Jump to content

Recommended Posts

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.

 

 

Link to post
Share on other sites
  • 3 weeks later...

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 by Country Boy
Link to post
Share on other sites
  • 3 months later...

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?

Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.