How does one show the below matrix is a linear transformations I know I need to multiply something by (0,1) and (1,0)

An operator f(...) is linear if f(A+B) = f(A) + f(B) and f(a*A) = a*f(A), with addition and multiplication being the addition of two vectors and their multiplication with a real number, respectively, in your case. Alternate form of the same statements for a matrix M, vectors x, y, and a scalar a: M(x+y) = Mx + My, M(a*x) = a*(Mx).

When interpreted as an operator V -> V, matrices are always linear. But it should be straightforward to explicitly show that for your given matrix by starting from one side of the two defining equations and rearranging until you get the other side.

 

 

Yes, the "natural basis" for [tex]R^2[/tex] is {(1, 0), (0, 1)}.  Rotating (1, 0) through $\pi/3$ radians counter-clockwise gives [math](cos(\pi/3), sin(\pi/3))= (1/2, \sqrt{3}/2)[/math] and rotating (0, 1) through $\pi/3$ radians counter clockwise gives $(cos(4\pi/3), sin(4\pi/3)= (-sin(\pi/3), cos(\pi/3))= (-\sqrt{3}/2, 1/2)$.  To represent that as a matrix, you need $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ so that $\begin{pmatrix}a & b \\ c & d \end{pmatrix}\begin{pmatrix}1 \\ 0 \end{pmatrix}= \begin{pmatrix}a \\ c \end{pmatrix}= \begin{pmatrix}1/2 \\ \sqrt{3}/2}\end{pmatrix}$ so a= 1/2 and $c= \sqrt{3}/2$.  And, similarly $\begin{pmatrix}a & b \\ c & d \end{pmatrix}\begin{pmatrix}0 \\ 1 \end{pmatrix}= \begin{pmatrix}b \\ c \end{pmatrix}= \begin{pmatrix}\sqrt{3}/2, 1/2\end{pmatrix}$ so $b= -sqrt{3}/2$ and $d= 1/2$.

Edited February 10, 20205 yr by Country Boy

In other words, ask yourself, is it true that,

\[\left[R_{\pi/3}\right]_{\mathcal{B}}\left(\lambda\boldsymbol{u}+\mu\boldsymbol{v}\right)=\lambda\left[R_{\pi/3}\right]_{\mathcal{B}}\boldsymbol{u}+\mu\left[R_{\pi/3}\right]_{\mathcal{B}}\boldsymbol{v}\]

for any u, v, lambda and mu?