Linear Transformations, Null Space, and Image

Prerequisites

• Vectors and Linear Combinations
• Vector Spaces, Subspaces, and Matrices
• Linear Dependence and Independence, Bases and Dimension
• Injection, surjection

Linear transformations

A special class of functions that preserve the structure of vector spaces are called linear transformations. They are fundamental across pure and applied mathematics, social and natural sciences, and engineering.

Definition
Let $\mathbb{V}$ and $\mathbb{W}$ be $F$-vector spaces. A function $T: \mathbb{V} \to \mathbb{W}$ is called a linear transformation from $\mathbb{V}$ to $\mathbb{W}$ if, for all $\mathbf{x}, \mathbf{y} \in \mathbb{V}$ and $c \in F$, the following hold:

1. $T(\mathbf{x}+\mathbf{y}) = T(\mathbf{x}) + T(\mathbf{y})$
2. $T(c\mathbf{x}) = cT(\mathbf{x})$

When $T$ is a linear transformation, we also simply say that $T$ is linear. A linear transformation $T: \mathbb{V} \to \mathbb{W}$ satisfies the following four properties.

1. $T$ linear $\quad \Rightarrow \quad T(\mathbf{0}) = \mathbf{0}$
2. $T$ linear $\quad \Leftrightarrow \quad T(c\mathbf{x} + \mathbf{y}) = cT(\mathbf{x}) + T(\mathbf{y}) \; \forall \, \mathbf{x}, \mathbf{y} \in \mathbb{V},\, c \in F$
3. $T$ linear $\quad \Rightarrow \quad T(\mathbf{x} - \mathbf{y}) = T(\mathbf{x}) - T(\mathbf{y}) \; \forall \, \mathbf{x}, \mathbf{y} \in \mathbb{V}$
4. $T$ linear $\quad \Leftrightarrow \quad T\left( \sum_{i=1}^n a_i \mathbf{x}_i \right) = \sum_{i=1}^n a_i T(\mathbf{x}_i)$

When proving that a function is linear, it is often convenient to use Property 2.

Linear algebra has wide and varied applications in geometry because many important geometric maps are linear. In particular, the three principal geometric transformations—rotation, reflection, and projection—are linear transformations.

Two linear transformations occur especially often:

Identity and zero transformations
For $F$-vector spaces $\mathbb{V}, \mathbb{W}$:

• Identity transformation: the function $I_\mathbb{V}: \mathbb{V} \to \mathbb{V}$ defined by $I_\mathbb{V}(\mathbf{x}) = \mathbf{x}$ for all $\mathbf{x} \in \mathbb{V}$
• Zero transformation: the function $T_0: \mathbb{V} \to \mathbb{W}$ defined by $T_0(\mathbf{x}) = \mathbf{0}$ for all $\mathbf{x} \in \mathbb{V}$

Many other familiar operations are linear transformations.

Examples of linear transformations

• Rotation
• Reflection
• Projection
• Transpose
• Differentiation of a differentiable function
• Integration of a continuous function

Null space and image

Definitions of the null space and the image

Definition
For vector spaces $\mathbb{V}, \mathbb{W}$ and a linear transformation $T: \mathbb{V} \to \mathbb{W}$:

• Null space (or kernel): the set of vectors $\mathbf{x} \in \mathbb{V}$ such that $T(\mathbf{x}) = \mathbf{0}$, denoted $\mathrm{N}(T)$

  \[\mathrm{N}(T) = \{ \mathbf{x} \in \mathbb{V}: T(\mathbf{x}) = \mathbf{0} \}\]

• Range (or image): the subset of $\mathbb{W}$ consisting of all values of $T$, denoted $\mathrm{R}(T)$

  \[\mathrm{R}(T) = \{ T(\mathbf{x}): \mathbf{x} \in \mathbb{V} \}\]

e.g. For vector spaces $\mathbb{V}, \mathbb{W}$, the identity $I: \mathbb{V} \to \mathbb{V}$ and the zero map $T_0: \mathbb{V} \to \mathbb{W}$ satisfy:

• $\mathrm{N}(I) = \{\mathbf{0}\}$
• $\mathrm{R}(I) = \mathbb{V}$
• $\mathrm{N}(T_0) = \mathbb{V}$
• $\mathrm{R}(T_0) = \{\mathbf{0}\}$

A key point going forward is that the null space and the image of a linear transformation are subspaces of the corresponding vector spaces.

Theorem 1
For vector spaces $\mathbb{V}, \mathbb{W}$ and a linear transformation $T: \mathbb{V} \to \mathbb{W}$, the sets $\mathrm{N}(T)$ and $\mathrm{R}(T)$ are subspaces of $\mathbb{V}$ and $\mathbb{W}$, respectively.

Proof
Denote the zero vectors of $\mathbb{V}$ and $\mathbb{W}$ by $\mathbf{0}_\mathbb{V}$ and $\mathbf{0}_\mathbb{W}$, respectively.

Since $T(\mathbf{0}_\mathbb{V}) = \mathbf{0}_\mathbb{W}$, we have $\mathbf{0}_\mathbb{V} \in \mathrm{N}(T)$. Moreover, for $\mathbf{x}, \mathbf{y} \in \mathrm{N}(T)$ and $c \in F$,

\[\begin{align*} T(\mathbf{x} + \mathbf{y}) &= T(\mathbf{x}) + T(\mathbf{y}) = \mathbf{0}_\mathbb{W} + \mathbf{0}_\mathbb{W} = \mathbf{0}_\mathbb{W}, \\ T(c\mathbf{x}) &= cT(\mathbf{x}) = c\mathbf{0}_\mathbb{W} = \mathbf{0}_\mathbb{W}. \end{align*}\]

$\therefore$ Since $\mathbf{0}_\mathbb{V} \in \mathrm{N}(T)$ and $\mathrm{N}(T)$ is closed under addition and scalar multiplication, $\mathrm{N}(T)$ is a subspace of $\mathbb{V}$.

Similarly, $T(\mathbf{0}_\mathbb{V}) = \mathbf{0}_\mathbb{W}$ implies $\mathbf{0}_\mathbb{W} \in \mathrm{R}(T)$. For all $\mathbf{x}, \mathbf{y} \in \mathrm{R}(T)$ and $c \in F$ (there exist $\mathbf{v}, \mathbf{w} \in \mathbb{V}$ with $T(\mathbf{v}) = \mathbf{x}$ and $T(\mathbf{w}) = \mathbf{y}$), we have

\[\begin{align*} T(\mathbf{v} + \mathbf{w}) &= T(\mathbf{v}) + T(\mathbf{w}) = \mathbf{x} + \mathbf{y}, \\ T(c\mathbf{v}) &= cT(\mathbf{v}) = c\mathbf{x}. \end{align*}\]

$\therefore$ Since $\mathbf{0}_\mathbb{W} \in \mathrm{R}(T)$ and $\mathrm{R}(T)$ is closed under addition and scalar multiplication, $\mathrm{R}(T)$ is a subspace of $\mathbb{W}$. $\blacksquare$

Furthermore, given a basis $\beta = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \}$ of $\mathbb{V}$, we can find a generating set of the image $\mathrm{R}(T)$ as follows.

Theorem 2
For vector spaces $\mathbb{V}, \mathbb{W}$, a linear transformation $T: \mathbb{V} \to \mathbb{W}$, and a basis $\beta = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \}$ of $\mathbb{V}$, we have

\[\mathrm{R}(T) = \mathrm{span}(\{T(\mathbf{v}): \mathbf{v} \in \beta \}) = \mathrm{span}(\{T(\mathbf{v}_1), T(\mathbf{v}_2), \dots, T(\mathbf{v}_n) \})\]

Proof

\[T(\mathbf{v}_i) \in \mathrm{R}(T) \quad \forall \mathbf{v}_i \in \beta.\]

Since $\mathrm{R}(T)$ is a subspace, by Theorem 2 of Vector Spaces, Subspaces, and Matrices,

\[\mathrm{span}(\{T(\mathbf{v}_1), T(\mathbf{v}_2), \dots, T(\mathbf{v}_n) \}) = \mathrm{span}(\{T(\mathbf{v}_i): \mathbf{v}_i \in \beta \}) \subseteq \mathrm{R}(T).\]

Also,

\[\forall \mathbf{w} \in \mathrm{R}(T) \ (\exists \mathbf{v} \in \mathbb{V} \ (\mathbf{w} = T(\mathbf{v}))).\]

Because $\beta$ is a basis of $\mathbb{V}$,

\[\mathbf{v} = \sum_{i=1}^n a_i \mathbf{v}_i \quad \text{(where } a_1, a_2, \dots, a_n \in F \text{)}.\]

Since $T$ is linear,

\[\mathbf{w} = T(\mathbf{v}) = \sum_{i=1}^n a_i T(\mathbf{v}_i) \in \mathrm{span}(\{T(\mathbf{v}_i): \mathbf{v}_i \in \beta \})\] \[\mathrm{R}(T) \subseteq \mathrm{span}(\{T(\mathbf{v}_i): \mathbf{v}_i \in \beta \}) = \mathrm{span}(\{T(\mathbf{v}_1), T(\mathbf{v}_2), \dots, T(\mathbf{v}_n) \}).\]

$\therefore$ Since both contain each other, $\mathrm{R}(T) = \mathrm{span}({T(\mathbf{v}): \mathbf{v} \in \beta })$. $\blacksquare$

This theorem remains valid even when the basis $\beta$ is infinite.

Dimension theorem

Because the null space and image are especially important subspaces, we give special names to their dimensions.

For vector spaces $\mathbb{V}, \mathbb{W}$ and a linear transformation $T: \mathbb{V} \to \mathbb{W}$, assume $\mathrm{N}(T)$ and $\mathrm{R}(T)$ are finite-dimensional.

• Nullity: the dimension of $\mathrm{N}(T)$, denoted $\mathrm{nullity}(T)$
• Rank: the dimension of $\mathrm{R}(T)$, denoted $\mathrm{rank}(T)$

For a linear transformation, the larger the nullity, the smaller the rank, and vice versa.

Theorem 3: Dimension theorem
For vector spaces $\mathbb{V}, \mathbb{W}$ and a linear transformation $T: \mathbb{V}\to \mathbb{W}$, if $\mathbb{V}$ is finite-dimensional, then

\[\mathrm{nullity}(T) + \mathrm{rank}(T) = \dim(\mathbb{V})\]

Proof

Let $\dim(\mathbb{V}) = n$ and $\mathrm{nullity}(T) = \dim(\mathrm{N}(T)) = k$, and let a basis of $\mathrm{N}(T)$ be $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \}$.

By “Linear Dependence and Independence, Bases and Dimension” — Corollary 6-1, we can extend $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \}$ to a basis $\beta = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \}$ of $\mathbb{V}$.

We now show that $S = \{T(\mathbf{v}_{k+1}), T(\mathbf{v}_{k+2}), \dots, T(\mathbf{v}_n) \}$ is a basis of $\mathrm{R}(T)$. First, for $1 \leq i \leq k$, $T(\mathbf{v}_i) = 0$, so by Theorem 2,

\[\begin{align*} \mathrm{R}(T) &= \mathrm{span}(\{T(\mathbf{v}_1), T(\mathbf{v}_2), \dots, T(\mathbf{v}_n) \}) \\ &= \mathrm{span}(\{T(\mathbf{v}_{k+1}), T(\mathbf{v}_{k+2}), \dots, T(\mathbf{v}_n) \}) \\ &= \mathrm{span}(S). \end{align*}\]

Thus $S$ generates $\mathrm{R}(T)$. By Corollary 5-2 of the replacement theorem, it suffices to show that $S$ is linearly independent.

Suppose $\sum_{i=k+1}^n b_i T(\mathbf{v}_i) = 0$ (with $b_{k+1}, b_{k+2}, \dots, b_n \in F$). Since $T$ is linear,

\[\sum_{i=k+1}^n b_i T(\mathbf{v}_i) = 0 \Leftrightarrow T\left(\sum_{i=k+1}^n b_i \mathbf{v}_i \right) = 0 \Leftrightarrow \sum_{i=k+1}^n b_i \mathbf{v}_i \in \mathrm{N}(T).\]

Therefore,

\[\begin{align*} &\exists c_1, c_2, \dots, c_k \in F, \\ &\sum_{i=k+1}^n b_i \mathbf{v}_i = \sum_{i=1}^k c_i \mathbf{v}_i \\ \Leftrightarrow &\sum_{i=1}^k (-c_i)\mathbf{v}_i + \sum_{i=k+1}^n b_i \mathbf{v}_i = 0. \end{align*}\]

Since $\beta$ is a basis of $\mathbb{V}$, the unique solution of $\sum_{i=1}^k (-c_i)\mathbf{v}_i + \sum_{i=k+1}^n b_i \mathbf{v}_i = 0$ is

\[c_1 = c_2 = \cdots = c_k = b_{k+1} = b_{k+2} = \cdots = b_n = 0\]

and hence

\[\sum_{i=k+1}^n b_i T(\mathbf{v}_i) = 0 \quad \Rightarrow \quad b_i = 0.\]

Thus $S$ is linearly independent and is a basis of $\mathrm{R}(T)$.

\[\therefore \mathrm{rank}(T) = n - k = \dim{\mathbb{V}} - \mathrm{nullity}(T). \blacksquare\]

Linear transformations and injections/surjections

For linear transformations, injectivity and surjectivity are closely tied to rank and nullity.

Theorem 4
For vector spaces $\mathbb{V}, \mathbb{W}$ and a linear transformation $T: \mathbb{V} \to \mathbb{W}$,

\[T \text{ is injective} \quad \Leftrightarrow \quad \mathrm{N}(T) = \{\mathbf{0}\}.\]

Theorem 5
If finite-dimensional vector spaces $\mathbb{V}, \mathbb{W}$ have the same dimension and $T: \mathbb{V} \to \mathbb{W}$ is linear, then the following four statements are equivalent.

1. $T$ is injective.
2. $\mathrm{nullity}(T) = 0$
3. $\mathrm{rank}(T) = \dim(\mathbb{V})$
4. $T$ is surjective.

Using the dimension theorem, Properties 1 and 3 of linear transformations, and “Linear Dependence and Independence, Bases and Dimension” — Theorem 6, one can prove Theorem 4 and Theorem 5.

These two theorems are useful when deciding whether a given linear transformation is injective or surjective.

For an infinite-dimensional vector space $\mathbb{V}$ and a linear transformation $T: \mathbb{V} \to \mathbb{V}$, injectivity and surjectivity are not equivalent.

If a linear transformation is injective, the following theorem can be useful in some cases for testing whether a subset of the domain is linearly independent.

Theorem 6
For vector spaces $\mathbb{V}, \mathbb{W}$, an injective linear transformation $T: \mathbb{V} \to \mathbb{W}$, and a subset $S \subseteq \mathbb{V}$,

\[S \text{ is linearly independent} \quad \Leftrightarrow \quad \{T(\mathbf{v}): \mathbf{v} \in S \} \text{ is linearly independent.}\]

Linear transformations and bases

A key feature of linear transformations is that their action is determined by their values on a basis.

Theorem 7
Let $\mathbb{V}, \mathbb{W}$ be $F$-vector spaces, let $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \}$ be a basis of $\mathbb{V}$, and let $\mathbf{w}_1, \mathbf{w}_2, \dots, \mathbf{w}_n \in \mathbb{W}$. Then there exists a unique linear transformation $T: \mathbb{V} \to \mathbb{W}$ such that

\[T(\mathbf{v}_i) = \mathbf{w}_i \quad (i = 1, 2, \dots, n).\]

Proof
For $\mathbf{x} \in \mathbb{V}$, the representation

\[\mathbf{x} = \sum_{i=1}^n a_i \mathbf{v}_i \text{ (}a_1, a_2, \dots, a_n \in F \text{)}\]

is unique. Define a linear transformation $T: \mathbb{V} \to \mathbb{W}$ by

\[T(\mathbf{x}) = T\left( \sum_{i=1}^n a_i \mathbf{v}_i \right) = \sum_{i=1}^n a_i \mathbf{w}_i.\]

i) For $i = 1, 2, \dots, n$, $T(\mathbf{v}_i) = \mathbf{w}_i$.

ii) Suppose another linear transformation $U: \mathbb{V} \to \mathbb{W}$ satisfies $U(\mathbf{v}_i) = \mathbf{w}_i$ for $i = 1, 2, \dots, n$. Then for $\mathbf{x} = \sum_{i=1}^n a_i \mathbf{v}_i \in \mathbb{V}$,

\[U(\mathbf{x}) = \sum_{i=1}^n a_i U(\mathbf{v}_i) = \sum_{i=1}^n a_i \mathbf{w}_i = T(\mathbf{x}_i)\] \[\therefore U = T.\]

From i) and ii), the linear transformation satisfying $T(\mathbf{v}_i) = \mathbf{w}_i$ for $i = 1, 2, \dots, n$ is unique and given by

\[T(\mathbf{x}) = T\left( \sum_{i=1}^n a_i \mathbf{v}_i \right) = \sum_{i=1}^n a_i \mathbf{w}_i. \ \blacksquare\]

Corollary 7-1
Let $\mathbb{V}, \mathbb{W}$ be vector spaces and suppose $\mathbb{V}$ has a finite basis $\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \}$. If two linear transformations $U, T: \mathbb{V} \to \mathbf{W}$ satisfy $U(\mathbf{v}_i) = T(\mathbf{v}_i)$ for $i = 1, 2, \dots, n$, then $U = T$.
In other words, if two linear transformations agree on a basis, they are equal.