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MathidiaMathematics Academy
Linear algebra II
Lesson 13 of 75
Standard practice

Eigenstructure

Objective. Find eigenspaces and algebraic multiplicities.

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  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

Eigenstructure describes how a linear operator stretches or shrinks vectors along special directions: v is an eigenvector with eigenvalue λ if Av = λv for v ≠ 0. Eigenvalues are roots of the characteristic polynomial det(A − λI) = 0, and the eigenspace for λ is ker(A − λI), whose dimension is the geometric multiplicity, bounded above by the algebraic multiplicity (the root's multiplicity in the polynomial). For A = [[2,0],[0,3]], eigenvalues are 2 and 3, each with one-dimensional eigenspaces spanned by the standard basis vectors. Comparing algebraic and geometric multiplicities reveals whether a matrix is diagonalisable.

Worked example

Find the eigenvalues and eigenspace structure of the matrix A = [[4,1],[0,4]].

4 … 1
Illustrated card showing 4 … 1
  1. 1Solve det(A − λI) = (4 − λ)² = 0, giving λ = 4 with algebraic multiplicity 2.
  2. 2Find ker(A − 4I): solve (A − 4I)v = 0, which gives the matrix [[0,1],[0,0]].
  3. 3This forces the second component to be zero, leaving v = (t, 0) for any t.
  4. 4So the eigenspace is one-dimensional, spanned by (1, 0), giving geometric multiplicity 1.

Answer: λ = 4, algebraic multiplicity 2, geometric multiplicity 1

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Common mistake. Students assume geometric multiplicity always equals algebraic multiplicity; for A = [[2,1],[0,2]], λ = 2 has algebraic multiplicity 2 but the eigenspace is only one-dimensional, so geometric multiplicity is 1.

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