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## 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(

**M**

_{1}+

**M**

_{2}) = T(

**M**

_{1}) + T(

**M**

_{2})

T(k

**M**) = 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.