Let $T:R^3 \rightarrow R^3$ be a linear transformation such that $T(1,1,1) = (2,0,-1), T(0,-1,2) = (-3,2,-1)$ and $T(1,0,1)=(1,1,0)$. Find the indicated image $T(2,-1,1)$
I used the rule that:
If $v=c_1v_1+c_2v_2+...+c_nv_n$, then $T(v) = T(c_1v_1+c_2v_2+...+c_nv_n) = c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$
So I did the following:
$T(2,-1,1) = 2(2,0,-1)-(-3,2,-1)+(1,1,0)=(8,-1,0)$
The book has this as the answer:
Because $(2,-1,1)$ can be written as
$(2,-1,1) = -\frac{3}{2}(1,1,1)-\frac{1}{2}(0,-1,2)+\frac{7}{2}(1,0,1)$, then
$T(2,-1,1)=-\frac{3}{2}T(1,1,1)-\frac{1}{2}T(0,-1,2)+\frac{7}{2}T(1,0,1)$
which gives a final answer of:
$\left(2,\frac{5}{2},2\right)$
Where did the fractions come from and why is my answer different if I'm using the same rule?
You need to write the vector $$\begin{pmatrix}2\\-1\\1\end{pmatrix}$$ as a combination of the vectors $$\begin{pmatrix}1\\1\\1\end{pmatrix},\begin{pmatrix}0\\-1\\2\end{pmatrix},\begin{pmatrix}1\\0\\1\end{pmatrix}$$ not as a combination of the standard basis of $\mathbb{R}^3$, which is what you did. (That's because you don't immediately know how $T$ acts on the standard basis, but you do know how $T$ acts on the three vectors listed above: that's exactly what you're given.)