Jordan Canonical Form

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

Solutions

Expert Solution

Solution

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

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

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

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

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

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

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

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

Answer

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

SUN BUNRA answered 2 years ago

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