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?


1 Answer 1


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$$\begin{bmatrix}2&2\\2&-2\end{bmatrix},\begin{bmatrix}2&1\\2&-1\end{bmatrix}\in T.$$


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