The Cayley-Hamilton Theorem

The Cayley-Hamilton Theorem is a remarkable result that states every square matrix satisfies its own characteristic equation. It has powerful applications for finding matrix inverses, computing matrix powers, and expressing higher powers of a matrix in terms of lower ones.

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Statement of the Theorem

Let $A$A be an $n \times n$n×n matrix with characteristic polynomial:

$p(\lambda) = \det(A - \lambda I)$p(λ)=det(A−λI)

The Cayley-Hamilton Theorem states:

$p(A) = O$p(A)=O

That is, when you substitute the matrix $A$A into its own characteristic polynomial (replacing $\lambda$λ with $A$A and the constant term with that constant times $I$I), the result is the zero matrix.

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2×2 Case

For a 2×2 matrix $A$A, the characteristic equation is:

$\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$λ2−tr(A)λ+det(A)=0

The Cayley-Hamilton Theorem gives:

$A^2 - \text{tr}(A)\cdot A + \det(A)\cdot I = O$A2−tr(A)⋅A+det(A)⋅I=O

Worked Example 1: Verify the Cayley-Hamilton Theorem for $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$A=(13​24​).

$\text{tr}(A) = 5$tr(A)=5, $\det(A) = 4 - 6 = -2$det(A)=4−6=−2.

Characteristic equation: $\lambda^2 - 5\lambda - 2 = 0$λ2−5λ−2=0.

Cayley-Hamilton says: $A^2 - 5A - 2I = O$A2−5A−2I=O.

$A^2 = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix}$A2=(13​24​)(13​24​)=(715​1022​)

$5A = \begin{pmatrix} 5 & 10 \\ 15 & 20 \end{pmatrix}$5A=(515​1020​), $2I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$2I=(20​02​)

$A^2 - 5A - 2I = \begin{pmatrix} 7-5-2 & 10-10-0 \\ 15-15-0 & 22-20-2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$A2−5A−2I=(7−5−215−15−0​10−10−022−20−2​)=(00​00​) ✓

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3×3 Case

For a 3×3 matrix with characteristic polynomial $\lambda^3 + a\lambda^2 + b\lambda + c = 0$λ3+aλ2+bλ+c=0:

$A^3 + aA^2 + bA + cI = O$A3+aA2+bA+cI=O

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Applications

1. Finding the Inverse

From the 2×2 Cayley-Hamilton result: $A^2 - \text{tr}(A)\cdot A + \det(A)\cdot I = O$A2−tr(A)⋅A+det(A)⋅I=O.

If $\det(A) \neq 0$det(A)=0, rearrange:

$\det(A)\cdot I = -A^2 + \text{tr}(A)\cdot A = A(-A + \text{tr}(A)\cdot I)$det(A)⋅I=−A2+tr(A)⋅A=A(−A+tr(A)⋅I)

$I = A \cdot \frac{-A + \text{tr}(A)\cdot I}{\det(A)}$I=A⋅det(A)−A+tr(A)⋅I​

So: $A^{-1} = \frac{\text{tr}(A)\cdot I - A}{\det(A)}$A−1=det(A)tr(A)⋅I−A​

Worked Example 2: Use Cayley-Hamilton to find the inverse of $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$A=(13​24​).

$\text{tr}(A) = 5$tr(A)=5, $\det(A) = -2$det(A)=−2.

$A^{-1} = \frac{5I - A}{-2} = \frac{1}{-2}\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$A−1=−25I−A​=−21​(4−3​−21​)=(−23/2​1−1/2​)

Check: $AA^{-1} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$AA−1=(13​24​)(−23/2​1−1/2​)=(10​01​) ✓

2. Computing Higher Powers

The Cayley-Hamilton Theorem lets you express $A^2$A2 (for a 2×2 matrix) in terms of $A$A and $I$I:

$A^2 = \text{tr}(A)\cdot A - \det(A)\cdot I$A2=tr(A)⋅A−det(A)⋅I

Then $A^3 = A \cdot A^2 = \text{tr}(A) \cdot A^2 - \det(A) \cdot A$A3=A⋅A2=tr(A)⋅A2−det(A)⋅A, and you substitute again. This means every power of a 2×2 matrix can be expressed as a linear combination of $A$A and $I$I.

Worked Example 3: Express $A^3$A3 for $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$A=(13​24​) in terms of $A$A and $I$I.

$A^2 = 5A + 2I$A2=5A+2I (from Cayley-Hamilton)