Thank you so much. Also quick question any idea how I’d draw the diagram that’s askedYou 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 $
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 butThe 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 $(-\sqrt{3}/2 , 1/2)$.
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