Null Spaces, Column Spaces and Linear Transformations

The null space of a matrix is simply the set of all inputs it sends to zero, and the column space is the set of all outputs it can produce. "Null" means zero, so the null space is everything the matrix wipes out. The column space gets its name because every output $A\vec x$ is a mix of $A$'s columns ($x_1$ times the first column, plus $x_2$ times the second, and so on), so "everything you can reach" and "every combination of the columns" are the same set. Together they tell you what a matrix destroys and what it can reach, which is exactly what decides whether $A\vec x = \vec b$ has a solution and whether it's the only one. All we're really saying with $\operatorname{Nul}A \subseteq \mathbb R^n$ and $\operatorname{Col}A \subseteq \mathbb R^m$ is that for an $m\times n$ matrix, inputs have $n$ entries (so that's where the null space lives) and outputs have $m$ entries (so that's where the column space lives). Two sets, one on each side of the matrix; bear with it.

Both are subspaces, and last lesson's three checks show it. If $A\vec u = \vec 0$ and $A\vec v = \vec 0$, then $A(\vec u + \vec v) = \vec 0 + \vec 0 = \vec 0$ and $A(c\vec u) = c\vec 0 = \vec 0$. And sums and multiples of combinations of the columns are still combinations of the columns.

Two different tests:

• Is $\vec v$ in $\operatorname{Nul}A$? Compute $A\vec v$ and see if it's $\vec 0$. That's quick.
• Is $\vec b$ in $\operatorname{Col}A$? Solve $A\vec x = \vec b$ and see if it's consistent. That takes row reduction.

A common mistake. It's tempting to think $\operatorname{Nul}A$ and $\operatorname{Col}A$ live in the same space, because they come from the same matrix. They don't: for a $3\times5$ matrix, the null space is made of vectors with 5 entries and the column space of vectors with 3 entries. Asking whether one vector is in both usually doesn't even make sense.

Linear transformations in general. None of this needs a matrix. For $T: V \to W$, the kernel is $\{\vec v : T(\vec v) = \vec 0\}$ and the range is $\{T(\vec v)\}$. Take differentiation $D(p) = p'$ on $\mathbb P_3$. Its kernel is the constant polynomials, since those are the only ones with derivative zero. Its range is $\mathbb P_2$, since every polynomial of degree at most 2 is the derivative of something in $\mathbb P_3$ (just integrate it).

There's an animation further down that shows both spaces as shadows on the ground.

The null space of a structural matrix reveals mechanisms (ways a frame can move with no force); the column space of a sensor matrix tells you which measurements are even possible.