Answer to Question #283601 in Linear Algebra for Dipu

Question #283601

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

1
Expert's answer
2022-02-02T08:20:11-0500

Transforming matrix tt is given by




t=(123045006)t=\begin{pmatrix} 1&2&3 \\ 0&4&5 \\ 0&0&6 \end{pmatrix}


Characteristic polynomial of tt is of form


λ3D1λ2+D2λD3=0\lambda^3-D_1\lambda^2+D_2\lambda-D_3=0



Where D1=D_1= Sum of main diagonal element

=1+4+6=11=1+4+6=11


D2=D_2= Sum of minors of the main diagonal element


D2=4506+1306+1204D_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=24+6+4=34




D3=dett=1(240)+2(00)+3(00)D_3=det\>|t|=1(24-0)+2(0-0)+3(0-0)


=24=24




Characteristic polynomial:

λ311λ2+34λ24=(λ1)(λ4)(λ6)\lambda^3-11\lambda^2+34\lambda-24=(\lambda-1)(\lambda-4)(\lambda-6)





The minimum polynomial will have roots

1,4and61,4\>and\>6


Putting the matrix in the characteristic polynomial


T1=(023035005)T-1=\begin{pmatrix} 0&2&3 \\ 0&3&5\\ 0&0&5 \end{pmatrix}\>



T4=(323005002)T-4=\begin{pmatrix} -3& 2&3 \\ 0&0 & 5\\ 0&0&2 \end{pmatrix}




T6=(523025000)T-6=\begin{pmatrix} -5&2& 3\\ 0&-2& 5\\ 0&0&0 \end{pmatrix}




(023035005)(323005002)(523025000)=(000000000)\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 tt is


Mtx=(x1)(x4)(x6)M_tx=(x-1)(x-4)(x-6)

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