<s>[INST] 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.
 [/INST] 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.
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