Answer to Question #105618 in Abstract Algebra for Sourav Mondal

Question #105618
M3[ Z ]has no nilpotent elements.
1
Expert's answer
2020-03-16T13:55:48-0400

"M_3[Z]=\\{\\begin{pmatrix}\n a_{11} & a_{12}& a_{13} \\\\\n a_{21} & a_{22}& a_{23} \\\\\n a_{31} & a_{32}& a_{33} \\\\\n\\end{pmatrix},\\\\\na_{ij}\\in Z\\}"

Сonsider the basis

"E_{11}=\\begin{pmatrix}\n 1 &0&0 \\\\\n 0&0&0\\\\\n0&0&0\n\\end{pmatrix}, \nE_{12}=\\begin{pmatrix}\n 0 &1&0 \\\\\n 0&0&0\\\\\n0&0&0\n\\end{pmatrix}, \\\\\nE_{13}=\\begin{pmatrix}\n 0 &0&1 \\\\\n 0&0&0\\\\\n0&0&0\n\\end{pmatrix}, E_{21}=\\begin{pmatrix}\n 0&0&0 \\\\\n 1&0&0\\\\\n0&0&0\n\\end{pmatrix}, \\\\\nE_{22}=\\begin{pmatrix}\n 0 &0&0 \\\\\n 0&1&0\\\\\n0&0&0\n\\end{pmatrix}, E_{23}=\\begin{pmatrix}\n 0 &0&0 \\\\\n 0&0&1\\\\\n0&0&0\n\\end{pmatrix}, \\\\\nE_{31}=\\begin{pmatrix}\n 0 &0&0 \\\\\n 0&0&0\\\\\n1&0&0\n\\end{pmatrix}, \nE_{32}=\\begin{pmatrix}\n 0 &0&0 \\\\\n 0&0&0\\\\\n0&1&0\n\\end{pmatrix}, \\\\\nE_{33}=\\begin{pmatrix}\n 0 &0&0 \\\\\n 0&0&0\\\\\n0&0&1\n\\end{pmatrix}"


"A=\\begin{pmatrix}\n a_{11} & a_{12}& a_{13} \\\\\n a_{21} & a_{22}& a_{23} \\\\\n a_{31} & a_{32}& a_{33} \\\\\n\\end{pmatrix}=a_{11}E_{11}+a_{12}E_{12}+\\\\\n+a_{13}E_{13}+a_{21}E_{21}+a_{22}E_{22}+\\\\\n+a_{23}E_{23}+a_{31}E_{31}+a_{32}E_{32}+\\\\\n+a_{33}E_{33}"


Let's put together a Kelly table and  we denote

"O=\\begin{pmatrix}\n 0 & 0&0 \\\\\n 0 & 0&0\\\\\n0&0&0\n\\end{pmatrix}"

"\\begin{matrix}\n &E_{11} &E_{12}&E_{13}&E_{21}&E_{22}&E_{23}&E_{31}&E_{32}&E_{33}\\\\\n E_{11}& E_{11}&E_{12}&E_{13}&O&O&O&O&O&O\\\\\nE_{12}& O&O&O&E_{11}&E_{12}&E_{13}&O&O&O\\\\\nE_{13}& O&O&O&O&O&O&E_{11}&E_{12}&E_{13}\\\\\nE_{21}& E_{21}&E_{22}&E_{23}&O&O&O&O&O&O\\\\\nE_{22}& O&O&O&E_{21}&E_{22}&E_{23}&O&O&O\\\\\nE_{23}& O&O&O&O&O&O&E_{21}&E_{22}&E_{23}\\\\\nE_{31}& E_{31}&E_{32}&E_{33}&O&O&O&O&O&O\\\\\nE_{32}& O&O&O&E_{31}&E_{32}&E_{33}&O&O&O\\\\\nE_{33}& O&O&O&O&O&O&E_{31}&E_{32}&E_{33}\n\\end{matrix}"


Let's show that "A^2\\neq0"

"A^2=(a_{11}E_{11}+a_{12}E_{12}+\\\\\n+a_{13}E_{13}+a_{21}E_{21}+a_{22}E_{22}+\\\\\n+a_{23}E_{23}+a_{31}E_{31}+a_{32}E_{32}+\\\\\n+a_{33}E_{33})(a_{11}E_{11}+a_{12}E_{12}+\\\\\n+a_{13}E_{13}+a_{21}E_{21}+a_{22}E_{22}+\\\\\n+a_{23}E_{23}+a_{31}E_{31}+a_{32}E_{32}+\\\\\n+a_{33}E_{33})=\\\\\n=a_{11}^2E_{11}+a_{11}a_{12}E_{12}+\\\\\n+a_{11}a_{13}E_{13}+a_{12}a_{21}E_{11}+a_{12}a_{22}E_{12}+\\\\\n+a_{12}a_{23}E_{13}+a_{13}a_{31}E_{11}+a_{13}a_{32}E_{12}+\\\\\n+a_{13}a_{33}E_{13}+"


"+a_{21}a_{11}E_{21}+a_{21}a_{12}E_{22}+\\\\\n+a_{21}a_{13}E_{23}+a_{22}a_{21}E_{21}+a_{22}^2E_{22}+\\\\\n+a_{22}a_{23}E_{23}+a_{23}a_{31}E_{21}+a_{23}a_{32}E_{22}+\\\\\n+a_{23}a_{33}E_{23}+\\\\\n+a_{31}a_{11}E_{31}+a_{31}a_{12}E_{32}+\\\\\n+a_{31}a_{13}E_{33}+a_{32}a_{21}E_{31}+a_{32}a_{22}E_{32}+\\\\\n+a_{32}a_{23}E_{33}+a_{33}a_{31}E_{31}+a_{33}a_{32}E_{32}+\\\\\n+a_{33}^2E_{33}=\\\\"


"=(a_{11}^2+a_{12}a_{21}+a_{13}a_{31})E_{11}+\\\\\n+(a_{11}a_{12}+a_{12}a_{22}+a_{13}a_{32})E_{12}+\\\\\n+(a_{11}a_{13}+a_{12}a_{23}+a_{13}a_{33})E_{13}="


"+(a_{21}a_{11}+a_{22}a_{21}+a_{23}a_{31})E_{21}+\\\\\n+(a_{21}a_{12}+a_{22}^2+a_{23}a_{32})E_{22}+\\\\\n+(a_{21}a_{13}+a_{22}a_{23}+a_{23}a_{33})E_{23}+\\\\\n+(a_{31}a_{11}+a_{32}a_{21}+a_{33}a_{31})E_{31}+\\\\\n+(a_{31}a_{12}+a_{32}a_{22}+a_{33}a_{32})E_{32}+\\\\\n+(a_{31}a_{13}+a_{32}a_{23}+a_{33}^2)E_{33}"

If "A^2=0" then

"a_{11}^2+a_{12}a_{21}+a_{13}a_{31}=0\\\\\na_{11}a_{12}+a_{12}a_{22}+a_{13}a_{32}=0\\\\\na_{11}a_{13}+a_{12}a_{23}+a_{13}a_{33}=0\\\\\na_{21}a_{11}+a_{22}a_{21}+a_{23}a_{31}=0\\\\\na_{21}a_{12}+a_{22}^2+a_{23}a_{32}=0\\\\\na_{21}a_{13}+a_{22}a_{23}+a_{23}a_{33}=0\\\\\na_{31}a_{11}+a_{32}a_{21}+a_{33}a_{31}=0\\\\\na_{31}a_{12}+a_{32}a_{22}+a_{33}a_{32}=0\\\\\na_{31}a_{13}+a_{32}a_{23}+a_{33}^2=0"


This system has zero solution only "a_{ij}\\in Z" .

So "A^2\\neq0" .

Then "M_3[Z]" has no nilpotent elements.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
24.03.20, 14:47

Dear Shiva, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!

Shiva
24.03.20, 14:33

Thanks

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS