4.2 Null spaces, column spaces, and linear transformations

Big idea

Every matrix carries two subspaces with it. One lives on the input side: the set of vectors the matrix sends to zero, called the null space. The other lives on the output side: the set of vectors the matrix can produce, called the column space. Between them they answer the two questions you keep asking about $A\mathbf{x} = \mathbf{b}$. Is there a solution, and is it unique.

They are described in opposite ways, and that difference drives the work. The null space comes with a condition to check and no list of members, so producing a spanning set takes row reduction. The column space comes with a list of members already, the columns, so producing a spanning set is free, and the hard question is whether a given vector belongs.

The same pair of subspaces exists for a linear transformation between any two vector spaces, where they are called the kernel and the range. For the transformation $\mathbf{x} \mapsto A\mathbf{x}$ they are exactly the null space and the column space. Working abstractly costs nothing extra and buys you the differentiation and evaluation examples, where no matrix is in sight.

Decoder

The null space of a matrix is the set of all solutions of the homogeneous equation, and it is a subspace of the domain.

In plain terms: collect every input vector that the matrix flattens to zero. Nothing new has to be computed; you already solved homogeneous systems, and the solution set you wrote in parametric vector form is this subspace, with the spanning vectors in front of the free variables.

“Subspace of the domain” is a size reminder. If $A$ is $m \times n$, its null space sits in $\mathbb{R}^n$, because that is where the inputs live. Its column space sits in $\mathbb{R}^m$, because that is where the outputs live. For a non-square matrix the two subspaces live in different spaces entirely and cannot be compared.

Definitions and results

Null space. For an $m \times n$ matrix $A$, the null space $\text{Nul}\,A$ is the set of all $\mathbf{x}$ in $\mathbb{R}^n$ with $A\mathbf{x} = \mathbf{0}$. It is a subspace of $\mathbb{R}^n$: the zero vector qualifies, and $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v} = \mathbf{0}$ and $A(c\mathbf{u}) = cA\mathbf{u} = \mathbf{0}$ give the two closure conditions.

Spanning set for the null space. Row reduce $A$, solve for the pivot variables in terms of the free variables, and write the general solution in parametric vector form. The vectors multiplying the free variables span $\text{Nul}\,A$, and they are automatically independent because each carries a $1$ in a slot where the others carry $0$. The number of them is the number of free variables.

Uniqueness test. $\text{Nul}\,A$ contains only the zero vector exactly when $A$ has a pivot in every column. In that case $A\mathbf{x} = \mathbf{b}$ has at most one solution for every $\mathbf{b}$.

Column space. The column space $\text{Col}\,A$ is the span of the columns of $A$, a subspace of $\mathbb{R}^m$. Equivalently it is the set of all vectors of the form $A\mathbf{x}$, so $A\mathbf{x} = \mathbf{b}$ is consistent exactly when $\mathbf{b}$ is in $\text{Col}\,A$.

Existence test. $\text{Col}\,A$ is all of $\mathbb{R}^m$ exactly when $A$ has a pivot in every row. In that case $A\mathbf{x} = \mathbf{b}$ has at least one solution for every $\mathbf{b}$.

The two are described in opposite ways. The null space is defined by a condition, so testing membership is easy and listing members is work. The column space is defined by a list, so listing members is easy and testing membership is work, since it means solving a system.

Linear transformation between vector spaces. A map $T$ from a vector space $V$ to a vector space $W$ is linear when $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ and $T(c\mathbf{u}) = cT(\mathbf{u})$ for all vectors and scalars. Linearity forces $T(\mathbf{0}) = \mathbf{0}$, which is a quick way to disqualify a candidate.

Kernel and range. The kernel of $T$ is the set of vectors in $V$ with $T(\mathbf{v}) = \mathbf{0}$; it is a subspace of $V$. The range of $T$ is the set of all outputs $T(\mathbf{v})$; it is a subspace of $W$. For $T(\mathbf{x}) = A\mathbf{x}$ the kernel is $\text{Nul}\,A$ and the range is $\text{Col}\,A$.

Examples beyond matrices. Differentiation from $\mathbb{P}_2$ to $\mathbb{P}_1$ is linear, with kernel the constant polynomials. Evaluating a polynomial at a fixed list of points is linear. Both have kernels and ranges computed with the same reasoning, and neither starts life as a matrix.

Worked examples

A spanning set for a null space

Take

$$ A = \begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 4 & -1 & 7 \end{bmatrix} $$

Subtract twice row one from row two to get $\begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix}$, then add that row to row one:

$$ \begin{bmatrix} 1 & 2 & 0 & 4 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$

Pivots sit in columns one and three, so $x_2$ and $x_4$ are free and the equations read $x_1 = -2x_2 - 4x_4$ and $x_3 = -x_4$. In parametric vector form,

$$ \mathbf{x} = x_2\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4\begin{bmatrix} -4 \\ 0 \\ -1 \\ 1 \end{bmatrix} $$

so those two vectors span $\text{Nul}\,A$, a subspace of $\mathbb{R}^4$.

Check both directly against the original matrix. For the first: row one gives $-2 + 2 = 0$ and row two gives $-4 + 4 = 0$. For the second: row one gives $-4 + 0 + 1 + 3 = 0$ and row two gives $-8 + 0 + 1 + 7 = 0$. Both are in the null space.

The column space of the same matrix

The columns of $A$ span $\text{Col}\,A$, so $(1,2)$, $(2,4)$, $(-1,-1)$ and $(3,7)$ is already a spanning set. It is wasteful: the reduction above put pivots in columns one and three, so $(1,2)$ and $(-1,-1)$ carry the whole span.

Since two pivots appear in a matrix with two rows, every row has a pivot, so $\text{Col}\,A = \mathbb{R}^2$ and $A\mathbf{x} = \mathbf{b}$ is consistent for every $\mathbf{b}$ in $\mathbb{R}^2$.

Check that claim on a target. Solve $c_1(1,2) + c_2(-1,-1) = (5, 7)$. From the first coordinate $c_1 = 5 + c_2$; substituting into the second gives $2(5 + c_2) - c_2 = 7$, so $c_2 = -3$ and $c_1 = 2$. Verify: $2(1,2) - 3(-1,-1) = (2 + 3,\; 4 + 3) = (5, 7)$.

Kernel and range with no matrix in sight

Let $T$ send a polynomial $p$ in $\mathbb{P}_2$ to the pair of numbers $(p(0), p(1))$ in $\mathbb{R}^2$. Evaluation respects sums and scalar multiples, so $T$ is linear.

For the kernel, write $p(t) = a_0 + a_1 t + a_2 t^2$. The condition $p(0) = 0$ says $a_0 = 0$. The condition $p(1) = 0$ then says $a_1 + a_2 = 0$, so $a_1 = -a_2$. Every kernel member is $a_2(t^2 - t)$, so the kernel is $\text{Span}\{t^2 - t\}$.

Check: at $t = 0$ the polynomial $t^2 - t$ gives $0$, and at $t = 1$ it gives $1 - 1 = 0$. Both required values come out zero.

For the range, note $T(1) = (1, 1)$ and $T(t) = (0, 1)$. Those two vectors are not multiples of each other, so they span $\mathbb{R}^2$ and the range is all of $\mathbb{R}^2$. The transformation is onto but not one-to-one, since its kernel is bigger than the zero subspace.

Practice

First the input-side subspace: reduce, identify free variables, and write the spanning vectors.

Then the output-side subspace: produce a spanning set and decide whether a given vector is reachable.

Last the same two ideas stated for a transformation, including ones defined on polynomials.

Videos

Watch the part where a transformation squashes space onto a line and a whole line of inputs collapses to the origin. That collapsing set is the null space, and its size is exactly the failure of uniqueness.

Quiz

Six items across null spaces, column spaces, and kernels and ranges of linear transformations.