Questions: Determine which of the following subsets of 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: [a b 0] b. Vectors of the form: [a b 1] c. Vectors of the form: [a b c] where a+b+c=0 d. Vectors of the form: [a b c] where ab=c^2 e. Vectors of the form: [a b c] where c >= 0 f. Vectors of the form: [a b c] where a-2b=0 and a+2c=0

Determine which of the following subsets of 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: [a b 0]
b. Vectors of the form: [a b 1]
c. Vectors of the form: [a b c] where a+b+c=0
d. Vectors of the form: [a b c] where ab=c^2
e. Vectors of the form: [a b c] where c >= 0
f. Vectors of the form: [a b c] where a-2b=0 and a+2c=0

Transcript text: 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: $\left[\begin{array}{l}a \\ b \\ 0\end{array}\right]$ b. Vectors of the form: $\left[\begin{array}{l}a \\ b \\ 1\end{array}\right]$ c. Vectors of the form: $\left[\begin{array}{l}a \\ b \\ c\end{array}\right]$ where $a+b+c=0$ d. Vectors of the form: $\left[\begin{array}{l}a \\ b \\ c\end{array}\right]$ where $|a b|=c^{2}$ e. Vectors of the form: $\left[\begin{array}{l}a \\ b \\ c\end{array}\right]$ where $c \geq 0$ f. Vectors of the form: $\left[\begin{array}{l}a \\ b \\ c\end{array}\right]$ where $a-2 b=0$ and $a+2 c=0$

Solution

Solution Steps

To determine if a subset of $\mathbf{R}^{3}$ is a subspace, we need to check three conditions:

The zero vector is in the subset.
The subset is closed under vector addition.
The subset is closed under scalar multiplication.

Let's analyze each subset:

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

The zero vector $\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ is in the subset.
Closed under addition: $\left[\begin{array}{l}a_1 \\ b_1 \\ 0\end{array}\right] + \left[\begin{array}{l}a_2 \\ b_2 \\ 0\end{array}\right] = \left[\begin{array}{l}a_1 + a_2 \\ b_1 + b_2 \\ 0\end{array}\right]$.
Closed under scalar multiplication: $k \left[\begin{array}{l}a \\ b \\ 0\end{array}\right] = \left[\begin{array}{l}ka \\ kb \\ 0\end{array}\right]$.

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

The zero vector $\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ is not in the subset.
Not closed under addition: $\left[\begin{array}{l}a_1 \\ b_1 \\ 1\end{array}\right] + \left[\begin{array}{l}a_2 \\ b_2 \\ 1\end{array}\right] = \left[\begin{array}{l}a_1 + a_2 \\ b_1 + b_2 \\ 2\end{array}\right]$.
Not closed under scalar multiplication: $k \left[\begin{array}{l}a \\ b \\ 1\end{array}\right] = \left[\begin{array}{l}ka \\ kb \\ k\end{array}\right]$.

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

The zero vector $\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ is in the subset.
Closed under addition: If $a_1 + b_1 + c_1 = 0$ and $a_2 + b_2 + c_2 = 0$, then $(a_1 + a_2) + (b_1 + b_2) + (c_1 + c_2) = 0$.
Closed under scalar multiplication: If $a + b + c = 0$, then $k(a + b + c) = 0$.

Step 1: Check Subset (a)

Subset: Vectors of the form $\left[\begin{array}{c}a \\ b \\ 0\end{array}\right]$
Zero vector: $\left[\begin{array}{c}0 \\ 0 \\ 0\end{array}\right]$ is in the subset.
Closed under addition: $\left[\begin{array}{c}a_1 \\ b_1 \\ 0\end{array}\right] + \left[\begin{array}{c}a_2 \\ b_2 \\ 0\end{array}\right] = \left[\begin{array}{c}a_1 + a_2 \\ b_1 + b_2 \\ 0\end{array}\right]$ is in the subset.
Closed under scalar multiplication: $k \left[\begin{array}{c}a \\ b \\ 0\end{array}\right] = \left[\begin{array}{c}ka \\ kb \\ 0\end{array}\right]$ is in the subset.

Since all conditions are satisfied, subset (a) is a subspace.

Step 2: Check Subset (b)

Subset: Vectors of the form $\left[\begin{array}{c}a \\ b \\ 1\end{array}\right]$
Zero vector: $\left[\begin{array}{c}0 \\ 0 \\ 0\end{array}\right]$ is not in the subset.
Not closed under addition: $\left[\begin{array}{c}a_1 \\ b_1 \\ 1\end{array}\right] + \left[\begin{array}{c}a_2 \\ b_2 \\ 1\end{array}\right] = \left[\begin{array}{c}a_1 + a_2 \\ b_1 + b_2 \\ 2\end{array}\right]$ is not in the subset.
Not closed under scalar multiplication: $k \left[\begin{array}{c}a \\ b \\ 1\end{array}\right] = \left[\begin{array}{c}ka \\ kb \\ k\end{array}\right]$ is not in the subset.

Since the zero vector is not in the subset and it is not closed under addition and scalar multiplication, subset (b) is not a subspace.

Step 3: Check Subset (c)

Subset: Vectors of the form $\left[\begin{array}{c}a \\ b \\ c\end{array}\right]$ where $a + b + c = 0$
Zero vector: $\left[\begin{array}{c}0 \\ 0 \\ 0\end{array}\right]$ is in the subset since $0 + 0 + 0 = 0$.
Closed under addition: If $a_1 + b_1 + c_1 = 0$ and $a_2 + b_2 + c_2 = 0$, then $(a_1 + a_2) + (b_1 + b_2) + (c_1 + c_2) = 0$.
Closed under scalar multiplication: If $a + b + c = 0$, then $k(a + b + c) = 0$.

Since all conditions are satisfied, subset (c) is a subspace.