Finding a linear transformation such that $T((4,1)) = (0,21)$ and $T((1,5))=(-19,10)$ Finding a linear transformation such that $T((4,1)) = (0,21)$ and $T((1,5))=(-19,10)$
So, i'm trying to find multiple different ways to do this. I know that every linear transformation can be represented by a matrix, so i'm letting $A$ be a $2 \times 2$ matrix and defining $T(x) = Ax$ and trying to figure out what $A$ has to be. Is this a good way to go about this problem?
I'm ending up with two seperate systems of linear equations, each with four variables and two equations, t hus I have four variables and four equations, so that's good I guess, but I feel that I'm getting lost in the weeds and there may be an easier way to do this...
Would appreciate multiple perspectives on this one... Thank you!!
 A: Let's do this the way you initially tried (and yes, it's a good way to go about the problem). Let
$$A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right].$$
We are given
$$A \left[\begin{array}{c}4\\1 \end{array} \right] = \left[\begin{array}{c} 0\\21 \end{array} \right], \text{ and}$$
$$A \left[\begin{array}{c} 1\\5 \end{array} \right] = \left[\begin{array}{c} -19\\10 \end{array} \right].$$
That is, we have
$$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[\begin{array}{c}4\\1 \end{array} \right] = \left[\begin{array}{c} 0\\21 \end{array} \right], \text{ and}$$
$$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[\begin{array}{c} 1\\5 \end{array} \right] = \left[\begin{array}{c} -19\\10 \end{array} \right],$$
giving us the system
$$4a+b=0$$
$$4c+d=21$$
$$a+5b=-19$$
$$c+5d=10.$$
Focussing on the system of two equations with the $a$ and $b$ variables, i.e. the system
$$4a+b=0$$
$$a+5b=-19,$$
we have an easy system of two simultaneous equations with two variables. Solving this, we have $a=1$ and $b=-4$.
Now do the same with the two equations involving the $c$ and $d$ variables, and you're done. Of course, the strategy outlined in the comments is equally valid, but you asked for multiple perspectives!
A: Since the vectors $(4,1)$ and $(1,5)$ are linearly independent (not multiples of each other, in dimension two), you can write the matrix rel the basis that they form simply as $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$, where $\begin{cases}a(4,1)+c(1,5)=(0,21)\\b(4,1)+d(1,5)=(-19,10)\end{cases}$.
If you wish to get the matrix in the standard basis, conjugate by the change of basis matrix $$\begin{pmatrix}4&1\\1&5\end{pmatrix}$$.
A: $A\begin{bmatrix} 4&1\\1&5\end{bmatrix} = \begin{bmatrix} 0&-19\\21&10\end{bmatrix}$
Multiply both on the right by $\begin{bmatrix} 4&1\\1&5\end{bmatrix}^{-1}$
A: Let have you change of base matrix $P=\begin{pmatrix}4&1\\1&5\end{pmatrix}$ and its inverse $P^{-1}=\frac 1{19}\begin{pmatrix}5&-1\\-1&4\end{pmatrix}$
Since $\det(P)=4\times 5-1\times 1=19$ and for $2\times 2$ matrices the inverse is obtained by permuting diagonal elements and negate the others.
Let's now have or result matrix $R=\begin{pmatrix}0&-19\\21&10\end{pmatrix}$
By hypothesis we have: $\quad TP=R$
Therefore $T=RP^{-1}=\frac 1{19}\begin{pmatrix}0&-19\\21&10\end{pmatrix}\begin{pmatrix}5&-1\\-1&4\end{pmatrix}=\frac 1{19}\begin{pmatrix}19&-76\\95&19\end{pmatrix}=\begin{pmatrix}1&-4\\5&1\end{pmatrix}$
A: $5(4,1)−(1,5)=(19,0);(4,1)−4(1,5)=(0,−19)$,
so $(1,0)=\frac1{19}[5(4,1)-(1,5)]$ and $(0,1)=-\frac1{19}[(4,1)-4(1,5)]$,
so $T((1,0))=\frac1{19}[5T((4,1))-T((1,5))]$ and $T((0,1))=-\frac1{19}[T((4,1))-4T((1,5))]$,
so $T((1,0))=\frac1{19}[5(0,21)-(-19,10)]=(1,5)$
and $T((0,1))=-\frac1{19}[(0,21)-4(-19,10)]=(-4,1),$
so the matrix is $\pmatrix{1&-4\\5&1}$.
