Let the be vectors
$w_1=(1,1,2),w_2=(2,1,1),w_3=(1,2,a)$ and $v_1=(1,2,1),v_2=(1,1,2),v_3=(2,a,1)$
Find for which values of $a$ the transformation $$T(w_1)=T(v_1),T(w_2)=T(v_2),T(w_3)=T(v_3)$$ is linear transformation
for $a=1$ find the explicit formula of the linear transformation $T(x,y,z)=...$
Find the inverse linear transformation of $2.$
- For a transformation to be linear we need $T(w_1+w_2+w_3)=v_1+v_2+v_3$
So $$T((4,4,a+3))=T((1,1,2)+(2,1,1)+(1,2,a))=((1,2,1)+(1,1,2)+(2,a,1))=(4,3+a,4)$$ So we need
$$T((4,4,a+3))=(4,3+a,4)$$ or $$a+3=4\iff a=1$$
- We will define the transformation on a basis and find how its work on a general vector in the span of the basis vectors.
$$\begin{pmatrix} 1 & 2 &1\\ 1 & 1 & 2\\ 2 &1 &1 \end{pmatrix}\Rightarrow \begin{pmatrix} 1 & 0 &0\\ 0 & 1 & 0\\ 0 &0 &1 \end{pmatrix}$$
So the vectors are linearly independent and are $3$ so they are a basis for $\mathbb{R}^3$
Now we will write $(x,y,z)\in Span(w_1,w_2,w_3)$
$$ \left(\begin{array}{rrr|r} 1 & 2 & 1 & x \\ 1 & 1 & 2 & y \\ 2 & 1 & 1 & z \end{array}\right)\Rightarrow \left(\begin{array}{rrr|r} 1 & 0 & 0 & \frac{1}{4}(-x-y+3z) \\ 0 & 1 & 0 & \frac{1}{4}(3x-y-z) \\ 0 & 0 & 1 & \frac{1}{4}(-x+3y-z) \end{array}\right) $$
So $$(x,y,z)=\frac{1}{4}(-x-y+3z)\cdot w_1+\frac{1}{4}(3x-y-z)\cdot w_2+\frac{1}{4}(-x+3y-z)\cdot w_3$$
Taking the linear transformation on both sides we get
$$T(x,y,z)=\frac{1}{4}(-x-y+3z)\cdot (1,2,1)+\frac{1}{4}(3x-y-z)\cdot (1,1,2)+\frac{1}{4}(-x+3y-z)\cdot (2,1,1)=(y,z,x)$$
- Taking the Transformation on the standard basis we get
$$\begin{pmatrix} 0 & 1 &0\\ 0 & 0 & 1\\ 1 &0 & 0 \end{pmatrix}$$ which is inverse is
$$\begin{pmatrix} 0 & 0 &1\\ 1 & 0 & 0\\ 0 &1 & 0 \end{pmatrix}$$
Is that correct?