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A linear transform $T : R^3 \rightarrow R^2$ has the matrix representation:

$$ T = \begin{bmatrix} 2 & -1 & 2 \\ 4 & 1 & 5 \end{bmatrix} $$

the $R^2$ basis is (4,3), and (3,2).

the $R^3$ basis is the standard basis: (1,0,0), (0,1,0), (0,0,1)

What's the procedure to convert matrix T into a linear transform function T(a,b,c)?

Book says the answer is:

$$ T(a,b,c) = a T(1,0,0) + b T(0,1,0) + c T(0,0,1)$$

$$ T(a,b,c) = (20a - b + 23c, 14a -b + 16c)$$

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Let $B$ be that basis of $\mathbb{R}^2$.

First, you compute $T(1,0,0)$, $T(0,1,0)$, and $T(0,0,1)$. These are equal to $(2,4)_B$, to $(-1,1)_B$ and to $(2,5)_B$. But

  • $(2,4)_B=(20,14)$;
  • $(-1,1)_B=(-1,-1)$;
  • $(2,5)_B=(23,16)$.

So\begin{align}T(a,b,c)&=aT(1,0,0)+bT(0,1,0)+cT(0,0,1)\\&=a(20,14)+b(-1,-1)+c(23,16)\\&=(20a-b+23c,14a-b+16c).\end{align}So, yes, that answer from your book is indeed correct.

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$$T: V \rightarrow W $$

T Matrix Representation: $$ T = \begin{bmatrix} 2 & -1 & 2 \\ 4 & 1 & 5 \end{bmatrix} $$

Basis Sets:

$$basis\{V\} = \left\{v_1=\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, v_2=\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}, v_3=\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right\} $$

$$basis\{W\} = \left\{w_1=\begin{bmatrix}4 \\ 3 \end{bmatrix}, w_2=\begin{bmatrix}3 \\2 \end{bmatrix}\right\} $$

A = matrix formed by concatenating the column vectors of V basis

B = matrix formed by concatenating the column vectors of W basis

$$B=\begin{bmatrix}4 & 3 \\ 3 & 2\end{bmatrix}$$

$$ \begin{aligned} T(v_1) &= B\ col(1, T) \\ T(v_2) &= B\ col(2, T) \\ T(v_3) &= B\ col(3, T) \end{aligned} $$

$$ v_x = \begin{bmatrix} a\\ b\\ c \end{bmatrix}$$

$$T(v_x) = aT(v_1) +bT(v_2)+cT(v_3)$$

or:

$$T(a,b,c) = aT(v_1) +bT(v_2)+cT(v_3)$$

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