0
$\begingroup$

T is a linear transformation represented as $\left(\begin{array}{ccc}1 & 1 & 0 \\0 & 2 & 0 \\3 & 1 & 0 \\0 & 1 & 1\end{array}\right)$ w.r.t the standard basis.

Now find a representation for $T$ w.r.t bases $(1,0,0)^t, (0,1,1)^t, (1,0,1)^t$ for $R^3$ and $(1,0,0,0)^t, (0,1,1,0)^t, (0,0,1,0)^t, (1,0,0,1)^t$ for $R^4$.

Work so far:

Let $M = \left(\begin{array}{ccc}1 & 0 & 1 \\0 & 1 & 0 \\0 & 1 & 1\end{array}\right)$.

New $T_n$ = $T*M$. If I'm not mistaken this gives us $T$ w.r.t a new basis in $R^3$ (please correct, if wrong). How do I get $T$ w.r.t new basis in $R^4$?

$\endgroup$
1
$\begingroup$

If you multiply $M$ with a vector's representation in the new basis you'll get the vector's representation in respect of the standard basis. Call $N$ the corresponding matrix of th new Basis of $R^4$. Given a the coordinates of a 3-dimensional vector $v$ in respect of the new basis then $Mv$ is it's representation in respect of the standard basis, further $TMv$ is the representation ofbit's image under $T$ in respect of the standard basis; finally $N^{-1}TMv$ is the image's representation in respect to the new basis. Hence $T_n=N^{-1}TM$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.