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Linear Transformation Isomorphism

  • Thread starter zwingtip
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  • #1
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I think I've solved this problem, but the examples in my textbook are not giving me any indication as to whether my reasoning is sound.

Homework Statement


Is the transformation

[tex]T(M) = M\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right][/tex]

from [tex]\mathbb{R}[/tex]2x2 to [tex]\mathbb{R}[/tex]2x2 linear? If it is, determine whether it is an isomorphism.

Homework Equations


T(f + g) = T(f) + T(g)
T (kf) = k T(f)
T-1(T(M)) = M

The Attempt at a Solution


T(M1+M2) = T(M1) + T(M2)

T(kM) = k T(M

Therefore, T(M) is a linear transformation.

[tex]\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right][/tex] is not invertible, so T is an isomorphism if Ker(T) = 0.

[tex]\left[ \begin{array}{cccc} m_1 & m_2 \\ m_3 & m_4\end{array} \right]\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right]=\left[ \begin{array}{cccc} m_1+3m_2 & 2m_1+6m_2 \\ m_3+3m_4 & 2m_3+6m_4\end{array} \right]=\left[ \begin{array}{cccc} 0 & 0 \\ 0 & 0\end{array} \right][/tex]

Then [tex]m_1 = -3m_2[/tex], [tex]m_3 = -3m_4[/tex] and

[tex]Ker(T)=\left[ \begin{array}{cccc} -3 & 1 \\ -3 & 1\end{array} \right]\neq \left[ \begin{array}{cccc} 0 & 0 \\ 0 & 0\end{array} \right][/tex]

Therefore, the transformation T(M) is linear, but is not an isomorphism.

So I guess my question is, have I done this correctly? Thanks for any help.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Yes, I believe you have done that correctly. T(0)=0 and T([-3,1],[-3,1])=0. Definitely not 1-1.
 
  • #3
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Awesome. Thanks!
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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I think I've solved this problem, but the examples in my textbook are not giving me any indication as to whether my reasoning is sound.

Homework Statement


Is the transformation

[tex]T(M) = M\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right][/tex]

from [tex]\mathbb{R}[/tex]2x2 to [tex]\mathbb{R}[/tex]2x2 linear? If it is, determine whether it is an isomorphism.

Homework Equations


T(f + g) = T(f) + T(g)
T (kf) = k T(f)
T-1(T(M)) = M

The Attempt at a Solution


T(M1+M2) = T(M1) + T(M2)

T(kM) = k T(M

Therefore, T(M) is a linear transformation.

[tex]\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right][/tex] is not invertible, so T is an isomorphism if Ker(T) = 0.
Because T is not invertible, it is not an isomorphism- every isomorphism is, by definition, invertible.

[tex]\left[ \begin{array}{cccc} m_1 & m_2 \\ m_3 & m_4\end{array} \right]\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right]=\left[ \begin{array}{cccc} m_1+3m_2 & 2m_1+6m_2 \\ m_3+3m_4 & 2m_3+6m_4\end{array} \right]=\left[ \begin{array}{cccc} 0 & 0 \\ 0 & 0\end{array} \right][/tex]

Then [tex]m_1 = -3m_2[/tex], [tex]m_3 = -3m_4[/tex] and

[tex]Ker(T)=\left[ \begin{array}{cccc} -3 & 1 \\ -3 & 1\end{array} \right]\neq \left[ \begin{array}{cccc} 0 & 0 \\ 0 & 0\end{array} \right][/tex]

Therefore, the transformation T(M) is linear, but is not an isomorphism.

So I guess my question is, have I done this correctly? Thanks for any help.
 

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