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