# Show that the following set is a vector space, and find its dimension.

I was given this problem on an exam review guide for my introductory linear algebra class.

I'm extremely bad with LaTeX, but I will try my best. This was the given problem:

$$\left\{\begin{bmatrix} a\\ b\\ c \end{bmatrix} : a − 2b = 4c, b = 3c \right\}$$

The answer was revealed in class, and my teacher outlined the process as follows:

• set each given equation equal to zero $$(a - 2b - 4c = 0; b - 3c = 0)$$
• arrange these coefficients into a matrix : $$\begin{bmatrix} 1 & -2 & -1\\ 0 & 1 & -3\\ \end{bmatrix}$$
• determine that $$x_3$$ (or $$c$$) is a free variable, meaning dim nul$$A = 1$$ and dim col$$A = 2$$

I understand the solution process completely up to this point

• my professor then stated that the answer was: "satisfies nul$$A$$, and nul$$A$$ is a vector space. dim nul$$A = 1$$"

What does it mean to "satisfy nul$$A$$"? And how am I able to make the conclusion that nulA is a vector space? What does the dimension of nulA have to do with the answer to the problem? If the dimension of nul$$A$$ wasn't $$1$$, would nul$$A$$ not be satisfied?

More generally, what's a sure-fire way to prove something is a vector space?

Any help is appreciated, thanks!

• $A$ represents the linear map $f:\Bbb R^3\to \Bbb R^2,(x,y,z)\mapsto (1x-2y-4z,0x+1y-3z)$; "satisfy nul$A$" means "looking for $\{( a, b, c) : a − 2b -4c=0, b - 3c=0 \}$, i.e. $\{(a,b,c):f((a,b,c))=(0,0)\}$ Commented Nov 9, 2022 at 5:54

You have a typo in your matrix. It should be $$\begin{bmatrix}1\quad-2\quad\color{blue}{-4}\\0\quad 1\quad-3\end{bmatrix}$$.

Next, call the matrix $$A$$. The space you are considering is actually the null space of the linear transformation represesented by this matrix.

Thus it's a vector space.

That's it's the set of vectors $$x=\begin{pmatrix}a\\b\\c\end {pmatrix}$$ satisfying $$Ax=0$$.

There's a couple ways to see that the dimension is $$1$$. One is that the rank of the matrix is $$2$$. But it represents a linear transformation from $$\Bbb R^3$$ to $$\Bbb R^2$$. The rank-nullity theorem then implies the conclusion.

Another way, in case you don't have the theorem yet, is to row-reduce. You get $$\begin{bmatrix}1\quad-2\quad\color{blue}{-4}\\0\quad 1\quad-3\end{bmatrix}\to\begin {bmatrix}1\quad 0\quad-10\\0\quad 1\quad-3\end {bmatrix}.$$

Then back-substitute. You can choose $$c$$ freely. Then $$b-3c=0\implies b=3c$$. And, $$a-10c=0\implies a=10c$$.

Thus the solutions are $$\begin {bmatrix}10c\\3c\\c\end {bmatrix}$$.

Finally, the dimension could have been different.

• I would have specified for OP : the solution is $\left\{c\begin {bmatrix}10\\3\\1\end {bmatrix}: c\in \Bbb R\right\}=\Bbb R\begin {bmatrix}10\\3\\1\end {bmatrix}$, which is a vector line, in other words of dimension $1$ (It's a detail.) Otherwise, I don't understand your last sentence "the dimension could have been different." I would have written : the dimension could'nt have been different. :) Commented Nov 9, 2022 at 10:18
• @StéphaneJaouen nice catch. I see your point. I may adjust it. Commented Nov 9, 2022 at 11:36

Here are a couple of ways of looking at it for vector spaces over a field $$F$$. You haven't specified what your field is but it could be the set of real numbers or the set of rational numbers, among many other possibilities.1st way: The set of $$n-$$ place column vectors over $$F$$ is a vector space $$V$$ of dimension $$n.$$ Let $$A$$ be an $$m \times n$$ matrix. Definition : row space of $$A$$=set of all $$F-$$ linear combinations of rows of $$A$$. Definition: row rank of $$A$$=dimension of row space of $$A$$. Similar definitition for column space and column rank of $$A$$. yet another definition : determinantal rank of $$A$$=largest $$r$$ such that $$A$$ contains an $$r \times r$$ sub-matrix with non-zero determinant. theorem: row rank of $$A$$=column rank of $$A$$=determinantal rank of $$A$$. From now on, we'll call any of these the rank of $$A$$, and denote it by $$r$$. The set of $$n-$$ place column vectors $$v$$ such that $$Av=\mathbf 0$$ is a vector space of dimension $$n-r.$$ 2nd way: Multiplication by $$A$$ is a linear transformation $$\psi :V \rightarrow W$$ where $$W$$ is the set of $$m-$$ place column-vectors over $$F$$. theorem: dim(ker( $$\psi$$)) + dim(im ($$\psi$$))=dim(domain($$\psi$$)).