# Transformation Matrix with respect to a basis and the General Linear Group

I need some help going over this problem because I'm not entirely sure if my solution is sound.

Let $$\phi:\mathbb{R^2} \rightarrow \mathbb{R^2},\begin{pmatrix} x\\y \end{pmatrix} \mapsto \begin{pmatrix} \frac{3}{2}x- \frac{1}{2}y \\ -\frac{1}{2}x+\frac{3}{2}y \end{pmatrix}$$

a) Show that there exists a basis $$B$$ of $$\mathbb{R^2}$$ such that the transformation matrix is equal to $${^B}A_\phi^B= \begin{pmatrix} 1&0 \\ 0&2 \end{pmatrix}$$.

b) Show that the set $$Z= \bigg\{T\in GL_2(\mathbb{R})| \ T \begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2} \end {pmatrix}T^{-1} = \begin{pmatrix} 1&0 \\ 0&2 \end{pmatrix} \bigg\}$$ is not empty.

c) Show that $$\forall \ T\in Z$$ and $$\forall \lambda\in \mathbb{R}- \{0 \}$$ also $$\lambda T \in Z$$ is.

d) For any $$T_1,T_2 \in Z$$ does a $$\lambda \in \mathbb{R}-\{0\}$$ exist such that $$T_2=\lambda T_1$$?

My work:

a) Let $$B=\{B_1,B_2\}$$ be a basis for $$\mathbb{R^2}$$ with $$B_1= \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}, B_2= \begin{pmatrix} b_3 \\ b_4 \end{pmatrix}$$. We find the image of $$B_1$$ and $$B_2$$ and then express them in terms of the basis with real coefficients $$p_i,q_i$$.

$$\phi(B_1)= \begin{pmatrix} \frac{3}{2}b_1-\frac{1}{2}b_2 \\ -\frac{1}{2}b_1+\frac{3}{2}b_2 \end{pmatrix}= p_1\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}+ p_2\begin{pmatrix} b_3 \\ b_4 \end{pmatrix}$$

$$\phi(B_2)= \begin{pmatrix} \frac{3}{2}b_3-\frac{1}{2}b_4 \\ -\frac{1}{2}b_3+\frac{3}{2}b_4 \end{pmatrix}= p_1\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}+ p_2\begin{pmatrix} b_3 \\ b_4 \end{pmatrix}$$

Since the transformation matrix has been given we find the coefficients as follows: $$p_1=1,p_2=0,q_1=0,q_2=2$$. Which simply means that $$\phi(B_1)=B_1, \phi(B_2)=2B_2$$. I then simply chose the vectors $$\begin{pmatrix} 2\\2 \end{pmatrix}$$, $$\begin{pmatrix} 2\\-2 \end{pmatrix}$$.

b) This question was much easier and I guessed, correctly, that the matrix $$\begin{pmatrix} 2 &2 \\2&-2 \end{pmatrix}$$ was an element of the set.

The last two parts of this problem is where I need help. I tried simply computing the results of $$\lambda T \begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{3}{2} \end {pmatrix}T^{-1}$$ but it got quickly very complicated and I was wondering if there was an easier way to show it?

c) If $$T\in Z$$ and $$\lambda\in\Bbb R\setminus\{0\}$$, then\begin{align}(\lambda T)\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}(\lambda T)^{-1}&=\lambda T\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}\lambda^{-1}T^{-1}\\&=\lambda\lambda^{-1}T\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}T^{-1}\\&=\begin{bmatrix}1&0\\0&2\end{bmatrix}.\end{align}
d) Not necessarily, since$$\begin{bmatrix}2&2\\2&-2\end{bmatrix},\begin{bmatrix}2&1\\2&-1\end{bmatrix}\in T.$$