I was studying the Rank Nullity theoreom and found this explanation that I think I'll be happy with if I really understood it, as it's easy to remember. They're supposing that the equation is on the form Ax = b, where A is our matrix.
How come that if we have more freedom in choosing our x, this gives us less freedom in choosing our b and vice versa?
Thanks!
This answer will expand on Gerry Myerson's comment.$\newcommand{\N}{\mathcal{N}}\DeclareMathOperator{\im}{im}$
Setup. $A$ is an $m \times n$ matrix, and $b$ is a $m \times 1$ column vector. We are wishing to solve $$Ax = b \tag{$*$}$$ for $n \times 1$ column vectors $x$.
$\N(A)$ denotes the null space of $A$. That is, it is the set of all those column vectors $x$ such that $Ax = 0$. In other words, the solution set to $(*)$ upon putting $b = 0$.
$\im(A)$ denotes the image of $A$. That is, it is the set of all those column vectors $b$ such that $(*)$ has at least one solution.
The key point is the following lemma:
Thus, $\N(A)$ measures exactly "how many" solutions there are to $(*)$, assuming that there is at least one solution.
Now, the rank-nullity theorem tells you that $$\dim(\N(A)) + \dim(\im(A)) = n \tag{$\dagger$}.$$ For the purpose of this question, assume that $A$ is varying over all $m \times n$ matrices. Thus, the right hand side of $(\dagger)$ is constant.
This means that if $\N(A)$ is "larger", then $\im(A)$ is "smaller". But note that $\N(A)$ measures how many $x$ there are which solve $(*)$ (assuming there is at least one solution), whereas $\im(A)$ measures how many $b$ there are for which $(*)$ can be solved.