Linear Transformation Isomorphism

[zwingtip]

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

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

from $$\mathbb{R}$$^2x2 to $$\mathbb{R}$$^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.

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

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

Then $$m_1 = -3m_2$$, $$m_3 = -3m_4$$ and

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

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.

 

[Dick]

Yes, I believe you have done that correctly. T(0)=0 and T([-3,1],[-3,1])=0. Definitely not 1-1.

 

[zwingtip]

Awesome. Thanks!

 

[HallsofIvy]

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

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

from $$\mathbb{R}$$^2x2 to $$\mathbb{R}$$^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.

$$\left[ \begin{array}{cccc} 1 & 2 \\ 3 & 6\end{array} \right]$$ 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.

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

Then $$m_1 = -3m_2$$, $$m_3 = -3m_4$$ and

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

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.