 # Let there be a linear transformation going from <mi mathvariant="double-struck">R 3 kokoszzm 2022-06-06 Answered
Let there be a linear transformation going from ${\mathbb{R}}^{3}$ to ${\mathbb{R}}^{2}$, defined by $T\left(x,y,z\right)=\left(x+y,2z-x\right)$. Find the transformation matrix if base 1:
$⟨\left(1,0,-1\right),\left(0,1,1\right),\left(1,0,0\right)⟩$,
base 2: $⟨\left(0,1\right),\left(1,1\right)⟩$
An attempt at a solution included calculating the transformation on each of the bases in ${\mathbb{R}}^{3}$, (base 1) and then these vectors, in their column form, combined, serve as the transformation matrix, given the fact they indeed span all of ${B}_{1}$ in ${B}_{2}$
Another point: if the basis for ${\mathbb{R}}^{3}$ and ${\mathbb{R}}^{2}$ are the standard basis for these spaces, the attempt at a solution is a correct answer.
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Let us call the basis of ${\mathbb{R}}^{3}$ $\left\{{\lambda }_{1},{\lambda }_{2},{\lambda }_{3}\right\}$ and the basis of ${\mathbb{R}}^{2}$ $\left\{{\gamma }_{1},{\gamma }_{2}\right\}$
So all you need to do now is the following:
$T\left({\lambda }_{1}\right)={\alpha }_{11}{\gamma }_{1}+{\alpha }_{21}{\gamma }_{1}$
$T\left({\lambda }_{2}\right)={\alpha }_{12}{\gamma }_{1}+{\alpha }_{22}{\gamma }_{1}$
$T\left({\lambda }_{3}\right)={\alpha }_{13}{\gamma }_{1}+{\alpha }_{23}{\gamma }_{1}$
while $\mathrm{\forall }{\alpha }_{i,j}\in \mathbb{R}$
And the matrix will look like:
$\left[T\right]=\left(\begin{array}{ccc}{\alpha }_{11}& {\alpha }_{12}& {\alpha }_{13}\\ {\alpha }_{21}& {\alpha }_{22}& {\alpha }_{23}\end{array}\right)$
And this is how it is done