2.2 - Using the characteristic equation

2.2 - Using the characteristic equation

A matrix satisfies its own characteristic equation, with the identity taking the place of scalar constant terms. You will turn that fact into a working relation, reduce high powers to lower ones, and find an inverse without constructing cofactors.

A matrix satisfies its equation

Could an equation for scalar eigenvalues also simplify the matrix itself? For B=(3113)\mathbf B=\begin{pmatrix}3&1\\1&3\end{pmatrix}, the characteristic equation is λ2−6λ+8=0\lambda^2-6\lambda+8=0. Direct multiplication gives B2=(106610)=6B−8I\mathbf B^2=\begin{pmatrix}10&6\\6&10\end{pmatrix}=6\mathbf B-8\mathbf I. The matrix obeys the same coefficients.

Cayley–Hamilton relation

A square matrix satisfies its own characteristic equation. Thus, if a two-by-two matrix has characteristic equation λ2+sλ+t=0\lambda^2+s\lambda+t=0, then

A2+sA+tI=0.\mathbf A^2+s\mathbf A+t\mathbf I=\mathbf0.

For a three-by-three matrix with equation λ3+sλ2+tλ+u=0\lambda^3+s\lambda^2+t\lambda+u=0,

A3+sA2+tA+uI=0.\mathbf A^3+s\mathbf A^2+t\mathbf A+u\mathbf I=\mathbf0.

Here 0\mathbf0 is the zero matrix of the appropriate size. Not in the formula booklet: learn it.

Why does this hold in the diagonalised cases? If pp is the characteristic polynomial and A=QDQ−1\mathbf A=\mathbf Q\mathbf D\mathbf Q^{-1}, then p(A)=Qp(D)Q−1p(\mathbf A)=\mathbf Qp(\mathbf D)\mathbf Q^{-1}. Each diagonal entry of p(D)p(\mathbf D) is p(λi)=0p(\lambda_i)=0, so the whole product is zero. The Cayley–Hamilton fact also holds beyond distinct-eigenvalue cases; this argument explains the cases already studied.

Worked example 1: replacing a scalar constant by an identity

Question. Find the characteristic equation of B=(3113)\mathbf B=\begin{pmatrix}3&1\\1&3\end{pmatrix}, write its matrix relation, and verify it directly.

Plan. Expand the characteristic determinant, replace powers of λ\lambda by powers of B\mathbf B, and use I\mathbf I for the constant term.

det⁡(λI−B)=(λ−3)2−1=λ2−6λ+8.\det(\lambda\mathbf I-\mathbf B)=(\lambda-3)^2-1 =\lambda^2-6\lambda+8.

Therefore B2−6B+8I=0\mathbf B^2-6\mathbf B+8\mathbf I=\mathbf0. Directly,

(106610)−(186618)+(8008)=(0000).\begin{pmatrix}10&6\\6&10\end{pmatrix} -\begin{pmatrix}18&6\\6&18\end{pmatrix} +\begin{pmatrix}8&0\\0&8\end{pmatrix} =\begin{pmatrix}0&0\\0&0\end{pmatrix}.

Answer. The characteristic equation is λ2−6λ+8=0\lambda^2-6\lambda+8=0, and its matrix relation is verified.

Check. The eigenvectors (1,1)T(1,1)^{\mathsf T}, (1,−1)T(1,-1)^{\mathsf T} have eigenvalues 4,24,2. Both scalars satisfy λ2−6λ+8=0\lambda^2-6\lambda+8=0, so the relation sends each basis vector to zero, independently checking the whole matrix identity.

Increase the dimension to three. The only change in the substitution rule is the extra power; the constant still multiplies the identity.

Worked example 2: a cubic matrix relation

Question. A matrix A=(310130105)\mathbf A=\begin{pmatrix}3&1&0\\1&3&0\\1&0&5\end{pmatrix} has characteristic equation (λ−2)(λ−4)(λ−5)=0(\lambda-2)(\lambda-4)(\lambda-5)=0. Write and verify its matrix relation.

Plan. Expand the cubic first, then compare its matrix terms using direct products.

The expansion is λ3−11λ2+38λ−40=0\lambda^3-11\lambda^2+38\lambda-40=0. Hence

A3−11A2+38A−40I=0.\mathbf A^3-11\mathbf A^2+38\mathbf A-40\mathbf I=\mathbf0.

Multiplication gives

A2=(106061008125),A3=(36280283605011125).\mathbf A^2=\begin{pmatrix}10&6&0\\6&10&0\\8&1&25\end{pmatrix},\qquad \mathbf A^3=\begin{pmatrix}36&28&0\\28&36&0\\50&11&125\end{pmatrix}.

For example, the bottom row of the relation is
(50,11,125)−11(8,1,25)+38(1,0,5)−(0,0,40)=(0,0,0)(50,11,125)-11(8,1,25)+38(1,0,5)-(0,0,40)=(0,0,0).
The other rows cancel in the same direct subtraction.

Answer. A3−11A2+38A−40I=0\mathbf A^3-11\mathbf A^2+38\mathbf A-40\mathbf I=\mathbf0, verified by the displayed products.

Check. Independently, the matrix's eigenvalues are 2,4,52,4,5, and substitution gives 8−44+76−40=08-44+76-40=0, 64−176+152−40=064-176+152-40=0, 125−275+190−40=0125-275+190-40=0. Its three independent eigenvectors therefore all give zero under the relation.

The identity does not mean that A\mathbf A equals a scalar root. It means a particular polynomial combination of matrices is the zero matrix.

Watch out

Writing A2−6A+8=0\mathbf A^2-6\mathbf A+8=0 leaves a scalar mixed into a matrix sum. Write 8I8\mathbf I, which adds 88 to each diagonal entry and zero elsewhere.

Translate another characteristic equation accurately.

Reducing higher powers

A characteristic relation replaces the highest power by lower ones. Multiply by the matrix to reach the next power, then substitute the relation again. A two-by-two relation reduces every positive power to a combination of A\mathbf A and I\mathbf I; a three-by-three relation reduces it to a combination of A2,A,I\mathbf A^2,\mathbf A,\mathbf I.

Only powers of the same matrix and scalar multiples of its identity are involved here, so these factors commute. That does not license rearranging arbitrary different matrices.

Worked example 3: reducing fourth powers in two dimensions

Question. Given B2=6B−8I\mathbf B^2=6\mathbf B-8\mathbf I, express B4\mathbf B^4 as a linear combination of B\mathbf B and I\mathbf I.

Plan. Find the third power, substitute for B2\mathbf B^2, and repeat once.

B3=6B2−8B=6(6B−8I)−8B=28B−48IB4=28B2−48B=28(6B−8I)−48B=120B−224I.\begin{aligned} \mathbf B^3&=6\mathbf B^2-8\mathbf B =6(6\mathbf B-8\mathbf I)-8\mathbf B =28\mathbf B-48\mathbf I\\ \mathbf B^4&=28\mathbf B^2-48\mathbf B =28(6\mathbf B-8\mathbf I)-48\mathbf B =120\mathbf B-224\mathbf I. \end{aligned}

Answer. B4=120B−224I\mathbf B^4=120\mathbf B-224\mathbf I.

Check. For the concrete B=(3113)\mathbf B=\begin{pmatrix}3&1\\1&3\end{pmatrix}, squaring B2=(106610)\mathbf B^2=\begin{pmatrix}10&6\\6&10\end{pmatrix} gives (136120120136)\begin{pmatrix}136&120\\120&136\end{pmatrix}, exactly 120B−224I120\mathbf B-224\mathbf I.

Now increase to a cubic relation. Keep the quadratic term instead of trying to eliminate it using an unrelated rule.

Continue this lesson.

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