Determine if there's a linear map such that... Today in the class we were solving a following problem:
Determine if there's a linear map $\phi : R^3 \rightarrow R^3$ such that $\phi(2,1,0)=(3,2,1), \phi(4,2,1) = (4,1,0), \phi(2,2,1) = (2,3,3), \phi(2,2,2) = (0,0,1)$. It was then solved by creating an augmented matrix with the inputs for $\phi$ written in rows in the left hand side of the matrix and outputs of $\phi$ in the right hand side, also in rows. Then the matrix has been row reduced until an identity appeared in the left hand side... it seemed to me like some black magic. I have no idea why or how it works. This whole method doesn't make any sense to me. Is there any better way of doing it or some explanation why something like this works?
 A: The first three input vectors are linearly independent,
so we can evaluate $\phi$ on the standard basis vectors by linearity:
$\phi(1,0,0)=\phi(\frac12((4,2,1)-(2,2,1))=\frac12((4,1,0)-(2,3,3))=(1,-1,-\frac32);$
$\phi(0,1,0)=\phi((2,2,0)+(2,2,1)-(4,2,1))=(3,2,1)+(2,3,3)-(4,1,0)=(1,4,4)$;
and $\phi(0,0,1)=\phi((2,2,2)-(2,2,1))=(0,0,1)-(2,3,3)=(-2,-3,-2).$
So   $\phi$ corresponds to $\pmatrix{1&1&-2\\-1&4&-3\\-\frac32&4&-2}$.
Check that this matrix applied to the fourth input vector $(2,2,2)$ yields $(0,0,1)$.
A: Another approach you may like is as follows:
Let the input vectors are $u_1,u_2,u_3,u_4$ and the corresponding output vectors are $v_1,v_2,v_3,v_4$ so that for a $3\times 3$ matrix $A=\left (x_{ij}\right )$, the system $Au_i=v_i, (i=1,2,3)$ gives you a set of $9$ equations involving $9$ variables $x_{ij}, (1\le i\le 3, 1\le j\le 3)$.
Solving the above you can obtain matrix $A$. Now just verify whether $Au_4=v_4$ holds or not. If it does then such linear transformation is possible.
A: $$A=\left(
\begin{array}{ccc}
 a & b & c \\
 d & e & f \\
 g & h & i \\
\end{array}
\right)$$
$$A\cdot (2,1,0)=(3,2,1),A\cdot (4,2,1)=(4,1,0),A\cdot (2,2,1)=(2,3,3),A\cdot (2,2,2)=(0,0,1)$$
$$\begin{cases}
2 a + b=3\\ 
2 d + e=2\\
2 g + h=1\\
4 a + 2 b + c=4\\
4 d + 2 e + f=2\\
4 g + 2 h + i=1\\
2 a + 2 b + c=2\\
2 d + 2 e + f=3\\
2 g + 2 h + i=3\\
\end{cases}
$$
which gives
$$a= 1,b= 1,c= -2,d= -1,e= 4,f= -3,g= -\frac{3}{2},h= 4,i= -2$$
These results satisfy the fourth set of conditions
$$\begin{cases}
2 a + 2 b + 2 c=0\\
2 d + 2 e + 2 f=0\\
2 g + 2 h + 2 i=1\\
\end{cases}
$$
Therefore the matrix is
$$
A=\left(
\begin{array}{ccc}
 1 & 1 & -2 \\
 -1 & 4 & -3 \\
 -\frac{3}{2} & 4 & -2 \\
\end{array}
\right)
$$
A: You must have studied some verson of the following:
Theorem: Any linear map $\;T:V\to W\;$ , with $\;V,\,W\;$ linear spaces over the same field, is uniquely and completely determined once we know the action of $\;T\;$ on any basis $\;A=\{v_1,...\}\;$ of $\;V\;$ , meaning: $\;\{Tv_1,...\}\;$ completely and uniquely determines $\;T\;$ .
The above theorem holds for both infinite and finite dimesnional spaces, and from here it is not hard to see that
For any basis $\;A=\{v_1,...\}\;$ of $\;V\;$ and any set $\;B=\{w_2,...\}\;$ of elements in $\;W\;$ , with $\;|A|=|B|\;$ , any function $\;t: A\to B\;,\;\;tv_i=w_i\;$ can uniquely be extended by linearity to a linear map $\;T: V\to W\;$ such that $\;Tv_i=w_i\;$ .
Thus, in your case, you only have to check that the set of vectors on which the action of $\;\phi\;$ is given is a basis of $\;\Bbb R^3\;$ ...
