Definition 1.1.5 (Subspace of $\R^n$). A nonempty subset $V\subset \R^n$ is a vector subspace if it is closed under addition and closed under multiplication by scalars; that is, $V$ is a vector subspace when
$$
\vec x, \vec y \in V \text{ and }a \in \R, \text{ then }\vec x + \vec y\in V \text{ and }a\vec x \in V.
$$
Kernels are a way of speaking about the uniqueness of solutions of linear equations. Images are a way of speaking about existence of solutions of linear equations.
Definition 2.5.1 (Kernel and image). Let $T: \R^n \to \R^m$ be a linear transformation.
- The kernel of $T$, denoted $\ker T$, is the set of vectors $\vec x \in \R^n$ such that $T(\vec x) = \vec 0$.
- The image of $T$, denoted $\text{img } T$, is the set of vectors $\vec w \in \R^m$ such that there is a vector $\vec v \in \R^n$ with $T(\vec v) = \vec w$.
Proposition 2.5.3. Let $T: \R^n \to \R^m$ be a linear transformation. The system of linear equations $T(\vec x) = \vec b$ has
- at most one solution for every $\vec b \in \R^m$ if and only if $\ker T = \{\vec 0\}$. ($T$ is injective).
- at least one solution for every $\vec b \in \R^m$ if and only if $\text{img } T = \R^m$. ($T$ is surjective).
Proof.
- If the kernel of $T$ is not $\{\vec 0\}$, then there is more than one solution to $T(\vec x) = \vec 0$. We use proof by contradiction for the other direction. If there exists a $\vec b$ for which $T(\vec x) = \vec b$ has more than one solution, i.e., $T(\vec x_1) = T(\vec x_2) = \vec b$ and $\vec x_1 \neq \vec x_2$, then $T(\vec x_1 - \vec x_2) = T(\vec x_1) - T(\vec x_2) = \vec b - \vec b = \vec 0$. So $(\vec x_1 - \vec x_2)$ must be a nonzero element of the kernel and $\ker T\neq \{\vec 0\}$.
- This is true by definition.
Finding bases for the image and kernel
Theorem 2.4.5 (A basis for the image). The pivotal columns of $[T]$ form a basis of $\text{img } T$.
Example. Consider the matrix $A$ that describes a linear transformation from $\R^5$ to $\R^4$.