Kernel, Image and Rank-Nullity

Definition Let $T \colon \mathbb{R}^n \to \mathbb{R}^m$ be a linear map. The kernel of $T$ is the set of inputs that $T$ sends to the zero vector,

It is a subspace of the input space $\mathbb{R}^n$. It contains $\mathbf{0}$ because $T(\mathbf{0}) = \mathbf{0}$, and if $\mathbf{u}$ and $\mathbf{v}$ lie in it, then $T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v}) = \mathbf{0}$ for all real $a$ and $b$. Its dimension $\dim \ker T$ is the nullity of $T$. When $T(\mathbf{x}) = A\mathbf{x}$ for a real $m \times n$ matrix $A$, the kernel is the null space, $\ker T = \operatorname{Null} A$.

Definition The image of a linear map $T \colon \mathbb{R}^n \to \mathbb{R}^m$ is the set of outputs it actually reaches,

It is a subspace of the codomain $\mathbb{R}^m$, since $\mathbf{0} = T(\mathbf{0})$ is reached and a combination $aT(\mathbf{u}) + bT(\mathbf{v}) = T(a\mathbf{u} + b\mathbf{v})$ of reached outputs is reached again. Its dimension is the rank of $T$, written $\operatorname{rank} T$. For $T(\mathbf{x}) = A\mathbf{x}$ the image is the column space, $\operatorname{im} T = \operatorname{Col} A$, so $\operatorname{rank} T = \operatorname{rank} A$. The codomain is the declared output space, and the image can be a proper subspace of it.

Definition Let $T \colon \mathbb{R}^n \to \mathbb{R}^m$ be a linear map. Then

the dimension of the input space. For a real $m \times n$ matrix $A$ this reads $\operatorname{rank} A + \dim \operatorname{Null} A = n$. The same statement, with the same proof, holds for a linear map from any finite-dimensional vector space, with $n$ replaced by the dimension of that space.

The proof uses an adapted basis. Choose a basis $\mathbf{u}_1, \dots, \mathbf{u}_k$ of $\ker T$ and extend it to a basis $\mathbf{u}_1, \dots, \mathbf{u}_k, \mathbf{v}_1, \dots, \mathbf{v}_{n-k}$ of $\mathbb{R}^n$, which is always possible in finite dimension. The vectors $T(\mathbf{v}_1), \dots, T(\mathbf{v}_{n-k})$ span $\operatorname{im} T$, because writing any $\mathbf{x}$ in the full basis and applying $T$ removes every $\mathbf{u}_i$ term. They are also independent. If $d_1T(\mathbf{v}_1) + \cdots + d_{n-k}T(\mathbf{v}_{n-k}) = \mathbf{0}$, then $d_1\mathbf{v}_1 + \cdots + d_{n-k}\mathbf{v}_{n-k}$ lies in $\ker T$, so it equals some combination of the $\mathbf{u}_i$, and independence of the full basis forces every $d_j$ to be $0$. Hence $\operatorname{rank} T = n - k$. In elimination terms, an $m \times n$ matrix with $r$ pivot columns has rank $r$ and $n - r$ free variables, one kernel basis vector for each.

Definition A map $T \colon \mathbb{R}^n \to \mathbb{R}^m$ is injective, or one-to-one, if $T(\mathbf{x}) = T(\mathbf{y})$ implies $\mathbf{x} = \mathbf{y}$. A linear map is injective exactly when $\ker T = \{\mathbf{0}\}$. If the kernel is $\{\mathbf{0}\}$ and $T(\mathbf{x}) = T(\mathbf{y})$, then $T(\mathbf{x} - \mathbf{y}) = \mathbf{0}$, so $\mathbf{x} = \mathbf{y}$. If some nonzero $\mathbf{k}$ lies in the kernel, then $\mathbf{k}$ and $\mathbf{0}$ have the same output.

The map is surjective onto $\mathbb{R}^m$, or onto, if every $\mathbf{b}$ in $\mathbb{R}^m$ equals $T(\mathbf{x})$ for some $\mathbf{x}$. That means $\operatorname{im} T = \mathbb{R}^m$, or equivalently $\operatorname{rank} T = m$. By rank-nullity, a linear $T$ is injective exactly when $\operatorname{rank} T = n$. Since the rank is at most $m$ and at most $n$, an injective linear map needs $n \le m$ and a surjective one needs $n \ge m$. When $n = m$, injective and surjective are equivalent, because each one says that the rank equals $n$.

Definition Let $T \colon \mathbb{R}^n \to \mathbb{R}^m$ be linear and let $\mathbf{b}$ be a vector in $\mathbb{R}^m$. The fiber of $T$ over $\mathbf{b}$ is the set of inputs with output $\mathbf{b}$,

which is the solution set of $T(\mathbf{x}) = \mathbf{b}$. It is empty when $\mathbf{b}$ is not in $\operatorname{im} T$. If $\mathbf{x}_p$ is any one vector in it, then the fiber is the translated kernel

Every such vector is a solution, since $T(\mathbf{x}_p + \mathbf{k}) = \mathbf{b} + \mathbf{0} = \mathbf{b}$. Conversely, any solution $\mathbf{x}$ satisfies $T(\mathbf{x} - \mathbf{x}_p) = \mathbf{b} - \mathbf{b} = \mathbf{0}$, so $\mathbf{x} - \mathbf{x}_p$ lies in the kernel. The fiber over $\mathbf{0}$ is the kernel itself. A fiber over $\mathbf{b} \ne \mathbf{0}$ is not a subspace, because it does not contain $\mathbf{0}$.

Definition Let $T \colon \mathbb{R}^n \to \mathbb{R}^m$ and $S \colon \mathbb{R}^m \to \mathbb{R}^p$ be linear, so that $S$ can receive the outputs of $T$ and the composition $S \circ T$ is defined. Then

The first statement holds because $T(\mathbf{k}) = \mathbf{0}$ gives $S(T(\mathbf{k})) = S(\mathbf{0}) = \mathbf{0}$. For the rank, every output of $S \circ T$ lies in $\operatorname{im} S$, so $\operatorname{rank}(S \circ T) \le \operatorname{rank} S$. If $\mathbf{w}_1, \dots, \mathbf{w}_r$ is a basis of $\operatorname{im} T$, then every output $S(T(\mathbf{x}))$ is a combination of $S(\mathbf{w}_1), \dots, S(\mathbf{w}_r)$, so $\operatorname{rank}(S \circ T) \le r = \operatorname{rank} T$. In matrix form, $\operatorname{rank}(MA) \le \min(\operatorname{rank} M, \operatorname{rank} A)$ whenever the product $MA$ is defined.