Matrix theory—Cayley-Hamilton theorem

 

Cayley-Hamilton theorem content

Typical examples of Cayley-Hamilton theorem 

Typical examples

        Let's first look at this question. The question requires A^{100}+2A^{50}that if matrix A is directly substituted into the calculation, it will be very complicated. Therefore, this path is impossible. We try to introduce the Cayley-Hamilton theorem
        we introduced today to solve this problem. Order , if we ask , just ask . Next we determine the characteristic polynomial of matrix A
g(\lambda )=\lambda^{100}+2\lambda^{50}A^{100}+2A^{50}g(A )
\phi (\lambda)
\phi(\lambda)=det(\lambda*IA)=\begin{vmatrix}\lambda-1&-1&1\\-1&\lambda-1&-1\\0&1&\lambda-2\ end{vmatrix}=\begin{vmatrix} \lambda & -\lambda& 2\\ -1& \lambda-1 &-1 \\ 0&1 &\lambda-2 \end{vmatrix}
=\lambda*(-1)^{1+1}\begin{vmatrix} \lambda-1&-1 \\ 1& \lambda-2 \\ \end{vmatrix}+(-1)*(-1)^{2+1}\begin{vmatrix} -\lambda &2 \\ 1& \lambda-2 \end{vmatrix}
=\lambda((\lambda-1)(\lambda-2)+1)+(-\lambda(\lambda-2)-2)
=\lambda(\lambda^{2}-3\lambda+3)-\lambda^{2}+2\lambda-2
=\lambda^{3}-4\lambda^{2}+5\lambda-2=\lambda^{3}-\lambda^{2}-3\lambda^{2}+5\lambda-2=\lambda^{2}(\lambda-1)-(3\lambda^{2}-5\lambda+2)
=\lambda^2(\lambda-1)-(\lambda-1)(3\lambda-2) =(\lambda-1)[\lambda^2-3\lambda+2]=(\lambda-1)^2(\lambda-2)

Assume g(\lambda )=\phi (\lambda)q(\lambda)+b_0+b_1\lambda+b_2\lambda^2, next determine the coefficients b_0,b_1,b_2:


Putting \lambda=1,2respectively into the above formula, we have:
g(1 )=\phi (1)q(1)+b_0+b_11+b_21^2 =b_0+b_1+b_2=3                          (1)
g(2)=\phi (2)q(2)+b_0+b_12+b_22^2 =b_0+b_12+b_24=2^{100}+2^{51}      (2)
To g(\lambda )find \lambdathe differential about, we have:
g^{'}(\lambda ) = [2(\lambda-1)(\lambda-2)+(\lambda-1)^2]q(\lambda)+ \phi(\lambda)q^{'}(\lambda)+b_1+2b_2\lambda
Bringing \lambda=1in g^{'}(\lambda ), we have:
g^{'}(1)=b_1+2b_2=200                                                                                   (3)

Solving (1), (2), and (3) simultaneously, we get:
b_0=2^{100}+2^{51}-400
b_1=606-2^{101}-2^{52}
b_2=-203+2^{100}+2^{51}

So there is:
A^{100}+2A^{50}=g(A)=b_0I+b_1A+b_2A^2

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Origin blog.csdn.net/m0_48241022/article/details/133044250