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Diagonalising symmetric matrices

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

A symmetric matrix can always be written in diagonal form using an orthogonal matrix of its normalised eigenvectors. This is the last topic in F3 and pulls together eigenvalues, eigenvectors, transposes and inverses.

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Orthogonal matrices

A square matrix P is orthogonal if PT P = I, so P-1 = PT. Its columns are unit vectors that are mutually perpendicular (orthonormal). Rotations and reflections have orthogonal matrices.

Symmetric matrices

If AT = A then:

So if you normalise the eigenvectors and use them as the columns of P, P is orthogonal.

The diagonal form

With P = (normalised eigenvectors as columns), PT A P = D, where D is diagonal with the eigenvalues on the diagonal in the same order as the columns of P.

Method: find the eigenvalues, find an eigenvector for each, check they are perpendicular, normalise them, write down P and D. To find a third eigenvector of a 3 × 3 symmetric matrix quickly, take the vector product of the other two.

Worked example

Find an orthogonal matrix P and a diagonal matrix D with PT A P = D, where A has rows (2, 1), (1, 2).

  1. λ2 - 4λ + 3 = 0, so λ = 3 or 1
  2. λ = 3: (1, 1); λ = 1: (1, -1); these are perpendicular
  3. Normalise: (1/sqrt(2))(1, 1) and (1/sqrt(2))(1, -1)

Answer: P = (1/sqrt(2)) × rows (1, 1), (1, -1); D = diag(3, 1)

Key idea: For symmetric A: eigenvectors for different eigenvalues are perpendicular. Normalise them, make them the columns of P, then PT A P = D with eigenvalues in matching order.

Check yourself

  1. Why can PT be used in place of P-1 when diagonalising a symmetric matrix?
    • P is symmetric
    • P has determinant 0
    • P is orthogonal, because its columns are perpendicular unit vectors
    • P is diagonal
    Show answer

    P is orthogonal, because its columns are perpendicular unit vectors. Orthonormal columns give PT P = I, so PT = P-1.

  2. A has rows (1, 1, 0), (1, 0, 1), (0, 1, 1). Which is an eigenvector for the eigenvalue 1?
    • (1, 0, -1)
    • (1, 1, 1)
    • (1, -2, 1)
    • (1, 0, 1)
    Show answer

    (1, 0, -1). A(1, 0, -1) = (1, 1 - 1, -1) = (1, 0, -1).

  3. The same A has eigenvectors (1, 1, 1) and (1, 0, -1). The third eigenvector is perpendicular to both and has the form (1, k, 1). Find k.
    Show answer

    -2. (1, 1, 1) × (1, 0, -1) = (-1, 2, -1), a multiple of (1, -2, 1). So k = -2. Check: A(1, -2, 1) = (-1, 2, -1), eigenvalue -1.

  4. Normalise (1, 1, 1). Give the first component to 3 decimal places.
    Show answer

    0.577. Length sqrt(3), so each component is 1/sqrt(3) = 0.577.

  5. For this A, P has columns (1/sqrt(3))(1, 1, 1), (1/sqrt(2))(1, 0, -1) and (1/sqrt(6))(1, -2, 1) in that order. Find the entry in row 3, column 3 of D = PT A P.
    Show answer

    -1. D has the eigenvalues in the same order as the columns of P: 2, 1, -1. So the (3, 3) entry is -1.

  6. If you swap two columns of P, what happens to D = PT A P?
    • D is unchanged
    • The matching diagonal entries of D swap
    • D is no longer diagonal
    • D changes sign
    Show answer

    The matching diagonal entries of D swap. The eigenvalues in D appear in the same order as the eigenvectors in P.

  7. Find the sum of the diagonal entries of D for this A.
    Show answer

    2. D = diag(2, 1, -1), sum 2. This equals the trace of A: 1 + 0 + 1 = 2.

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Last updated 8 November 2026. Spotted a mistake? Tell us.