Question

{F} LA. find the minimal polynomial of the linear operator t : R³ "- R³" define by t (x,y,z) =(x+2y+3z, 4y+5z,6 z).is t

Answer 1

Transforming matrix $t$ is given by

$$t=\begin{pmatrix} 1&2&3 \\ 0&4&5 \\ 0&0&6 \end{pmatrix}$$

Characteristic polynomial of $t$ is of form

$\lambda^3-D_1\lambda^2+D_2\lambda-D_3=0$

Where $D_1=$ Sum of main diagonal element

$=1+4+6=11$

$D_2=$ Sum of minors of the main diagonal element

$D_2= \begin{vmatrix} 4 & 5 \\ 0& 6 \end{vmatrix}+\begin{vmatrix} 1& 3\\ 0& 6 \end{vmatrix}+\begin{vmatrix} 1& 2\\ 0& 4 \end{vmatrix}$

$=24+6+4=34$

$D_3=det\>|t|=1(24-0)+2(0-0)+3(0-0)$

$=24$

Characteristic polynomial:

$\lambda^3-11\lambda^2+34\lambda-24=(\lambda-1)(\lambda-4)(\lambda-6)$

The minimum polynomial will have roots

$1,4\>and\>6$

Putting the matrix in the characteristic polynomial

$T-1=\begin{pmatrix} 0&2&3 \\ 0&3&5\\ 0&0&5 \end{pmatrix}\>$

$T-4=\begin{pmatrix} -3& 2&3 \\ 0&0 & 5\\ 0&0&2 \end{pmatrix}$

$T-6=\begin{pmatrix} -5&2& 3\\ 0&-2& 5\\ 0&0&0 \end{pmatrix}$

$\begin{pmatrix} 0&2 & 3\\ 0&3& 5\\ 0&0&5 \end{pmatrix}\begin{pmatrix} -3& 2&3\\ 0&0&5\\ 0&0&2 \end{pmatrix}\begin{pmatrix} -5&2&3 \\ 0&-2&5\\ 0&0&0 \end{pmatrix}=\begin{pmatrix} 0&0&0\\ 0&0& 0\\ 0&0&0 \end{pmatrix}$

$\therefore$ The minimal polynomial of $t$ is

$M_tx=(x-1)(x-4)(x-6)$