\left[\begin{array}{c}ca\\ cb\\ 0\end{array}\right]$, which is still in the set.
• Contains the zero vector: $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ is in the set.

Thus, this is a subspace.

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b. Vectors of the form: $\left[\begin{array}{l}a\\ b\\ 1\end{array}\right]$

Answer:
This is not a subspace.

• Does not contain the zero vector: The zero vector $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ is not in the set because the third component is always $1$.
• Closure under addition: If $\mathbf{u}=\left[\begin{array}{c}{a}_1\\ b_1\\ 1\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{c}{a}_2\\ b_2\\ 1\end{array}\right]$, then $\mathbf{u}+\mathbf{v}=\left[\begin{array}{c}{a}_1+a_2\\ b_1+b_2\\ 2\end{array}\right]$, which is not in the set.
• Closure under scalar multiplication: If $\mathbf{u}=\left[\begin{array}{c}a\\ b\\ 1\end{array}\right]$ and $c\in \mathbb{R}$, then $c\mathbf{u}=\left[\begin{array}{c}ca\\ cb\\ c\end{array}\right]$, which is not in the set unless $c=1$.

Thus, this is not a subspace.

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c. Vectors of the form: $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]$ where $a+b+c=0$

Answer:
This is a subspace.

• Closure under addition: If $\mathbf{u}=\left[\begin{array}{c}{a}_1\\ b_1\\ c_1\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{c}{a}_2\\ b_2\\ c_2\end{array}\right]$ satisfy $a_1+b_1+c_1=0$ and $a_2+b_2+c_2=0$, then $\mathbf{u}+\mathbf{v}=\left[\begin{array}{c}{a}_1+a_2\\ b_1+b_2\\ c_1+c_2\end{array}\right]$ satisfies $(a_1+a_2)+(b_1+b_2)+(c_1+c_2)=0$.
• Closure under scalar multiplication: If $\mathbf{u}=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ satisfies $a+b+c=0$ and $k\in \mathbb{R}$, then $k\mathbf{u}=\left[\begin{array}{c}ka\\ kb\\ kc\end{array}\right]$ satisfies $ka+kb+kc=k(a+b+c)=0$.
• Contains the zero vector: $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ satisfies $0+0+0=0$.

Thus, this is a subspace.

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d. Vectors of the form: $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]$ where $|ab|=c^2$

Answer:
This is not a subspace.

• Does not contain the zero vector: The zero vector $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ satisfies $|0\cdot 0|=0^2$, so it is in the set. However, the set fails other conditions.
• Closure under addition: Let $\mathbf{u}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right]$. Both satisfy $|ab|=c^2$, but $\mathbf{u}+\mathbf{v}=\left[\begin{array}{c}0\\ 0\\ 2\end{array}\right]$ does not satisfy $|0\cdot 0|=2^2$.
• Closure under scalar multiplication: Let $\mathbf{u}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and $k=2$. Then $k\mathbf{u}=\left[\begin{array}{c}2\\ 2\\ 2\end{array}\right]$ does not satisfy $|2\cdot 2|=2^2$.

Thus, this is not a subspace.

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e. Vectors of the form: $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]$ where $c\ge 0$

Answer:
This is not a subspace.

• Closure under scalar multiplication: Let $\mathbf{u}=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ and $k=-1$. Then $k\mathbf{u}=\left[\begin{array}{c}0\\ 0\\ -1\end{array}\right]$, which is not in the set because $c<0$.

Thus, this is not a subspace.

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f. Vectors of the form: $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]$ where $a-2b=0$ and $a+2c=0$

Answer:
This is a subspace.

• Closure under addition: If $\mathbf{u}=\left[\begin{array}{c}{a}_1\\ b_1\\ c_1\end{array}\right]$ and $\mathbf{v}=\left[\begin{array}{c}{a}_2\\ b_2\\ c_2\end{array}\right]$ satisfy $a_1-2b_1=0$, $a_1+2c_1=0$, $a_2-2b_2=0$, and $a_2+2c_2=0$, then $\mathbf{u}+\mathbf{v}=\left[\begin{array}{c}{a}_1+a_2\\ b_1+b_2\\ c_1+c_2\end{array}\right]$ satisfies $(a_1+a_2)-2(b_1+b_2)=0$ and $(a_1+a_2)+2(c_1+c_2)=0$.
• Closure under scalar multiplication: If $\mathbf{u}=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ satisfies $a-2b=0$ and $a+2c=0$, and $k\in \mathbb{R}$, then $k\mathbf{u}=\left[\begin{array}{c}ka\\ kb\\ kc\end{array}\right]$ satisfies $ka-2kb=0$ and $ka+2kc=0$.
• Contains the zero vector: $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$ satisfies $0-2\cdot 0=0$ and $0+2\cdot 0=0$.

Thus, this is a subspace.

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2. Determine if the shaded regions of ${\mathbf{R}}^2$ 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.

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3. Find the null space of $A$.

a. $A=\left[\begin{array}{ccc}1& -2& 0\\ 1& 0& 2\end{array}\right]$

Answer:
The null space of $A$ is the set of all vectors $\mathbf{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$ such that $A\mathbf{x}=0$.
We solve the system:

$$\{\begin{array}{l}{x}_1-2x_2=0\\ x_1+2x_3=0\end{array}$$

From the first equation, $x_1=2x_2$. Substituting into the second equation:

$$2x_2+2x_3=0⟹x_2+x_3=0⟹x_3=-x_2.$$

Thus, the null space is:

$$\mathrm{null}(A)=\left\{\left[\begin{array}{c}2x_2\\ x_2\\ -x_2\end{array}\right]\mid x_2\in \mathbb{R}\right\}.$$

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b. $B=\left[\begin{array}{cccc}1& 3& 4& 0\\ 0& 2& 4& 4\\ 1& 1& 0& -4\end{array}\right]$

Answer:
The null space of $B$ is the set of all vectors $\mathbf{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]$ such that $B\mathbf{x}=0$.
We solve the system:

$$\{\begin{array}{l}{x}_1+3x_2+4x_3=0\\ 2x_2+4x_3+4x_4=0\\ x_1+x_2-4x_4=0\end{array}$$

From the second equation, $x_2=-2x_3-2x_4$. Substituting into the first equation:

$$x_1+3(-2x_3-2x_4)+4x_3=0⟹x_1-6x_3-6x_4+4x_3=0⟹x_1=2x_3+6x_4.$$

Substituting into the third equation:

$$(2x_3+6x_4)+(-2x_3-2x_4)-4x_4=0⟹0=0.$$

Thus, the null space is:

$$\mathrm{null}(B)=\left\{\left[\begin{array}{c}2x_3+6x_4\\ -2x_3-2x_4\\ x_3\\ x_4\end{array}\right]\mid x_3,x_4\in \mathbb{R}\right\}.$$

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4. Check whether or not $\mathbf{u}$ is in $\ker (T)$ and $\mathbf{v}$ is in range $(T)$ where $T(\mathbf{x})=A\mathbf{x}$ with:

$$A=\left[\begin{array}{ccc}1& 3& -1\\ 2& 4& 0\end{array}\right],\mathbf{u}=\left[\begin{array}{c}-2\\ 1\\ 1\end{array}\right],\text{ and }\mathbf{v}=\left[\begin{array}{l}5\\ 2\end{array}\right].$$

Answer:

• Check if $\mathbf{u}\in \ker (T)$:
  Compute $A\mathbf{u}$:

  $$A\mathbf{u}=\left[\begin{array}{ccc}1& 3& -1\\ 2& 4& 0\end{array}\right]\left[\begin{array}{c}-2\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}(1)(-2)+(3)(1)+(-1)(1)\\ (2)(-2)+(4)(1)+(0)(1)\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right].$$

  Since $A\mathbf{u}=0$, $\mathbf{u}\in \ker (T)$.

• Check if $\mathbf{v}\in \mathrm{range}(T)$:
  We need to find $\mathbf{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$ such that $A\mathbf{x}=\mathbf{v}$.
  Solve the system:

  $$\{\begin{array}{l}{x}_1+3x_2-x_3=5\\ 2x_1+4x_2=2\end{array}$$

  From the second equation, $x_1=1-2x_2$. Substituting into the first equation:

  $$(1-2x_2)+3x_2-x_3=5⟹1+x_2-x_3=5⟹x_2-x_3=4.$$

  Let $x_2=t$, then $x_3=t-4$ and $x_1=1-2t$. Thus, a solution exists:

  $$\mathbf{x}=\left[\begin{array}{c}1-2t\\ t\\ t-4\end{array}\right].$$

  Therefore, $\mathbf{v}\in \mathrm{range}(T)$.

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5. Classify the subspaces of ${\mathbb{R}}^2$.

Answer:
The subspaces of ${\mathbb{R}}^2$ are:

1. The zero subspace: $\{0\}$.
2. Lines through the origin: Any set of the form $\{k\mathbf{v}\mid k\in \mathbb{R}\}$, where $\mathbf{v}\ne 0$ is a fixed vector.
3. The entire space ${\mathbb{R}}^2$.

These are the only subspaces of ${\mathbb{R}}^2$.