Question

In: Advanced Math

Jordan Canonical Form

Let AM6(R) A\in M_6(\mathbb{R}) be an invertible matrix satisfies A34A2+3A=0 A^3-4A^2 + 3A = 0 and tr(A)=8. tr(A) = 8. Find the characteristics polynomial of A.

 

Solutions

Expert Solution

Solution

AM6(R)    deg(Pλ) A\in M_6(\mathbb{R}) \implies deg(P\lambda)

we have A34A2+3A=0 \hspace{2mm}A^3-4A^2+3A=0

    A(A24A+3I)=0 \iff A\bigg(A^2-4A+3I\bigg)=0

    (AI)(A3I)=0 \iff\bigg(A-I\bigg)\bigg(A-3I\bigg)=0

Let f(λ)=(λI)(λ3I) f(\lambda)=\bigg(\lambda-I\bigg)\bigg(\lambda-3I\bigg) Then f(A)=0 f(A)=0

    P(λ)=(λ1)n1(λ3)n2 \implies P(\lambda)=\bigg(\lambda-1\bigg)^{n_1}\bigg(\lambda-3\bigg)^{n_2} \hspace{2mm} where n1+n2=6(1) n_1+n_2=6 \hspace{2mm}(1)

we have tr(A)=8    n1(1)+n2(3)=8 tr(A)=8\iff n_1(1)+n_2(3)=8     n1+3n2=8 \iff n_1+3n_2=8

from.(1) {n1+n2=6n1+3n2=8    2n2=2    n2=1,n1=5 \begin{cases} n_1+n_2=6 & \quad\\ n_1+3n_2=8 & \quad \end{cases} \implies 2n_2=2\implies n_2=1,n_1=5

 


Answer

Therefore. P(λ)=(λ1)5(λ3) P(\lambda)=\bigg(\lambda-1\bigg)^5\bigg(\lambda-3\bigg)

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