Linear Algebra - Linear transformation quesiton

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SUMMARY

The linear operator T on R² is represented by the matrix \begin{bmatrix}4 & -1\\-4 & 3 \end{bmatrix} relative to the basis A = B = {(1,2), (0, 1)}. Applying T to the vector u, which has coordinates \begin{bmatrix}1 \\1\end{bmatrix}, results in T(u) = \begin{bmatrix}3 \\-1\end{bmatrix}. The final output in component form is T(u) = {(3,6), (0,-1)}, although the answer was initially noted as being in the wrong form.

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Homework Statement



The linear operator T on R2 has the matrix
[tex] \begin{bmatrix}4 & -1\\-4 & 3 \end{bmatrix} [/tex] relative to the basis A = B = {(1,2), (0, 1)}. A vector u has coordinates [tex] <br /> <br /> \begin{bmatrix}1 \\1\end{bmatrix}<br /> [/tex] relative to this basis. Find T(u) in component form (x, y)

Homework Equations





The Attempt at a Solution



I apply T to u = [tex] \begin{bmatrix}4 & -1\\-4 & 3 \end{bmatrix} \begin{bmatrix}1 \\1\end{bmatrix} = \begin{bmatrix} 3 \\-1\end{bmatrix}[/tex]

= relative to the given basis

Then T(u) = {(3,6), (0,-1)}

Does that look right?
 
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Your answer is in the wrong form.

[tex]\begin{bmatrix} 3 \\-1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\0 \end{bmatrix} + (-1) \begin{bmatrix} 0 \\1 \end{bmatrix} = 3 (1,2) + (-1) (0,1)[/tex]
 

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