512 views
1 votes
1 votes

1 Answer

Best answer
1 votes
1 votes

ghGiven that  $A=\begin{bmatrix} 3&3 &0 \\ 3& 3& 0\\0 &0 &6 \end{bmatrix}$

We can write characteristic equation $|A-\lambda I|=0$

                                                        $\begin{vmatrix} 3-\lambda&3&0 \\ 3&3-\lambda &0 \\ 0& 0& 6-\lambda\end{vmatrix}=0$

$(3-\lambda)[(3-\lambda)(6-\lambda)-0]-3[3(6-\lambda)-0]+0=0$

 $(3-\lambda)(3-\lambda)(6-\lambda)-9(6-\lambda)=0$

 $(3-\lambda)^{2}(6-\lambda)-9(6-\lambda)=0$

 $(6-\lambda)[(3-\lambda)^{2}-9]=0$

 $(\lambda-6)[(\lambda-3)^{2}-9]=0$

 $(\lambda-6)[\lambda^{2}+9-6\lambda-9]=0$

 $(\lambda-6)[\lambda^{2}-6\lambda]=0$

   $\lambda(\lambda-6)(\lambda-6)=0$

  So, we got characteristic roots  $\lambda=0,6,6$

Now,$f(x)=x(x-6)^{2}$

   So$,g(x)=x(x-6)$

By Cayley Hamilton theorem:-Every square matrix satisfy its characteristic equation.

        $f(x)=0$

            $(or)$

           $f(A)=0$

Lets verify $g(A)$ is zero or not$?$

    $g(x)=x(x-6)=x^{2}-6x$

     $g(A)=A^{2}-6A$

    $g(A)=\begin{bmatrix} 18&18 &0 \\18 & 18& 0\\0 &0 &36 \end{bmatrix}-6\begin{bmatrix} 3&3 &0 \\ 3& 3& 0\\0 &0 &6 \end{bmatrix}$

    $g(A)=\begin{bmatrix} 18&18 &0 \\18 & 18& 0\\0 &0 &36 \end{bmatrix}-\begin{bmatrix} 18&18 &0 \\ 18& 18& 0\\0 &0 &36 \end{bmatrix}$

    $g(A)=0$

    $A^{2}-6A=0$

also $h(x)=x$  $;A\neq0$

        $p(x)=(x-6)$  $;A-6\neq0$

g(x) is the monic polynomial of lowest degree such that $g(x)=0$

So,$g(x)=x(x-6)$ is the minimal polynomial.

edited by

Related questions

0 votes
0 votes
0 answers
1
Balaji Jegan asked Dec 11, 2018
292 views
0 votes
0 votes
0 answers
2
Balaji Jegan asked Dec 11, 2018
213 views
1 votes
1 votes
1 answer
3
Balaji Jegan asked Dec 11, 2018
503 views
Let $J$ be a $2\times 2$ complex matrix such that $trace(J)=1$ and $det(J)=6.$ Then, $trace(J^4 – J^3)$ is ___________
1 votes
1 votes
1 answer
4
Balaji Jegan asked Dec 11, 2018
3,551 views