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Column space, null space, row space and left null space: which space each lives in, how one reduction gives a basis for each, and why row rank equals column rank.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to name the four fundamental subspaces of a matrix and say which ambient space each sits in, produce a basis for each of them from a single reduction, take the column-space basis from the original matrix rather than from the reduced one and say why that distinction exists, give all four dimensions from the shape and the rank alone, and state the two questions about a linear system that the column space and the null space respectively answer.
You can reduce a matrix, read its rank from the pivots, extract a basis from a spanning set and give the dimension of a span. This lesson spends all of that at once: the same reduction that answered those questions answers four more, and the only new work is knowing where each answer is written.
For an $m \times n$ matrix $A$: the column space $C(A)$ is the span of its columns, a subspace of $\mathbb{R}^m$. The null space $N(A) = \{x : Ax = 0\}$ is a subspace of $\mathbb{R}^n$. The row space $C(A^{\mathsf{T}})$ is the span of its rows, a subspace of $\mathbb{R}^n$. The left null space $N(A^{\mathsf{T}}) = \{y : A^{\mathsf{T}}y = 0\}$ is a subspace of $\mathbb{R}^m$. A special solution is the null-space vector got by setting one free variable to $1$ and the rest to $0$.
A matrix acts between two spaces: it takes an $n$-entry column in and gives an $m$-entry column out. So its subspaces come in two pairs, one pair at each end.
| Subspace | Lives in | Basis from | Dimension |
|---|---|---|---|
| column space $C(A)$ | $\mathbb{R}^m$ | the pivot columns of $A$ | $r$ |
| left null space $N(A^{\mathsf{T}})$ | $\mathbb{R}^m$ | special solutions of $A^{\mathsf{T}}y = 0$ | $m - r$ |
| row space $C(A^{\mathsf{T}})$ | $\mathbb{R}^n$ | the non-zero rows of the reduced form | $r$ |
| null space $N(A)$ | $\mathbb{R}^n$ | special solutions of $Ax = 0$ | $n - r$ |
The single number $r$, the rank, appears twice — and that is the theorem hiding in the table. Row rank equals column rank. The number of independent rows and the number of independent columns of any matrix are the same, which is not obvious and is visible in the reduced form: each pivot sits in a row of its own and in a column of its own, so counting pivots counts both.
The two questions about $Ax = b$ are now questions about two of the four. Consistency: is $b$ in the column space? Uniqueness: is the null space $\{0\}$? They are independent questions, asked at opposite ends of the matrix, which is why a system can be consistent with many solutions, inconsistent with none, or either with a unique one.
Another way: picture
Draw two blobs: the input space $\mathbb{R}^n$ on the left and the output space $\mathbb{R}^m$ on the right. Inside the left blob, the null space and the row space; inside the right, the column space and the left null space. The matrix squashes the null space onto the single point $0$ on the right, and carries the row space one-to-one onto the column space — which is why those two have the same dimension. The left null space is the part of the output space that $A$ never reaches, seen from the side: it is exactly the directions orthogonal to everything $A$ produces.
Another way: steps
The whole method rests on one asymmetry, and it is worth stating exactly.
Row operations preserve the row space. Each new row is a combination of old rows and the step is reversible, so the span of the rows is untouched. That is why the non-zero rows of the reduced form are a basis of the row space of $A$ — and a very convenient one, since they are already in echelon form and obviously independent.
Row operations preserve the null space. The solution set of $Ax = 0$ is what reduction is designed to leave alone.
Row operations destroy the column space. Combining rows changes every column, so $C(R)$ and $C(A)$ are generally different subspaces. In
$$A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \longrightarrow R = \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}$$
the column space of $A$ is the line through $(1, 2)$ and the column space of $R$ is the line through $(1, 0)$. Different lines.
What survives is the pattern of dependence among the columns. If $c_1a_1 + \dots + c_na_n = 0$ then the same coefficients work for $R$, because both statements say $Ac = 0$ and $Rc = 0$ and reduction preserves that solution set. So the pivot positions transfer even though the vectors do not.
That is the whole of the rule: positions from $R$, vectors from $A$.
Take $A = \begin{pmatrix} 1 & 2 & 1 & 3 \\ 2 & 4 & 3 & 8 \\ 1 & 2 & 2 & 5 \end{pmatrix}$, which reduces to $R = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. Pivots in columns 1 and 3, so $r = 2$; $m = 3$, $n = 4$.
Column space ($\subseteq \mathbb{R}^3$, dimension $2$): columns 1 and 3 of $A$, namely $(1, 2, 1)$ and $(1, 3, 2)$. Not $(1,0,0)$ and $(0,1,0)$, which is what $R$ would have offered.
Row space ($\subseteq \mathbb{R}^4$, dimension $2$): the non-zero rows of $R$, namely $(1, 2, 0, 1)$ and $(0, 0, 1, 2)$.
Null space ($\subseteq \mathbb{R}^4$, dimension $4 - 2 = 2$): free columns are 2 and 4. Setting $x_2 = 1, x_4 = 0$ gives $(-2, 1, 0, 0)$; setting $x_2 = 0, x_4 = 1$ gives $(-1, 0, -2, 1)$.
Left null space ($\subseteq \mathbb{R}^3$, dimension $3 - 2 = 1$): the third row of $R$ became zero, and tracing the operations back, row 3 of $A$ minus row 2 plus row 1 is zero — so $(1, -1, 1)$ spans it.
Four subspaces, two of dimension $2$ and two of dimensions $2$ and $1$; every basis came from the one reduction, and only the column space had to be fetched from the original matrix.
Taking the pivot columns of $R$ for the column space. The single commonest error in the unit. Positions from $R$, vectors from $A$.
Taking the rows of $A$ for the row space. Not wrong, but wasteful: they span it and are usually dependent, so they are not a basis. The reduced rows are already independent.
Putting the null space in $\mathbb{R}^m$. A null-space vector is something the matrix eats, so it has $n$ entries. If the matrix is not square, getting this wrong makes the dimensions fail to add up and the mistake at least announces itself.
Thinking equal dimension means equal subspace. The row space and the column space have the same dimension for every matrix in existence, and for a non-square matrix they are not even in the same ambient space.
Expecting the left null space to be free. It needs its own reduction, of $A^{\mathsf{T}}$, or a record of which row combinations died. Its dimension is free, from $m - r$; a basis for it is not.
$A = \begin{pmatrix} 1 & 3 \\ 2 & 6 \end{pmatrix}$ reduces to $\begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix}$: one pivot, in column 1, so $r = 1$ and $m = n = 2$.
One reduction.
Column space: column 1 of $A$, so the line through $(1, 2)$. Row space: the line through $(1, 3)$. Both of dimension $1$, both in $\mathbb{R}^2$, and different lines.
Same dimension, different subspaces.
Null space: $x_1 + 3x_2 = 0$ gives $(-3, 1)$, dimension $2 - 1 = 1$. Left null space: row 2 is twice row 1, so $(2, -1)$ kills the rows; dimension $2 - 1 = 1$.
Nullities from the shape.
$A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \\ 3 & 3 \end{pmatrix}$, rank $1$, so the left null space has dimension $3 - 1 = 2$.
Dimension before basis.
A vector $y$ is in it when $y_1(1,1) + y_2(2,2) + y_3(3,3) = 0$, that is $y_1 + 2y_2 + 3y_3 = 0$: one equation in three unknowns.
It is the relations among the rows.
Special solutions $(-2, 1, 0)$ and $(-3, 0, 1)$. Each one is a recipe for a combination of the rows of $A$ that comes out zero — which is exactly what elimination discovers when a row goes to nothing.
A dead row is a left-null-space vector.
Column space and left null space live in $\mathbb{R}^5$; row space and null space in $\mathbb{R}^3$.
Place them first.
Both ranks are $3$; the nullity is $3 - 3 = 0$ and the left nullity is $5 - 3 = 2$.
Rank twice, then the two subtractions.
So $3, 0, 3, 2$. A null space of dimension $0$ means $Ax = b$ has at most one solution, and a left null space of dimension $2$ means two independent conditions on $b$ must hold for there to be any.
$A$ is $5 \times 6$ with rank $2$. For each of its four fundamental subspaces, give the dimension of the space it sits inside and the dimension of the subspace itself.
| Dimension of the space it sits in | Dimension of the subspace | |
|---|---|---|
| Column space | ||
| Null space | ||
| Row space | ||
| Left null space |
For $A = \begin{pmatrix} 1 & 2 & 6 \\ 4 & 8 & 1 \end{pmatrix}$ the pivots lie in columns $1$ and $3$. Enter the $2 \times 2$ matrix whose columns are a basis of the column space of $A$, in that order.
This task has no paper form; do it on a device.
$A$ is $3 \times 4$ and has been reduced. Match each subspace to the place a basis for it is read from.
| The pivot columns of $A$ itself | The special solutions of $Ax = 0$, one per free column | The non-zero rows of the reduced form | The special solutions of $A^{\mathsf{T}}y = 0$ | |
|---|---|---|---|---|
| The column space | ||||
| The null space | ||||
| The row space | ||||
| The left null space |
$A$ is $7 \times 8$ with rank $2$. What is the dimension of its left null space?
Answer:
$A$ is a $3 \times 4$ matrix that row-reduces to $R$. Which vectors form a basis of the column space of $A$?
$A$ is $3 \times 6$. Select every statement below that follows from the definitions of the four subspaces.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$A$ is $4 \times 3$ with rank $1$. For each of its four fundamental subspaces, give the dimension of the space it sits inside and the dimension of the subspace itself.
| Dimension of the space it sits in | Dimension of the subspace | |
|---|---|---|
| Column space | ||
| Null space | ||
| Row space | ||
| Left null space |
You can compute a basis and a dimension for each of the four subspaces of a given matrix. Say in your own words why row operations may be trusted about pivot positions and not about columns. Next: the counting theorem that ties the two dimensions at the input end together.
10. Your turn: $A$ is $5 \times 3$ with rank $3$. Give all four dimensions., step 3
Read the two questions off the two ends.