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Mathematics · University

Eigenvectors and diagonalization

Av=λv requires a nonzero vector v.

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Prerequisites

Matrix multiplication and characteristic polynomials.

Study notes

Eigenpair

Av=λv requires a nonzero vector v.

Diagonalization

A=PDP⁻¹ exists when there is a basis of eigenvectors over the chosen field.

Repeated eigenvalues

Algebraic multiplicity does not guarantee enough independent eigenvectors.

Symmetric matrices

A real symmetric matrix has an orthonormal eigenbasis over the reals.

Worked examples

For A=[[3,1],[1,3]], find the eigenvalue of (1,1).

A(1,1)=(4,4)=4(1,1).

4

For A=[[3,1],[1,3]], find the eigenvalue corresponding to (1,−1).

A(1,−1)=(2,−2)=2(1,−1).

2

Check your understanding

  1. Why is the zero vector excluded from eigenvectors?

    1. Its entries are undefined
    2. It satisfies Av=λv for every λ and gives no eigendirection
    3. It cannot be multiplied by a matrix
    4. It is never in a vector space

  2. Is [[1,1],[0,1]] diagonalizable over the reals?

    1. No; its eigenspace has dimension 1
    2. No matrix with eigenvalue 1 is diagonalizable
    3. Yes; repeated eigenvalues always suffice
    4. Yes; every triangular matrix is diagonalizable

  3. A diagonal matrix has diagonal entries 2,3,5. Find the trace of its square.

  4. Show that eigenvectors for two distinct eigenvalues are linearly independent.

Answers and marking guidance

Answers and marking notes
  1. It satisfies Av=λv for every λ and gives no eigendirection

    The definition needs a nonzero vector to encode a meaningful eigenvalue relation.

  2. No; its eigenspace has dimension 1

    The only eigenvalue is 1, and eigenvectors have second component zero.

  3. 38

    Squaring gives diagonal entries 4,9,25; their sum is 38.

  4. If v₂=cv₁ with c≠0, applying A gives λ₂cv₁=cλ₁v₁, forcing λ₂=λ₁, a contradiction.

    Credit use of nonzero vectors and the distinct-eigenvalue contradiction.

Further reading and review references

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