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Algebra
Linear Algebra

Proof of the Cayley-Hamilton Theorem

A comprehensive proof that every square matrix satisfies its own characteristic polynomial.

AAdmin
July 6, 202620 views1 comments

Step 4: Equate Coefficients

From

(λI−A)(λn−1Bn−1+λn−2Bn−2+⋯+λB1+B0)=(λn+cn−1λn−1+⋯+c0)I,(\lambda I-A) \left( \lambda^{n-1}B_{n-1} +\lambda^{n-2}B_{n-2} +\cdots +\lambda B_1 +B_0 \right) = (\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_0)I,

comparing coefficients of powers of (\lambda) gives

λn:Bn−1=I,λn−1:Bn−2−ABn−1=cn−1I,λn−2:Bn−3−ABn−2=cn−2I,⋮λ1:B0−AB1=c1I,λ0:−AB0=c0I.\begin{aligned} \lambda^n &: &&B_{n-1}=I,\\ \lambda^{n-1} &: &&B_{n-2}-AB_{n-1}=c_{n-1}I,\\ \lambda^{n-2} &: &&B_{n-3}-AB_{n-2}=c_{n-2}I,\\ & &&\vdots\\ \lambda^1 &: &&B_0-AB_1=c_1I,\\ \lambda^0 &: &&-AB_0=c_0I. \end{aligned}

Step 5: Evaluate at (A)

Multiply the equations respectively by (A^n,A^{n-1},\ldots,A,I). This gives

AnBn−1=An,An−1Bn−2−AnBn−1=cn−1An−1,An−2Bn−3−An−1Bn−2=cn−2An−2,  ⋮AB0−A2B1=c1A,−AB0=c0I.\begin{aligned} A^nB_{n-1} &= A^n,\\ A^{n-1}B_{n-2}-A^nB_{n-1} &=c_{n-1}A^{n-1},\\ A^{n-2}B_{n-3}-A^{n-1}B_{n-2} &=c_{n-2}A^{n-2},\\ &\ \ \vdots\\ AB_0-A^2B_1 &=c_1A,\\ -AB_0 &=c_0I. \end{aligned}

Adding all these equations, every term involving the matrices (B_i) cancels:

AnBn−1−AnBn−1+An−1Bn−2−An−1Bn−2+⋯+AB0−AB0=0.A^nB_{n-1} -A^nB_{n-1} +A^{n-1}B_{n-2} -A^{n-1}B_{n-2} +\cdots +AB_0 -AB_0 =0.

Hence,

An+cn−1An−1+cn−2An−2+⋯+c1A+c0I=0,A^n +c_{n-1}A^{n-1} +c_{n-2}A^{n-2} +\cdots +c_1A +c_0I =0,

that is,

pA(A)=0.p_A(A)=0.

This completes the proof.

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A
Admin1h ago(edited)

first proof in the website