Kernel and Range

Precise definition

For a linear map $T:V\to W$, the kernel is $\ker T=\{v\in V:T(v)=0\}$ and the range or image is $\operatorname{range}T=\{T(v):v\in V\}\subseteq W$. Both are subspaces: the kernel of the domain and range of the codomain.

Notation and mathematical language

For a matrix map $T(x)=Ax$, $\ker T$ is the null space of $A$, while $\operatorname{range}T$ is the column space of $A$. Nullity is $\dim\ker T$ and rank is $\dim\operatorname{range}T$. Rank–nullity gives $\dim V=\operatorname{nullity}T+\operatorname{rank}T$.

Conceptual picture

The kernel measures inputs erased by the map, so $T$ is injective exactly when $\ker T=\{0\}$. The range records reachable outputs, so $T$ is surjective exactly when its range equals the codomain.

Conditions and key results

A column-space basis uses pivot columns from the original matrix. Row operations preserve the null space equations but change the literal column space. The codomain may be larger than the range; these terms are not interchangeable.

A reliable strategy

1. Represent the map in coordinates and row-reduce its matrix.
2. Solve $Ax=0$ in parametric vector form to obtain a kernel basis.
3. Locate pivot columns and take the corresponding original columns for a range basis.
4. Check rank–nullity and interpret injectivity or surjectivity relative to the declared spaces.

Fully worked example

Interpretation and application

Kernel and range describe information loss, attainable outputs, identifiability, and constraints. A large kernel means many inputs produce the same output; that can be a designed compression or an obstacle to reconstruction.

Common mistakes

Verification and reasonableness

• Apply $A$ to every proposed kernel basis vector and get zero.
• Express every original column as a combination of the chosen range basis.
• Confirm rank plus nullity equals the number of input coordinates.

Practice

1. What kernel condition characterizes injectivity?
2. A $3\times5$ matrix has rank 3. Find nullity.
3. Where does the matrix range live?

Further deduction

The first isomorphism theorem states that $V/\ker T$ is naturally isomorphic to $\operatorname{range}T$. Informally, once inputs differing only by an invisible kernel vector are identified, the remaining information matches exactly the reachable outputs. In finite dimensions, its dimension statement is rank–nullity.

For composition $S\circ T$, every vector in $\ker T$ is automatically in $\ker(S\circ T)$, while $\operatorname{range}(S\circ T)\subseteq\operatorname{range}S$. These containments explain rank loss under composition and help locate which stage erases information or restricts outputs.

The fundamental orthogonality relations connect all four subspaces: $\operatorname{range}(A)^\perp=\ker(A^T)$ and $\operatorname{range}(A^T)^\perp=\ker A$ over real inner-product spaces. A vector is orthogonal to every column of $A$ exactly when $A^Ty=0$. These identities explain the consistency condition for $Ax=b$: the right-hand side must be orthogonal to every vector in the left null space.

Related topics

• Linear Transformations
• Subspaces
• Dimension