# General form of a linear transformation

Let v1=[2−1]v_1 = \begin{bmatrix} 2 \\ -1 \end{bmatrix} and v2=[1−1]v_2=\begin{bmatrix} 1 \\ -1 \end{bmatrix} and let A=[32−21]A= \begin{bmatrix} 3 & 2 \\ -2 & 1 \end{bmatrix} be a matrix for T:R2→R2T\colon \Bbb R^2\to \Bbb R^2 relative to the basis B={v1,v2}B = \{v_1, v_2\}.

From this I found that: T(v1)=[4−1]T(v_1) = \begin{bmatrix} 4 \\ -1 \end{bmatrix} and T(v2)=[5−3]T(v_2) = \begin{bmatrix} 5 \\ -3 \end{bmatrix}

How would I find a formula for T([x1x2])T\begin{pmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} \end{pmatrix}. The answer in my book is [−x1−6x22x1+5×2]\begin{bmatrix} -x_1-6x_2 \\ 2x_1+5x_2 \end{bmatrix}

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2

Welcome! Is your question about math in general or about Mathematica?
– Yves Klett
Nov 16 ’13 at 20:26

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1 Answer
1

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Given vectors v1,v2,Tv1,Tv2\;v_1, v_2, Tv_1, Tv_2:

{ v1, v2} = {{2, -1}, {1, -1}};
{Tv1, Tv2} = {{4, -1}, {5, -3}};

This is a direct way of solving underlying linear system:

With[{ m = Array[a, Dimensions[{v1, Tv1}]]},
m.Subscript @@@ {{x, 1}, {x, 2}} /.
Solve[ m.#1 == #2 & @@@ {{v1, Tv1}, {v2, Tv2}}, Join @@ m] //
Transpose // TraditionalForm]

Since it might seem to be an overkill, let’s provide another simpler way, obvious for those about to pass a linear algebra exam:

Transpose[{Tv1, Tv2}].Inverse @ Transpose @ {v1, v2}.{x, y} // MatrixForm

which can be written in the front-end this way:

where we put {x, y} instead of Subscript @@@ {{x, 1}, {x, 2}}.
These both ways are easily applicable to higher dimensional problems involving appropriate numbers of vectors, such that v1,…,vn\;v_1, \dots, v_n\; and Tv1,…,Tvn\;Tv_1, \dots, Tv_n satisfy:
Det @ { v1, …, vn} != 0 and Det @ { Tv1, …, Tvn} != 0