# Homework Help: Linear Transformation Isomorphism

1. Mar 1, 2010

### 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.

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
T(f + g) = T(f) + T(g)
T (kf) = k T(f)
T-1(T(M)) = M

3. 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.

2. Mar 1, 2010

### Dick

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

3. Mar 1, 2010

### zwingtip

Awesome. Thanks!

4. Mar 2, 2010

### HallsofIvy

Because T is not invertible, it is not an isomorphism- every isomorphism is, by definition, invertible.