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Let: $T:\mathcal M_{2\times2 }(\mathbb R)\to \mathbb R^3$ be a linear map that has the following transformation matrix in relation to the following bases.

pic Calculate $T\begin{pmatrix} 1 &2 \\ 3& 4\end{pmatrix}$ and find $X\in \mathcal M_{2\times2 }(\mathbb R)$ such that: $T(X)=\begin{pmatrix} 1 \\ 2\\3\end{pmatrix}$

I'm not sure on how to use the transformation matrix in order to calculate the $T$.

Any directions please ?

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Hint

You should express the matrix $A=\begin{pmatrix} 1 &2 \\ 3& 4\end{pmatrix}$ in the basis $\mathfrak F$ so assume that its components in this basis is $$X=\begin{pmatrix} x \\y \\z\\ t\end{pmatrix}$$ so solve the system of equations $A=xu_1+yu_2+zu_3+tu_4$ for $x,y,z$ and $t$ and then calculate:

$$TA=TX$$

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  • $\begingroup$ Why we need to find the variables with the basis to get to $A$ in order to find it's linear map ? And why we don't use the transformation matrix ? $\endgroup$ – GinKin Jan 8 '14 at 14:37
  • $\begingroup$ We can not apply a matrix of $T$ which $3\times 4$ on a matrix $2\times 2$, the number of columns of $T$ must be equal to the number of rows of $X$ in the product $TX$. $\endgroup$ – user63181 Jan 8 '14 at 14:42
  • $\begingroup$ I see. So then I get a $4\times 1 $ matrix $X$ that still isn't what I was asked to calculate right ? Now I can multiply it by the transformation matrix and the product is what I was asked to find. $\endgroup$ – GinKin Jan 8 '14 at 14:53
  • $\begingroup$ Yes you got it. $\endgroup$ – user63181 Jan 8 '14 at 14:54
  • $\begingroup$ I got the second part also. Thanks for the help. $\endgroup$ – GinKin Jan 8 '14 at 14:59

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