Determining whether a subset is a subspace The problems are as follows:

Determine whether the following subsets of $\mathbb{R}^3$ are subspaces of $\mathbb{R}^3$.

*

*$A = \{(u^2, v^2, w^2) \,|\, u, v, w \in \mathbb{R} \}$,

*$B = \left\{(a, b, c) \,|\,
\begin{pmatrix}
a & b & c\\
1 & 2 & 0\\
0 & 1 & 2
\end{pmatrix} \text{is not invertible}\right\}$,

*$C = \left\{(x, y, z) \,|\, \begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\right\}$.


Here's what I've tried so far:


*

*$A$ is a subspace of $\mathbb{R}^3$ as it contains the $0$ vector (?).


*The matrix is not invertible, meaning that the determinant is equal to $0$. With this in mind, computing the determinant of the matrix yields $4a - 2b + c = 0$. The original subset can thus be represented as $B = \left\{\left(\frac{2s - t}{4}, s, t\right) \,|\, s, t \in \mathbb{R}\right\}$; i.e. $B = \text{span}\left\{(\frac{1}{2}, 1, 0), (-\frac{1}{4}, 0, 1)\right\}$, a plane in $\mathbb{R}^3$.


*Solving for the linear system,
$$
\begin{aligned}
\begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} &=
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
5\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
9
\end{pmatrix}
&= 
\begin{pmatrix}
2x\\
2y\\
2z
\end{pmatrix}\\
x
\begin{pmatrix}
-1\\
4\\
7
\end{pmatrix}
+ y
\begin{pmatrix}
2\\
3\\
8
\end{pmatrix}
+ z
\begin{pmatrix}
3\\
6\\
7
\end{pmatrix}
&= 
\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}
\end{aligned}$$
Converting to row echelon form gives the trivial solution $x = 0, y = 0, \text{ and } z = 0$; $C$ only contains the $0$ vector.

Is my reasoning correct? Also, quite a bit of the computations have been omitted in this post for brevity. Regardless of this, are my answers well justified?
Thank you.
 A: You proof seems fine to me but for case $A$ we should observe that in general
$$(u_1^2, v_1^2, w_1^2) + (u_2^2, v_2^2, w_2^2)\neq ((u_1+u_2)^2, (v_1+v_2)^2, (w_1+w_2)^2)$$
and you can easily find by yourself an example where equality doesn't hold which is a necessary step to complete the proof.
A: It's important to remember that

Let $(V,\mathbb{F},+,\cdot)$ be a vector space and $(W,\mathbb{F},+,\cdot)$ a subset of $V$. Then $W$ is a subspace of $V$ if and only if the following three conditions hold for the operations defined in $V$.
i) $\vec{0}\in W$.
ii) $\forall x,y \in W: x+y\in W$.
iii) $\forall c \in \mathbb{F},\forall x\in W: c\cdot x\in W$.

Now, note that in $\color{blue}{a)}$ $A=\{(u^{2},v^{2},w^{2}): u,v,w\in \mathbb{R}\}$ we can see that $(0,0,0) \in A$. Now, let $v_{1}=(u_{1}^{2},v_{1}^{2},w_{1}^{2}), v_{2}=(u_{2}^{2},v_{2}^{2},w_{2}^{2})\in A$ so, we can see that in general $$v_{1}+v_{2}=(u_{1}^{2},v_{1}^{2},w_{1}^{2})+(u_{2}^{2},v_{2}^{2},w_{2}^{2})\not=((u_{1}+u_{2})^{2},(v_{1}+v_{2})^{2},(w_{1}+w_{2})^{2})$$similary if $\alpha \in \mathbb{R}$ and $v=(u^{2},v^{2},w^{2})\in A$ so, we can see that in general $$\alpha \cdot v=\alpha\cdot (u^{2},v^{2},w^{2})\not=((\alpha u)^{2},(\alpha v)^{2},(\alpha w)^{2})$$Then, what do you think about $A$? do you think $A$ is a vector subspace of $\mathbb{R}^{3}$?
Answer: $A$ is not a vector subspace of $\mathbb{R}^{3}$. Thinking about it.
Now, for $\color{blue}{b)}$ note that using your analysis we can see that $B=\{(a,b,c)\in \mathbb{R}^{3}: 4a-2b+c=0\}$. It's a vector subspace of $\mathbb{R}^{3}$ because:
i) $(0,0,0) \in \mathbb{R}^{3}$ since $4(0)-2(0)+0=0$.
ii) Let $x=(a,b,c), y=(a',b',c')\in B$, so by hypothesis $4a-2b+c=0$ and $4a'-2b'+c'=0$. Therefore $$v_{1}+v_{2}=(a,b,c)+(a',b',c')=(a+a',b+b',c+c') \in B$$ because $4(a+a')-2(b+b')+(c+c')=0$.
iii) Let $c \in \mathbb{R}$ and $(a,b,c)\in B$, so by hypothesis we can see that $4a-2b+c=0$. Therefore $$c\cdot(a,b,c)=(ca,cb,cd)\in B$$ because $4(ca)-2(cb)+(cd)=0$.
So, $B$ is a vector subspace of $\mathbb{R}^{3}$.
Finally, for $\color{blue}{c)}$ using your analysis we can  that $C=\{(x,y,z)\in \mathbb{R}^{3}: x=y=z=0\}=\{(0,0,0)\}$ and $C$ is a trivial subspace of $\mathbb{R}^{3}$.
A: Most of it is great.
In 1 the vectors in $A$ can not be multiplied by negative scalar, so it is not a subspace.
In 2 and 3 your reasoning is perfect, but maybe you can state it in a more concise manner, depending on what your learned already. Once you get that the subset is defined by linear equations you are done - it immediately implies that it is in fact a subspace. You don't have to actually solve the equations.
