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21. Null Spaces, Column Spaces, and Linear Transformations

Dated: 23-04-2025

In applications of linear algebra, subspaces of \(\mathbb R^n\) usually arise in one of the two ways

  • As a set1 of all solutions to a system of homogeneous linear equations2
  • As a set1 of all linear combinations of certain specified vectors.3

Null Space

Intuition

\[ \begin{aligned} x_1 - 3x_2 - 2x_3 = 0 \\ -5x_1 + 9x_2 + x_3 = 0 \end{aligned} \]

This can be written in \(A \vec x = \vec 0\) form.

\[ A = \begin{bmatrix} 1 & -3 & -2 \\ -5 & 9 & 1 \end{bmatrix} \]

Here \(\vec x\) can be thought of solution set1 for this homogeneous linear system.2
\(A\) being a linear operator, transforms \(\vec x\) (the whole set1) into \(\vec 0\).
Therefore, \(\vec x\) is called the null space.

Definition

The null space of \(A_{m \times n}\) is the solution set1 of the homogeneous linear equation2 \(A \vec x = \vec 0\).

\[Nul(A) = \{x : x \in \mathbb R^n, A \vec x = \vec 0\}\]

Example

\[ A = \begin{bmatrix} 1 & -3 & -2 \\ -5 & 9 & 1 \end{bmatrix} \]
\[ \vec u = \begin{bmatrix} 5 \\ 3 \\ -2 \end{bmatrix} \]

Determine if \(\vec u \in Nul(A)\)

Solution

\[\vec u \in Nul(A) \implies A \vec u = \vec 0\]
\[ Au = \begin{bmatrix} 1 & -3 & -2 \\ -5 & 9 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \\ -2 \end{bmatrix} = \begin{bmatrix} 5 - 9 + 4 \\ -25 + 27 - 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

Therefore, \(\vec u\) is in \(Nul(A)\).

Example

Determine the null space of the following matrix4

\[ A = \begin{bmatrix} 4 & 0 \\ -8 & 20 \end{bmatrix} \]

Solution

\[ \begin{bmatrix} 4 & 0 \\ -8 & 20 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \implies \begin{bmatrix} 4x_1 + 0x_2 \\ -8x_1 + 20x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]
\[ \begin{aligned} \implies 4x_1 + 0x_2 = 0 \implies x_1 = 0 \\ \implies -8x_1 + 20x_2 = 0 \implies x_2 = 0 \end{aligned} \]

Therefore, the null space of \(A\) is \(\{\vec 0\}\).

Theorem

Elementary row operations5 do not change the null space of a matrix.4

Theorem

The null space of \(A_{m \times n}\) is a subspace of \(\mathbb R^n\).

Example

Find spanning set for the null space of the matrix.4

\[ A = \begin{bmatrix} -3 & 6 & -1 & 1 & -7 \\ 1 & -2 & 2 & 3 & -1 \\ 2 & -4 & 5 & 8 & -4 \end{bmatrix} \]

Solution

After row reducing6 to echelon form6 of \([A \quad \vec 0]\), we have

\[ \begin{bmatrix} 1 & -2 & 0 & -1 & 3 & 0 \\ 0 & 0 & 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]
\[ \begin{array}{cccccccccccccccc} x_1 &- &2&x_2 &&& &- &&x_4 &+ &3&x_5 &= &0 \\ &&&&&&x_3 &+ &2&x_4 &- &2&x_5 &= &0 \\ &&&&&&&&&&&0& &= &0 \end{array} \]

The general solution is

\[x_1 = 2x_2 + x_4 - 3x_5\]
\[x_2 = \text{free variable}\]
\[x_3 = -2x_4 + 2x_5\]
\[x_4 = \text{free variable}\]
\[x_5 = \text{free variable}\]
\[ \begin{array}{cccccccccccc} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} &= &\begin{bmatrix} 2x_2 + x_4 - 3x_5\\ x_2\\ -2x_4 + 2x_5\\ x_4\\ x_5 \end{bmatrix} &= &x_2 &\begin{bmatrix} 2\\ 1\\ 0\\ 0\\ 0\\ \end{bmatrix} &+ &x_4 &\begin{bmatrix} 1\\ 0\\ -2\\ 1\\ 0\\ \end{bmatrix} &+ &x_5 &\begin{bmatrix} -3\\ 0\\ 2\\ 0\\ 1 \end{bmatrix}\\ &&&&&\Big\uparrow &&&\Big\uparrow &&&\Big\uparrow\\ &&&&&\vec u &&& \vec v &&& \vec w \end{array} \]
\[= x_2 \vec u + x_4 \vec v + x_5 \vec w\]

Every linear combination2 of \(\vec u, \vec v, \vec w\) is an element of \(Nul(A)\).
Thus \(\{\vec u, \vec v, \vec w\}\) is a spanning set1 for \(Nul(A)\).

  • The spanning set1 is automatically linearly independent7 because the free variables are the weights of spanning vectors3
  • When \(Nul(A)\) contains non zero vector,3 the number of vectors3 in the spanning set1 equals to the number of free variables in equation \(A \vec x = \vec 0\).

The Column Space of a Matrix

Column space is defined explicitly via linear combinations.8

Definition

If \(A_{m \times n} = [a_1 \quad a_2 \quad \ldots \quad a_n]\) written as \(Col \, A\) is the set1 of all linear combinations8 of the columns of \(A\).

\[Col \, (A) = Span\{a_1, a_2, \ldots, a_n\}\]

The column space of a matrix4 is that subspace spanned by the columns of the matrix4 (columns viewed as vectors3).
It is that space defined by all linear combinations8 of the column of the matrix.4

Theorem

The column space of \(A_{m \times n}\) is a subspace of \(\mathbb R^m\).

\[Col \, (A) = \{b : b = A \vec x \text{ for some } \vec x \in \mathbb R^n\}\]

Example

Find a matrix4 \(A\) such that \(\mathbb W = Col \, (A)\)

\[ \mathbb W = \left\{ \begin{bmatrix} 6a - b\\ a + b\\ -7a \end{bmatrix} : a, b \in \mathbb R \right\} \]
\[ W = \left\{ a \begin{bmatrix} 6 \\ 1 \\ -7 \end{bmatrix} + b \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} : a, b \in \mathbb{R} \right\} = \text{Span} \left\{ \begin{bmatrix} 6 \\ 1 \\ -7 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} \right\} \]
\[ A = \begin{bmatrix} 6 & -1 \\ 1 & 1 \\ -7 & 0 \end{bmatrix} \]

Theorem

A system of linear equations8 \(A \vec x = \vec b\) is consistent if and only if \(\vec b\) is in column space of \(A\).

Theorem

If \(x_0\) is a particular solution to linear combination8 and \(\vec {v_1}, \vec{v_2}, \ldots, \vec{v_k}\) form the solution set1 of the homogeneous linear system2 \(A \vec x = \vec 0\) then general solution can be written as \(\vec x = \vec {x_0} + c_1 \vec {v_1} + c_2 \vec {v_2} + \ldots + c_n \vec {v_n}\).

The Contrast between \(Nul \, (A)\) and \(Col \, (A)\)

\[ A = \begin{bmatrix} 2 & 4 & -2 & 1 \\ -2 & -5 & 7 & 3 \\ 3 & 7 & -8 & 6 \end{bmatrix} \]
  • The columns of \(A\) each have three entries,4 so Col \(A\) is a subspace of \(R^k\), where \(k = 3\)
  • A vector3 \(x\) such that \(A \vec x\) is defined must have 4 entries,4 so \(Nul \, (A)\) is a subspace of \(R^k\), where \(k = 4\).

When \(A\) is rectangular, both \(Nul (A)\) and \(Col(A)\) exist in completely different universes.

# Nul \(A\) Col \(A\)
1 Nul \(A\) is a subspace of \(\mathbb{R}^n\). Col \(A\) is a subspace of \(\mathbb{R}^m\).
2 Nul \(A\) is implicitly defined; i.e., we are given only a condition (\(Ax = 0\)) that vectors in Nul \(A\) must satisfy. Col \(A\) is explicitly defined; that is, we are told how to build vectors in Col \(A\).
3 It takes time to find vectors in Nul \(A\). Row operations on \([A \ \ 0]\) are required. It is easy to find vectors in Col \(A\). The columns of \(A\) are displayed; others are formed from them.
4 There is no obvious relation between Nul \(A\) and the entries in \(A\). There is an obvious relation between Col \(A\) and the entries in \(A\), since each column of \(A\) is in Col \(A\).
5 A typical vector \(v\) in Nul \(A\) has the property that \(Av = 0\). A typical vector \(v\) in Col \(A\) has the property that the equation \(Ax = v\) is consistent.
6 Given a specific vector \(v\), it is easy to tell if \(v\) is in Nul \(A\). Just compute \(Av\). Given a specific vector \(v\), it may take time to tell if \(v\) is in Col \(A\). Row operations on \([A \ \ v]\) are required.
7 Nul \(A = \{0\}\) if and only if the equation \(Ax = 0\) has only the trivial solution. Col \(A = \mathbb{R}^m\) if and only if the equation \(Ax = b\) has a solution for every \(b\) in \(\mathbb{R}^m\).
8 Nul \(A = \{0\}\) if and only if the linear transformation \(x \mapsto Ax\) is one-to-one. Col \(A = \mathbb{R}^m\) if and only if the linear transformation \(x \mapsto Ax\) maps \(\mathbb{R}^n\) onto \(\mathbb{R}^m\).

Kernel9 And Range of a Linear Transformation10

Subspaces of vector spaces other than \(R^n\) are often described in terms of a linear transformation10 instead of a matrix.4

Definition

A linear transformation10 \(T\) from a vector space \(\mathbb V\) into a vector space \(\mathbb W\) is a rule that assigns to each vector3 \(\vec x \in \mathbb V\) a unique vector3 \(T(\vec x) \in \mathbb W\) such that

  • \(T(\vec u + \vec v) = T(\vec u) + T(\vec v) \quad \forall \vec u, \vec v \in \mathbb V\)
  • \(T(c \vec u) = c T(\vec u) \quad \forall \vec u \in \mathbb V, c \in \mathbb R\)

Definition

If \(T: \mathbb V \to \mathbb W\) is a linear transformation,10 then the set1 of vectors3 in \(\mathbb V\) that \(\mathbb T\) maps into \(\vec 0\) is called the kernel of \(T\).
It is denoted by \(ker(T)\).
The set1 of all vectors3 in \(\mathbb W\) that are images under \(T\) of at least one vector3 in \(\vec V\) is called the range of \(T\).
It is denoted by \(R(T)\).

Example

If \(T_A : \mathbb R^n \to \mathbb R^m\) is multiplication by the \(m \times n\) matrix4 \(A\), then \(kert(T_A)\) is the null space of \(A\) and the range of \(T_A\) is the column space of \(A\).

Remarks

The kernel of \(T\) is a subspace of \(\mathbb V\) and the range of \(T\) is a subspace of \(\mathbb W\).

Example

Let \(\mathbb V\) be the vector space of all real valued functions11 defined on the interval12 \([a, b]\) with property that they are continuous13 and differentiable.14
Let \(\mathbb W\) be the vector space of all continuous functions13 on \([a, b]\) and let \(D : \mathbb V \to \mathbb W\) be the transformation10 that changes \(f\) in \(\mathbb V\) into its derivative14 \(f^\prime\).

\[D(f + g) = D(f) + D(g)\]
\[D(c f) = c D(f)\]

\(Kert(D)\) is the set1 of the continuous functions11 of \([a, b]\) and the range of \(D\) is the set1 \(\mathbb W\) of all continuous functions13 on \([a, b]\).

References

Read more about notations and symbols.

  1. Read more about sets. ↩↩↩↩↩↩↩↩↩↩↩↩↩↩

  2. Read more about homogeneous linear equations. ↩↩↩↩↩

  3. Read more about vectors. ↩↩↩↩↩↩↩↩↩↩↩

  4. Read more about matrices. ↩↩↩↩↩↩↩↩↩↩↩

  5. Read more about row operations. ↩

  6. Read more about echelon form. ↩↩

  7. Read more about linear independence. ↩

  8. Read more about linear combinations and systems. ↩↩↩↩↩

  9. Read more about kernel. ↩

  10. Read more about linear transformations. ↩↩↩↩↩

  11. Read more about functions. ↩↩

  12. Read more about intervals. ↩

  13. Read more about continuity. ↩↩↩

  14. Read more about differentiation. ↩↩