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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 $R^4$?

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1 Answer 1

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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 $T_n=N^{-1}TM$.

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