Answer to Question #283916 in Linear Algebra for Chang

Question #283916

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




1
Expert's answer
2022-02-02T08:11:09-0500

Transforming matrix "t" is given by




"t=\\begin{pmatrix}\n 1&2&3 \\\\\n 0&4&5 \\\\\n0&0&6\n\\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}\n 4 & 5 \\\\\n 0& 6\n\\end{vmatrix}+\\begin{vmatrix}\n 1& 3\\\\\n 0& 6\n\\end{vmatrix}+\\begin{vmatrix}\n 1& 2\\\\\n 0& 4\n\\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}\n 0&2&3 \\\\\n 0&3&5\\\\\n0&0&5\n\\end{pmatrix}\\>"



"T-4=\\begin{pmatrix}\n -3& 2&3 \\\\\n 0&0 & 5\\\\\n0&0&2\n\\end{pmatrix}"




"T-6=\\begin{pmatrix}\n -5&2& 3\\\\\n 0&-2& 5\\\\\n0&0&0\n\\end{pmatrix}"




"\\begin{pmatrix}\n 0&2 & 3\\\\\n 0&3& 5\\\\\n0&0&5\n\\end{pmatrix}\\begin{pmatrix}\n -3& 2&3\\\\\n 0&0&5\\\\\n0&0&2\n\\end{pmatrix}\\begin{pmatrix}\n -5&2&3 \\\\\n 0&-2&5\\\\\n0&0&0\n\\end{pmatrix}=\\begin{pmatrix}\n 0&0&0\\\\\n 0&0& 0\\\\\n0&0&0\n\\end{pmatrix}"



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


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

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