Find the standard matrix of the linear transformation $T$

Question

A linear transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined by $$T((2,11))=(8,5)$$ and $$T((1,5))=(5,3)$$ Find the matrix for this linear transformation.

I did this in a very clunky way - since $T$ is linear, I can write that

$$T((2, 11))=T(2e_1+11e_2)=2T(e_1)+11T(e_2)=(8,5)$$

and

$$T((1,5))=T(e_1+5e_2)=T(e_1)+5T(e_2)=(5,3)$$

I solved this system of two equations for $T(e_1)$ and $T(e_2)$ to get $$T(e_1)=(15,8)$$ $$T(e_2)=(-2,-1)$$

Thus, we get $$ \begin{pmatrix} 15 & -2 \\ 8 & -1 \\ \end{pmatrix} $$

However, is there a simpler way to get the answer?

Answer 1

What you did in solving the system corresponds to solve $TA=B$ as $T=BA^-1$.
Whether it is more convenient your way or that of finding the inverse of $A$ is a matter of .. what constraints, instruments, etc. you have.

Answer 2

Here is a bit of an elaboration on top of @G Cab's answer.

Consider the linear transformation $T_1$, which maps components of the identity matrix to vectors specified below.

$$T_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 11 \end{pmatrix}, \, T_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$$

Next, we define a separate linear transformation that maps components of the identity matrix to another set of vectors.

$$T_2 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \end{pmatrix}, \, T_2 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$$

Then, there should be some linear transformation $T$ that represents the relationship between these two linear transformations. In other words,

$$T\left(T_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right) = \begin{pmatrix} 8 \\ 5 \end{pmatrix}$$

and

$$T\left(T_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$$

We can package the basis vectors into a matrix to formulate a more compact expression:

$$T\left(T_1 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$

Because we intentionally chose very easy basis vectors, namely the components of the identity matrix, we know what $T_1$ and $T_2$ are:

$$T_1 = \begin{pmatrix} 2 & 1 \\ 11 & 5 \end{pmatrix}, \, T_2 = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$

Therefore, the equation above simplifies to the following:

$$T \begin{pmatrix} 2 & 1 \\ 11 & 5 \end{pmatrix} = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}$$

Can you find the inverse of $T_1$ and go from there?