Maths › Further Pure 2 › Diagonalisation and the Cayley-Hamilton theorem
Diagonalisation and the Cayley-Hamilton theorem
Change coordinates so the eigenvectors become the axes and the matrix turns diagonal, making high powers trivial. And every matrix, it turns out, satisfies its own characteristic equation.
Builds on Eigenvalues and eigenvectors and Determinants and inverses.
IN THIS TOPIC
- Build P from eigenvectors and D from eigenvalues so that P⁻¹MP = D.
- Use diagonalisation to compute high powers of a matrix.
- Diagonalise a symmetric matrix orthogonally, with P⁻¹ equal to the transpose.
- Apply the Cayley-Hamilton theorem to find powers and inverses.
COMMON MISCONCEPTION
Finding the tenth power of a matrix means multiplying it out ten times.
New axes, diagonal matrix
Set the eigenvectors as the columns of P and the matching eigenvalues down the diagonal of D. Then MP = PD, and P is invertible whenever the eigenvectors are independent.
In eigenvector coordinates the transformation is a stretch along each axis and nothing more. Powers become cheap, because Mn = PDnP-1 and a diagonal matrix takes its power entry by entry. Ten multiplications become two, whatever the exponent happens to be.
WORKED EXAMPLE
A tenth power without ten multiplications
For the matrix with rows (4, 1) and (2, 3), describe how to find M¹⁰.
The eigenvalues are 5 and 2, with eigenvectors (1, 1) and (1, −2), so P has columns (1, 1) and (1, −2), and D is diagonal with entries 5 and 2.
D¹⁰ is diagonal with 5¹⁰ = 9765625 and 2¹⁰ = 1024.
M¹⁰ = PD¹⁰P⁻¹, which is two multiplications and a pair of powers. The eigenvalue 5 dominates completely, so at large powers the matrix behaves as a stretch along y = x and very little else.
A real symmetric matrix always has an orthonormal set of eigenvectors, so P can be built orthogonal and P⁻¹ becomes the transpose of P. Eigenvectors belonging to different eigenvalues come out perpendicular on their own and only need normalising. A repeated eigenvalue is the case to watch, since its eigenvectors fill a whole plane and two chosen from it need not be perpendicular at all, so pick a perpendicular pair there, by Gram-Schmidt if the obvious choice fails, and normalise those. The inverse then costs nothing at all, and the whole manoeuvre is called orthogonal diagonalisation.
Every matrix solves its own equation
The Cayley-Hamilton theorem says a matrix satisfies its own characteristic equation. For a 2 × 2 with trace t and determinant d, that reads M² − tM + dI = 0. Rearrange it and you have M² written in terms of M and I, so every higher power collapses down to a combination of the two.
WORKED EXAMPLE
Two uses of one identity
For the matrix with rows (4, 1) and (2, 3), verify Cayley-Hamilton and use it to find M⁻¹.
M² has rows (18, 7) and (14, 11). Then M² − 7M + 10I has rows (18 − 28 + 10, 7 − 7) and (14 − 14, 11 − 21 + 10), which is the zero matrix, as promised.
Multiply the identity by M⁻¹: M − 7I + 10M⁻¹ = 0, so M⁻¹ = (7I − M)/10, which is one tenth of the matrix with rows (3, −1) and (−2, 4).
Multiplying M by that returns the identity. The inverse came out of the characteristic equation with no adjugate anywhere in sight.
GUIDED PRACTICE
Reducing a cube
A 2 × 2 matrix has trace 5 and determinant 6. Use Cayley-Hamilton to express M³ in terms of M and I.
Show the working
The theorem gives M² = 5M − 6I.
Multiplying by M: M³ = 5M² − 6M = 5(5M − 6I) − 6M = 19M − 30I.
Every power reduces to aM + bI for some numbers a and b, however high the exponent climbs.
ASSESSMENT FOCUS
- State P and D explicitly, with the eigenvector order matching the eigenvalue order. A mismatch invalidates everything after it.
- For symmetric matrices, make the eigenvectors perpendicular first and then normalise, so P is orthogonal and the inverse is the transpose.
- Quote Cayley-Hamilton by name, and show the characteristic equation it comes from.
- To get an inverse from the theorem, multiply the identity through by M inverse and rearrange.
CHECK YOURSELF
A 2 × 2 matrix M has trace 6 and determinant 5. Write down the equation Cayley-Hamilton gives, and hence M⁻¹ in terms of M and I.
Show a hint
The characteristic equation is λ² − 6λ + 5 = 0.
Show the answer
M² − 6M + 5I = 0. Multiplying by M⁻¹ gives M − 6I + 5M⁻¹ = 0, so M⁻¹ = (6I − M)/5.
P from eigenvectors, D from eigenvalues: P⁻¹MP = D, and Mⁿ = PDⁿP⁻¹ makes powers cheap.
For a symmetric matrix, an orthonormal set of eigenvectors makes P orthogonal and P⁻¹ is simply the transpose.
Cayley-Hamilton: a matrix satisfies its own characteristic equation, which yields inverses and collapses powers.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the diagonalisation and the cayley-hamilton theorem questions page.
CHECK YOUR PROGRESS
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- Build P from eigenvectors and D from eigenvalues so that P⁻¹MP = D.
- Use diagonalisation to compute high powers of a matrix.
- Diagonalise a symmetric matrix orthogonally, with P⁻¹ equal to the transpose.
- Apply the Cayley-Hamilton theorem to find powers and inverses.
Open the full revision checklist to see every objective in the course in one place.