Linear Map Kernel or Image Calculator

Math Linear Algebra • Linear Transformations and Eigenvalues

Written by STEM Calculators Team

Compute the kernel (null space) and image (column space) of a linear map \(T(\mathbf{x})=A\mathbf{x}\). The calculator row-reduces \(A\) to RREF, extracts pivot columns for \(\mathrm{Im}(T)\), and builds a basis for \(\ker(T)\) from the free variables. Includes a rank–nullity check.

Codomain dim (rows)

Domain dim (cols)

Tolerance

Display decimals

Steps

Show 2D visual (when possible) Show injective/surjective checks

Matrix \(A\)

A is 2×2

Inputs accept -3.5, 2e-4, fractions like 7/3, and constants pi, e.

Ready

Results

Rank \(=\dim\mathrm{Im}(T)\)

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Nullity \(=\dim\ker(T)\)

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Rank–nullity \(\text{rank}+\text{nullity}=n\)

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Kernel basis \(\ker(T)\subseteq\mathbb{R}^n\)

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Image basis \(\mathrm{Im}(T)\subseteq\mathbb{R}^m\)

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RREF of \(A\)

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Step-by-step

Enter a matrix and click “Compute kernel & image”.

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Theory

Kernel and image

A linear transformation \(T:\mathbb{R}^n\to\mathbb{R}^m\) can be written as \(T(\mathbf{x})=A\mathbf{x}\) for some matrix \(A\in\mathbb{R}^{m\times n}\).

The kernel (or null space) is the set of inputs sent to zero: \[ \ker(T)=\{\mathbf{x}\in\mathbb{R}^n: A\mathbf{x}=\mathbf{0}\}. \] The image (or column space) is the set of attainable outputs: \[ \mathrm{Im}(T)=\{A\mathbf{x}:\mathbf{x}\in\mathbb{R}^n\}=\mathrm{Col}(A)\subseteq\mathbb{R}^m. \]

Intuition: \(\ker(T)\) measures “how many directions collapse to zero”, while \(\mathrm{Im}(T)\) measures “how many independent output directions are reachable”.

How row reduction finds both subspaces

Row-reducing \(A\) to reduced row echelon form (RREF) does two key things:

The pivot columns identify a basis for the column space: the corresponding columns of the original matrix \(A\) form a basis of \(\mathrm{Im}(T)=\mathrm{Col}(A)\).
The free variables in the system \(A\mathbf{x}=\mathbf{0}\) generate a basis for the null space. Set one free variable to 1 (others to 0) and solve for the pivot variables to get one kernel basis vector.

This works because elementary row operations preserve the solution set of \(A\mathbf{x}=\mathbf{0}\).

Rank–nullity theorem

Let \(\mathrm{rank}(A)=\dim\mathrm{Col}(A)\) and \(\mathrm{nullity}(A)=\dim\ker(A)\). Then \[ \mathrm{rank}(A)+\mathrm{nullity}(A)=n. \] The right-hand side is the dimension of the domain \(\mathbb{R}^n\).

Quick checks: \(T\) is injective iff \(\ker(T)=\{\mathbf{0}\}\) (nullity 0), and surjective iff \(\mathrm{rank}(A)=m\).

Example

If \(A=\begin{bmatrix}1&0\\0&0\end{bmatrix}\), then \(A\mathbf{x}=\mathbf{0}\) gives \(x_1=0\) and \(x_2\) free, so \(\ker(T)=\mathrm{span}\{\begin{bmatrix}0\\1\end{bmatrix}\}\). The pivot column is the first column, so \(\mathrm{Im}(T)=\mathrm{span}\{\begin{bmatrix}1\\0\end{bmatrix}\}\).

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