Are the following sets a subspace of $\mathbb{R}^3$?
$$ \begin{array}{l}{\text { 3) }\left\{(a, b, c)^{T} \in \mathbb{R}^{3} | a+2 b+c=0\right\}} \\ {\text { 4) }\left\{(a, b, c)^{T} \in \mathbb{R}^{3} | a^{2}=b^{2}\right\}}\end{array} $$

Let one of the sets be S. Then I know that I need to show the following:

a) That the zero-element 0$\in S$
b) $\forall u,v\in S| u+v\in S$
c) $\forall u\in S$ and $\alpha \in\mathbb{R} | \alpha\cdot u\in S$

My attampt on a solution.
For 3: (a) Let a=b=c=0, the zero-vector in the set, and equation: $a+2b+c=0$ is valid.
(b) $\left( \begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)+\left( \begin{array}{c}{a^{\prime}} \\ {b^{\prime}} \\ {c}\end{array}\right)=\left( \begin{array}{c}{a+a^{\prime}} \\ {b+b^{\prime}} \\ {c+c^{\prime}}\end{array}\right)$, $a+a^{\prime}+2\left(b+b^{\prime}\right)+c+c^{\prime}=0$
How can i see that given the equation, that the sum of the two elements from the set are still in the set?
The sames goes for (c) multiplying with a scalar?

What about for set 4: (b) $\left( \begin{array}{l}{a} \\ {b} \\ {c}\end{array}\right)+\left( \begin{array}{c}{a^{\prime}} \\ {b^{\prime}} \\ {c^{\prime}}\end{array}\right)=\left( \begin{array}{c}{a+a^{\prime}} \\ {b+b^{\prime}} \\ {c+c^{\prime}}\end{array}\right)$ , $(a+a)^{2}=\left(b+b^{\prime}\right)^{2}$
Is that a subspace?

I'm not excactly sure how to tell if they're still cointained in the set. Or if approaching this question right. Could somebody show me or tell me, how you would do it?

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  • 1
    $\begingroup$ If $a^2=b^2$ and $a'^2=b'^2$, does it automatically follow that $(a+a')^2=(b+b')^2$? $\endgroup$ – Arthur May 23 at 19:40
  • $\begingroup$ For (3), remember that you're assuming the original vectors are in the set, so $a+2b+c=0$ and $a'+2b'+c'=0$. What can you conclude about $a+a'+2(b+b')+(c+c')$? About $k(a+2b+c)$? $\endgroup$ – Karl May 23 at 19:44
  • $\begingroup$ For (4), you just need one counterexample to show that the set is not closed under addition. Note that $(1, 1, 0)+(1, -1,0)=(1,0,0)$. $\endgroup$ – Karl May 23 at 19:54

For the first you are on the right track. The sum of two vectors of the subset is in the subset precisely because the sum satisfies the equation $a+2b+c=0$, which you have proved fir $(a+a',b+b',c+c')$. Likewise for the product by a scalar.

For the second one, the equation $a^2=b^2$ is satisfied by both $(a,a,c)$ and $(a,-a,c)$. But if the subset $S=\left\{(a, b, c)^{T} \in \mathbb{R}^{3} | a^{2}=b^{2}\right\}$ is a subspace, the sume is also in it, that is $(2a,0,2c)$ for all $a,c$. But for $a\ne0$, the equation can't be satisfied. For instance $(1,1,0)\in S$ and $(1,-1,0)\in S$ but $(2,0,0)\notin S$, so $S$ can't be a subspace.


3b. The vector $\begin{pmatrix}a+a'\\b+b'\\c+c'\end{pmatrix}$ satisfies $(a+a')+2(b+b')+(c+c')=0$ because $(a+a')+2(b+b')+(c+c')=(a+2b+c)+(a'+2b'+c')=0+0=0$.

3c. For $\begin{pmatrix}a\\b\\c\end{pmatrix}\in S$ and $\alpha\in\mathbb{R}$ we see that $\begin{pmatrix}\alpha a\\\alpha b\\\alpha c\end{pmatrix}\in S$ because we have $a+2b+c=0$ so we must have $\alpha a+2\alpha b+\alpha c = \alpha(a+2b+c)=0$.

For 4, that set is not a subspace of $\mathbb{R}^3$ because it violates condition b, closure under addition. A simple counter example would be that we have $\begin{pmatrix}1\\-1\\0\end{pmatrix},\begin{pmatrix}1\\1\\0\end{pmatrix}\in S$, but their sum, $\begin{pmatrix}2\\0\\0\end{pmatrix}$, is not in $S$.


In regards to (a): You have that $$a+2b+c=0$$ and $$a'+2b'+c'=0$$ thus $a + 2b+c=a'+2b'+c'$, rearranging and making use of the axioms of a vector space yields: $(a-a')+2(b-b')+(c-c')=0$, as required.

For scalar multiplication the required condition also holds. Let $k$ be an element of $\mathbb{R}$: $$k \cdot (a+2b+c)=k\cdot0 \Leftrightarrow ak + 2(bk) + (ck) = 0.$$

As for the other set, as others have mentioned, it's sufficient to note that $(1,1,0)^T$ and $(1, -1,0)^T$ are both elements of it, yet their sum $(2,0,0)^T$ is not since $2^2 \neq 0^2$.


To show that $U$ is a subspace of $V$, we need to show that

$$\alpha + k \beta \in U , \quad \forall \alpha, \beta \in U \text{ and scalar } k $$

So let's pick two arbitrary elements $x = (x_1,x_2,x_3)$ and $y= (y_1, y_2, y_3)$ in $U$ (as described in part 4 above) such that

$$x_1^2 = x_2^2 $$ $$\implies x_1 = \pm x_2 \quad (1)$$ $$y_1^2 = y_2^2$$ $$\implies y_1 = \pm y_2 \quad (2)$$ Now for $k \in \mathbb{R}$, we consider the vector,

$$x+ky = (x_1+ky_1, x_2+ky_2, x_3+ky_3)$$

Now using $(1)$ and $(2)$ we have $4$ cases on our hands.

$$x_1 = x_2, y_1=y_2 \quad (3)$$ $$x_1 = -x_2, y_1 = y_2 \quad (4) $$ $$x_1 = x_2, y_1= -y_2 \quad (5) $$ $$x_1 = -x_2, y_1 = -y_2 \quad (6)$$

It is easy to see that in case (4) and case (5) $$x+ky$$ is not in $U$. Therefore, $U$ is not a subspace of $R^3$.


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