📂Linear Algebra

Linear Transformation: Kernel and Range

Definition¹

Let’s say $T : V \to W$ is a linear transformation. The set of elements of $V$ that are mapped to $\mathbf{0}$ by $T$ is called the kernel or null space, and is denoted as follows.

$$\text{ker}(T) = N(T) := \left\{ \mathbf{v} \in V : T( \mathbf{v} ) = \mathbf{0} \right\}$$

The set of images under $\mathbf{v} \in V$ by $T$ is called the range or image of $T$, and is denoted as follows.

$$R(T) := \left\{ T(\mathbf{v}) : \forall \mathbf{v} \in V \right\}$$

Explanation

If $T : V \to W$ is a linear transformation and $V, W$ is finite-dimensional, $T$ is essentially the same as a matrix, and $N(T)$ is the null space of the matrix representing $T$.

Theorem

Let’s consider $T : V \to W$ to be a linear transformation. Then,

• (a) The kernel of $T$ is a subspace of $V$.
• (b) The range of $T$ is a subspace of $W$.

Proof

To show that it’s a subspace, we need to prove that it’s nonempty, and closed under addition and scalar multiplication.

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(a)

If $T$ is a linear transformation, then according to $T(\mathbf{0})=\mathbf{0}$, $N(T)$ is not empty. Now, let $\mathbf{v}_{1}, \mathbf{v}_{2} \in N(T)$ and consider $k$ to be any scalar. Then, the following holds.

$$\begin{align*} T( \mathbf{v}_{1} + \mathbf{v}_{2} ) &= T(\mathbf{v}_{1}) + T(\mathbf{v}_{2}) = \mathbf{0} + \mathbf{0} = \mathbf{0} \\ T( k\mathbf{v}_{1}) &= kT(\mathbf{v}_{1}) = k\mathbf{0} = \mathbf{0} \end{align*}$$

Hence, $N(T)$ is a subspace of $V$.

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(b)

If $T$ is a linear transformation, then, as per $T(\mathbf{0})=\mathbf{0}$, $R(T)$ is not empty. Now, let $\mathbf{w}_{1}, \mathbf{w}_{2} \in R(T)$ and consider $k$ to be any scalar. Then, it suffices to show that there exists $\mathbf{a}, \mathbf{b} \in V$ that satisfies the following.

$$T(\mathbf{a}) = \mathbf{w}_{1} + \mathbf{w}_{2} \quad \text{and} \quad T(\mathbf{b}) = k\mathbf{w}_{1}$$

But the statement $\mathbf{w}_{1}, \mathbf{w}_{2} \in R(T)$ means that there exists $\mathbf{v}_{1}, \mathbf{v}_{2} \in V$ that satisfies the following.

$$T(\mathbf{v}_{1}) = \mathbf{w}_{1} \quad \text{and} \quad T(\mathbf{v}_{2}) = \mathbf{w}_{2}$$

Therefore, the following equation holds.

$$\begin{align*} \mathbf{w}_{1} + \mathbf{w}_{2} &= T(\mathbf{v}_{1}) + T(\mathbf{v}_{2}) = T(\mathbf{v}_{1} + \mathbf{v}_{2}) = T(\mathbf{a}) \\ k\mathbf{w}_{1} &= kT(\mathbf{v}_{1}) = T(k\mathbf{v}_{1})= T(\mathbf{b}) \end{align*}$$

Hence, $R(T)$ is a subspace of $W$.

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