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!

  • 1
    $\begingroup$ $A$ 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. $\{(a,b,c):f((a,b,c))=(0,0)\}$ $\endgroup$ Commented Nov 9, 2022 at 5:54

2 Answers 2


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.

  • $\begingroup$ 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. :) $\endgroup$ Commented Nov 9, 2022 at 10:18
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    $\begingroup$ @StéphaneJaouen nice catch. I see your point. I may adjust it. $\endgroup$
    – i can try
    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($\psi$)).


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