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Any help on this linear transformation question is very much appreciated.

Let $V$ denote the real vector space $R^2$ and $\psi : V \rightarrow V$ be a real linear transformation such that $\psi ((1, 0)) = (11, 8)$ and $\psi ((0, 1)) = (4, 3)$. Express the image $\psi ((x, y))$ of $(x, y)$ in terms of $x$ and $y$. Assume that $w_1 = (4, 5)$ and $w_2 = (9,11)$ form an ordered basis $B$ for $V$ . Working from the denition determine the matrix $M^B_B (\psi)$ with respect to the basis $B$.

does $ M^B_B(\psi)= \begin{bmatrix} -469 & -1048 \\ 388 & 867 \end{bmatrix}$ ?

thanks in advance for any help

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what does $M_B^B(\psi)$ mean? –  chaohuang Jul 7 '12 at 15:06
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Denote $E$ the standard basis $(1,0)$, $(0,1)$. By definition, the matrix $M^E_E(\psi)$ (using your notation) is given by $$M^E_E(\psi)= \begin{bmatrix} 11 & 4 \\ 8 & 3 \end{bmatrix}.$$ Now, the transition matrix from $B$ to $E$ is given by $$M^E_B(id)= \begin{bmatrix} 4 & 9 \\ 5 & 11 \end{bmatrix}.$$ Hence the transition matrix from $E$ to $B$ is given by $$M^B_E(id)= \begin{bmatrix} 4 & 9 \\ 5 & 11 \end{bmatrix}^{-1}=\frac{1}{-1}\begin{bmatrix} 11 & -9 \\ -5 & 4 \end{bmatrix}=\begin{bmatrix} -11 & 9 \\ 5 & -4 \end{bmatrix}.$$ Combining all these, the matrix $M^B_B(\psi)$ is given by $$M^B_B(\psi)=M^B_E(id)\cdot M^E_E(\psi)\cdot M^E_B(id)=\begin{bmatrix} -11 & 9 \\ 5 & -4 \end{bmatrix}\cdot\begin{bmatrix} 11 & 4 \\ 8 & 3 \end{bmatrix}\cdot\begin{bmatrix} 4 & 9 \\ 5 & 11 \end{bmatrix}=...$$

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that explains alot. thank you very much –  user34742 Jul 8 '12 at 12:12
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The answer to the first part of the question is implicit in Paul's answer, but I'll bring it out explicitly.

$(x,y)=x(1,0)+y(0,1)$, and $\psi$ is linear, so $$\psi(x,y)=x\psi(1,0)+y\psi(0,1)=x(11,8)+y(4,3)=(11x+4y,8x+3y)$$

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