Question

Prove Caley-Hamilton Theorem​

Answer 1

Answer:

In linear algebra, the Cayley–Hamilton theorem states that every square matrix over a commutative ring satisfies its own characteristic equation.

Answer 2

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Answer:

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Proof: Let

• p(λ) = p0 + p1λ + ... + pn-1λn-1 + pnλn.

Let Q(λ) be the adjugate matrix of the square matrix A - λI, which may be considered as a polynomial in λ and with matrix coefficients

Q(λ) = Q0 + λQ1 + ... + λq-1Qq-1 + λqQq, where Qq are constant matrices.

By the formula (adjA)A = (detA)I, we have

Q(λ)(A - λI) = p(λ)I = p0I + p1λI + ... + pn-1λn-1I + pnλnI.

On the other hand, we have

Q(λ)(A - λI) = Q0A + λ(Q1A - Q0) + ... + λq(QqA - Qq-1) - λq+1Qq.

Thus we get q = n -1 and

Q0A = p0I,

Q0A = p0I, Q1A - Q0 = p1I,

Q0A = p0I, Q1A - Q0 = p1I, :

Q0A = p0I, Q1A - Q0 = p1I, :Qn-1A - Qn-2 = pn-1I,

Q0A = p0I, Q1A - Q0 = p1I, :Qn-1A - Qn-2 = pn-1I, - Qn-1 = pnI.

Multiplying powers of A on the right sides, we get

Q0A = p0I,

Q0A = p0I, Q1A2 - Q0A = p1A,

Q0A = p0I, Q1A2 - Q0A = p1A, :

Q0A = p0I, Q1A2 - Q0A = p1A, :Qn-1An - Qn-2An-1 = pn-1An-1,

Q0A = p0I, Q1A2 - Q0A = p1A, :Qn-1An - Qn-2An-1 = pn-1An-1, - Qn-1An = pnAn.

Adding all the equalities together, we get

p(A) = p0I + p1A + ... + pn-1An-1 + pnAn = 0.

By this discussion, the characteristic polynmomial of a 2 by 2 matrix A is

det(A - λI) = detA - (trA)λ + λ2.

Therefore, the Cayley-Hamilton Theorem tells us:

• (detA)I - (trA)A + A2 = O

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