Question #286738

Let T: R^3 to R^3 be the linear transformation defined by


T(x,y,z)= (-x,x-y, 3x+2y+z)


Check whether T satifies the polynomial (x-1)(x+1)^2. Also the find of minimal polynomial of T.



1
Expert's answer
2022-01-12T18:18:21-0500



T=x(113)+y(012)+z(001)T=x\begin{pmatrix} -1 \\ 1\\ 3 \end{pmatrix}+y\begin{pmatrix} 0 \\ -1\\ 2 \end{pmatrix}+z\begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}


T=(100110321)T=\begin{pmatrix} -1&0 & 0 \\ 1&-1 & 0\\ 3&2&1 \end{pmatrix}




T1=(200120320)T-1=\begin{pmatrix} -2&0& 0 \\ 1&-2 & 0\\ 3&2&0 \end{pmatrix}



(T+1)2=(000100322)2=(000000844)(T+1)^2=\begin{pmatrix} 0&0&0 \\ 1&0&0\\ 3&2&2 \end{pmatrix}^2=\begin{pmatrix} 0&0&0\\ 0&0& 0\\ 8&4&4 \end{pmatrix}



(T1)(T+1)2=(200120320)(000000844)=(000000000)(T-1)(T+1)^2=\begin{pmatrix} -2&0& 0 \\ 1&-2 & 0\\ 3&2&0 \end{pmatrix}\begin{pmatrix} 0&0&0 \\ 0&0 & 0\\ 8&4&4 \end{pmatrix}=\begin{pmatrix} 0&0& 0 \\ 0&0& 0\\ 0&0&0 \end{pmatrix}




T satisfies the polynomial (x1)(x+1)2(x-1)(x+1)^2


(T1)(T+1)=(200120320)(000100322)=(000200200)(000000000)(T-1)(T+1)=\begin{pmatrix} -2&0&0 \\ 1&-2 & 0\\ 3&2&0 \end{pmatrix}\begin{pmatrix} 0&0& 0\\ 1&0&0 \\ 3&2&2 \end{pmatrix}=\begin{pmatrix} 0&0 & 0\\ -2&0&0\\ 2&0&0 \end{pmatrix}\ne\begin{pmatrix} 0&0 & 0 \\ 0&0& 0\\ 0&0&0 \end{pmatrix}




\therefore Minimal polynomial is (x1)(x+1)2(x-1)(x+1)^2


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