Eigenspace Basis Finder

Math Linear Algebra • Linear Transformations and Eigenvalues

Written by STEM Calculators Team

Compute a basis for the eigenspace \(\mathcal{E}_\lambda = \{\mathbf{v}\neq \mathbf{0}\mid (A-\lambda I)\mathbf{v}=\mathbf{0}\}\). The calculator builds \(B=A-\lambda I\), row-reduces \(B\) to RREF, and returns a null-space basis. It also estimates algebraic vs geometric multiplicity and warns when the matrix is defective.

Matrix size

Eigenvalue \(\lambda\)

Tolerance

Display decimals

Steps

Show 2D eigenvector visual (when possible) Estimate algebraic multiplicity (characteristic polynomial)

Matrix \(A\)

A is 2×2

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

Ready

Results

Geometric multiplicity \(\dim \mathcal{E}_\lambda\)

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Estimated algebraic multiplicity

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Defect \((\text{alg}-\text{geo})\)

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Status

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Eigenspace basis (null space of \(A-\lambda I\))

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Matrix \(B=A-\lambda I\)

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RREF of \(B\)

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Characteristic polynomial (numeric)

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Step-by-step

Enter \(A\) and \(\lambda\), then click “Calculate”.

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Theory

What an eigenspace is

Let \(A\in\mathbb{R}^{n\times n}\) represent the linear map \(T(\mathbf{x})=A\mathbf{x}\). An eigenvector is a nonzero vector \(\mathbf{v}\neq\mathbf{0}\) whose direction is preserved by \(A\): \[ A\mathbf{v}=\lambda\mathbf{v}. \] The scalar \(\lambda\) is the corresponding eigenvalue. Geometrically, \(A\) may stretch, shrink, or flip vectors, but along an eigenvector direction it only scales by \(\lambda\).

Fix an eigenvalue \(\lambda\). The set of all vectors (including \(\mathbf{0}\)) that satisfy the eigenvector equation forms a subspace: \[ \mathcal{E}_\lambda=\{\mathbf{v}\in\mathbb{R}^n : A\mathbf{v}=\lambda\mathbf{v}\}. \] This subspace is called the eigenspace of \(\lambda\). We usually describe an eigenspace by giving a basis: a small list of vectors whose linear combinations generate all eigenvectors in that eigenspace.

Key identity: \(A\mathbf{v}=\lambda\mathbf{v}\) is equivalent to \((A-\lambda I)\mathbf{v}=\mathbf{0}\), so eigenspaces are null spaces.

How to compute a basis: \((A-\lambda I)\mathbf{v}=\mathbf{0}\)

Start from \[ A\mathbf{v}=\lambda\mathbf{v}\quad\Longleftrightarrow\quad (A-\lambda I)\mathbf{v}=\mathbf{0}. \] Define \(B=A-\lambda I\). Then the eigenspace is exactly the kernel (null space) of \(B\): \[ \mathcal{E}_\lambda=\ker(B). \] To find a basis, solve the homogeneous system \(B\mathbf{v}=\mathbf{0}\).

The standard workflow is:

Compute \(B=A-\lambda I\) (subtract \(\lambda\) from the diagonal of \(A\)).
Row-reduce \(B\) to RREF. Row operations preserve the solution set of \(B\mathbf{v}=\mathbf{0}\).
Identify pivot variables and free variables from the RREF.
For each free variable, set it to 1 (and the other free variables to 0) and solve for the pivot variables. Each choice gives one basis vector.

\[ \text{Eigenspace basis} \quad\Longleftarrow\quad \text{null-space basis of }(A-\lambda I). \]

If the only solution is \(\mathbf{v}=\mathbf{0}\), then \(\mathcal{E}_\lambda=\{\mathbf{0}\}\) and \(\lambda\) is not an eigenvalue of \(A\).

Geometric multiplicity

The geometric multiplicity of \(\lambda\) is the dimension of its eigenspace: \[ \mathrm{geo}(\lambda)=\dim\mathcal{E}_\lambda=\dim\ker(A-\lambda I). \] In RREF language, it equals the number of free variables (the nullity of \(A-\lambda I\)). If \(\mathrm{geo}(\lambda)=1\), the eigenspace is a line through the origin (one independent eigenvector direction). If \(\mathrm{geo}(\lambda)=2\) in \(\mathbb{R}^2\), then every vector is an eigenvector (this happens when \(A=\lambda I\)).

Geometric multiplicity matters because it counts how many independent eigenvectors you get for that eigenvalue. Diagonalization needs a full set of \(n\) linearly independent eigenvectors in total (across all eigenvalues).

Algebraic multiplicity and the defect warning

The algebraic multiplicity of \(\lambda\) is its multiplicity as a root of the characteristic polynomial \[ p(t)=\det(tI-A). \] For example, if \(p(t)=(t-2)^2(t+1)\), then \(\lambda=2\) has algebraic multiplicity 2. In general, when \(\lambda\) is an eigenvalue, \[ 1\le \mathrm{geo}(\lambda)\le \mathrm{alg}(\lambda). \]

If \(\mathrm{geo}(\lambda) < \mathrm{alg}(\lambda)\), the matrix is defective at \(\lambda\). The difference \[ \mathrm{defect}(\lambda)=\mathrm{alg}(\lambda)-\mathrm{geo}(\lambda) \] counts how many independent eigenvectors are “missing” compared to the algebraic repetition of the eigenvalue. A defective eigenvalue is a common reason a matrix fails to be diagonalizable.

Example (a classic defective case): \[ A=\begin{bmatrix}2&1\\0&2\end{bmatrix},\quad \lambda=2. \] Then \[ A-2I=\begin{bmatrix}0&1\\0&0\end{bmatrix} \] and solving \((A-2I)\mathbf{v}=\mathbf{0}\) forces \(v_2=0\), with \(v_1\) free. So the eigenspace is \(\mathcal{E}_2=\mathrm{span}\left\{\begin{bmatrix}1\\0\end{bmatrix}\right\}\), which has geometric multiplicity 1 even though \((t-2)^2\) shows algebraic multiplicity 2.

Practical note: for numeric matrices, an “algebraic multiplicity estimate” can be sensitive to rounding. The null space test \((A-\lambda I)\mathbf{v}=\mathbf{0}\) is the most direct confirmation that \(\lambda\) is an eigenvalue.

Advanced note: generalized eigenspaces

In more advanced linear algebra, defective eigenvalues are handled using generalized eigenvectors. Instead of requiring \((A-\lambda I)\mathbf{v}=\mathbf{0}\), one allows vectors that satisfy \[ (A-\lambda I)^k\mathbf{v}=\mathbf{0} \] for some integer \(k\ge 1\). The union of these solution spaces is the generalized eigenspace. This is the idea behind Jordan form: even when ordinary eigenvectors are not enough to form a basis, generalized eigenvectors can complete one (over an appropriate field).

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