# Linear isomorphism

#### Lauren1231

How do I show the below is a linear transformations . I’ve shown it’s 1-1

#### Greens

$[R_{\pi/3}]_{\mathcal{B}}= \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}$

Since this is the standard basis representation, multiplying by $[R_{\pi/3}]_{\mathcal{B}}$ is the same as applying the transformation. Choose two arbitrary elements $x,y \in \mathbb{R}^2$ and try to prove $T(cx+y) = cT(x)+T(y)$ which is equivalent to $[R_{\pi/3}]_{\mathcal{B}}(cx+y) = c[R_{\pi/3}]_{\mathcal{B}}x + [R_{\pi/3}]_{\mathcal{B}}y$

#### Lauren1231

I know I want to chose (1,0) and (0,1) but I think I have to perform some form of transformations first but idk how to do that?

#### Greens

You can't only choose $(1,0)$ and $(0,1)$. These are two vectors in $\mathbb{R}^2$, but you need vectors that can represent the entirety of $\mathbb{R}^2$. $x$ and $y$ need to represent ANY vector in $\mathbb{R}^2$. $x= (x_1 , x_2)$ and $y=(y_1 , y_2)$ are good to use.

Multiplying $[R_{\pi/3}]_{\mathcal{B}}$ to the vectors $x$ and $y$ is the transformation you need.

Try to show that for any $c \in \mathbb{R}$

$[R_{\pi/3}]_{\mathcal{B}} \cdot (cx+y) = [R_{\pi/3}]_{\mathcal{B}} \cdot cx + [R_{\pi/3}]_{\mathcal{B}} \cdot y$

Lauren1231

#### Lauren1231

You can't only choose $(1,0)$ and $(0,1)$. These are two vectors in $\mathbb{R}^2$, but you need vectors that can represent the entirety of $\mathbb{R}^2$. $x$ and $y$ need to represent ANY vector in $\mathbb{R}^2$. $x= (x_1 , x_2)$ and $y=(y_1 , y_2)$ are good to use.

Multiplying $[R_{\pi/3}]_{\mathcal{B}}$ to the vectors $x$ and $y$ is the transformation you need.

Try to show that for any $c \in \mathbb{R}$

$[R_{\pi/3}]_{\mathcal{B}} \cdot (cx+y) = [R_{\pi/3}]_{\mathcal{B}} \cdot cx + [R_{\pi/3}]_{\mathcal{B}} \cdot y$
Thank you so much. Also quick question any idea how I’d draw the diagram that’s asked

#### Greens

The notation $[ R_{\pi /3}]_{\mathcal{B}}$ means "The matrix representation of $R_{\pi /3}$ with respect to the basis $\mathcal{B}$."

It's mentioned that $\mathcal{B}$ is the natural basis (also sometimes called the standard basis) of $\mathbb{R}^2$, which is $\mathcal{B} = \{ (1,0) , (0,1) \}$

The form for $[ R_{\pi /3}]_{\mathcal{B}}$ is

$[ R_{\pi /3}]_{\mathcal{B}} = [ R_{\pi /3}(1,0) \; \; \; R_{\pi /3}(0,1)\;]$

Where $R_{\pi /3}(1,0)$ and $R_{\pi /3}(0,1)$ are the columns of the matrix.

The diagram should therefore show that rotating $(1,0)$ by $\pi /3$ gives $(1/2 , \sqrt{3} / 2)$ and rotating $(0,1)$by $\pi /3$ gives $(-\sqrt{3}/2 , 1/2)$.

Lauren1231

#### Lauren1231

g
The notation $[ R_{\pi /3}]_{\mathcal{B}}$ means "The matrix representation of $R_{\pi /3}$ with respect to the basis $\mathcal{B}$."

It's mentioned that $\mathcal{B}$ is the natural basis (also sometimes called the standard basis) of $\mathbb{R}^2$, which is $\mathcal{B} = \{ (1,0) , (0,1) \}$

The form for $[ R_{\pi /3}]_{\mathcal{B}}$ is

$[ R_{\pi /3}]_{\mathcal{B}} = [ R_{\pi /3}(1,0) \; \; \; R_{\pi /3}(0,1)\;]$

Where $R_{\pi /3}(1,0)$ and $R_{\pi /3}(0,1)$ are the columns of the matrix.

The diagram should therefore show that rotating $(1,0)$ by $\pi /3$ gives $(1/2 , \sqrt{3} / 2)$ and rotating $(0,1)$by $\pi /3$ gives $(-\sqrt{3}/2 , 1/2)$.
thanks so much. Bit of a different question but do you know I got to this answer below im trying find $[T]_B$. I think this is the answer but idk how I got to it. Not sure you may need more info but

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