91Ó°ÊÓ

Let \(A=\left[\begin{array}{lll}4 & -2 & 2 \\ 6 & -3 & 4 \\ 3 & -2 & 3\end{array}\right]\) and \(B=\left[\begin{array}{lll}3 & -2 & 2 \\ 4 & -4 & 6 \\\ 2 & -3 & 5\end{array}\right] .\) The characteristic polynomial of both matrices is \(\Delta(t)=(t-2)(t-1)^{2} .\) Find the minimal polynomial \(m(t)\) of each matrix. The minimal polynomial \(m(t)\) must divide \(\Delta(t) .\) Also, each factor of \(\Delta(t)\) (i.e., \(t-2\) and \(t-1)\) must also be a factor of \(m(t) .\) Thus, \(m(t)\) must be exactly one of the following: $$ f(t)=(t-2)(t-1) \quad \text { or } \quad g(t)=(t-2)(t-1)^{2} $$ (a) By the Cayley-Hamilton theorem, \(g(A)=\Delta(A)=0\), so we need only test \(f(t) .\) We have $$ f(A)=(A-2 I)(A-I)=\left[\begin{array}{lll} 2 & -2 & 2 \\ 6 & -5 & 4 \\ 3 & -2 & 1 \end{array}\right]\left[\begin{array}{lll} 3 & -2 & 2 \\ 6 & -4 & 4 \\ 3 & -2 & 2 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $$ Thus, \(m(t)=f(t)=(t-2)(t-1)=t^{2}-3 t+2\) is the minimal polynomial of \(A\). (b) Again \(g(B)=\Delta(B)=0\), so we need only test \(f(t)\). We get $$ f(B)=(B-2 I)(B-I)=\left[\begin{array}{lll} 1 & -2 & 2 \\ 4 & -6 & 6 \\ 2 & -3 & 3 \end{array}\right]\left[\begin{array}{lll} 2 & -2 & 2 \\ 4 & -5 & 6 \\ 2 & -3 & 4 \end{array}\right]=\left[\begin{array}{lll} -2 & 2 & -2 \\ -4 & 4 & -4 \\ -2 & 2 & -2 \end{array}\right] \neq 0 $$ Thus, \(m(t) \neq f(t) .\) Accordingly, \(m(t)=g(t)=(t-2)(t-1)^{2}\) is the minimal polynomial of \(B .[\mathrm{We}\) emphasize that we do not need to compute \(g(B)\); we know \(g(B)=0\) from the Cayley-Hamilton theorem.]

Step by step solution

01

Write down the possible minimal polynomials

We have two possible minimal polynomials, which are the factors of the characteristic polynomial: $$ f(t)=(t-2)(t-1) \quad \text { or } \quad g(t)=(t-2)(t-1)^{2} $$

02

Test the first minimal polynomial for matrix A

Calculate \(f(A)\) and check if it equals the zero matrix: $$ f(A)=(A-2 I)(A-I)=\left[\begin{array}{lll} 2 & -2 & 2 \\\ 6 & -5 & 4 \\\ 3 & -2 & 1 \end{array}\right]\left[\begin{array}{lll} 3 & -2 & 2 \\\ 6 & -4 & 4 \\\ 3 & -2 & 2 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \end{array}\right] $$ Thus, \(m(t)=f(t)=(t-2)(t-1)=t^{2}-3 t+2\) is the minimal polynomial of \(A\).

03

Test the first minimal polynomial for matrix B

Calculate \(f(B)\) and check if it equals the zero matrix: $$ f(B)=(B-2 I)(B-I)=\left[\begin{array}{lll} 1 & -2 & 2 \\\ 4 & -6 & 6 \\\ 2 & -3 & 3 \end{array}\right]\left[\begin{array}{lll} 2 & -2 & 2 \\\ 4 & -5 & 6 \\\ 2 & -3 & 4 \end{array}\right]=\left[\begin{array}{lll} -2 & 2 & -2 \\\ -4 & 4 & -4 \\\ -2 & 2 & -2 \end{array}\right] \neq 0 $$ Since \(f(B) \neq 0\), \(m(t) \neq f(t) .\)

04

Determine the minimal polynomial for matrix B

As \(f(B) \neq 0\) and \(g(B)=\Delta(B)=0\) by the Cayley-Hamilton theorem, the minimal polynomial for matrix B is \(g(t)\): $$ m(t)=g(t)=(t-2)(t-1)^{2} $$ Therefore, the minimal polynomial of \(A\) is \(t^{2}-3 t+2\) and the minimal polynomial of \(B\) is \((t-2)(t-1)^{2}\).

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