$$T : \mathbb R^3 → \mathbb R^2$$ such that
$$T(4, 0, −1) = (1, 10);$$
$$T(1, 2, 3) = (14, 32);$$
$$T(3, −1, 1) = (4, 13)$$
Determine whether the transformation is linear.
Now I know that a transformation is linear once it meets the following conditions:
$L(x+y) = L(x) + L(y)$ and $L(kx) = kL(x)$, where $k$ can be any real number.
But if I'm not mistaken – another way of proving its linearity/non-linearity would be trying to find a transformation matrix such that:
$A\times (4, 0, −1) = (1, 10)$; $A\times (1,2,3) = (14,32)$; $A\times (3,-1,1) = (4,13)$.
If that would be the case, how do I find a transformation matrix from the given information? I don't seem to get the right idea, my textbook doesn't cover this type of problems either.
Can anyone guide me to solving this? Thank you very much in advance.
Best Answer
You can package up Emilio Novati’s answer into a single matrix equation: $$ \begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ \end{bmatrix} \begin{bmatrix} 4&1&3\\0&2&-1\\-1&3&1 \end{bmatrix}= \begin{bmatrix} 1&14&4\\10&32&13 \end{bmatrix} $$ which can be solved by inverting the $3\times3$ matrix: $$ \begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ \end{bmatrix}= \begin{bmatrix} 1&14&4\\10&32&13 \end{bmatrix} \begin{bmatrix} 4&1&3\\0&2&-1\\-1&3&1 \end{bmatrix}^{-1}\tag{*}. $$ Here’s another way of approaching the problem. As you might already know, the columns of a transformation matrix are the images of the basis for its domain. The three vectors in $\mathbb R^3$ that you were given are linearly independent, so they form a basis for this space. Now, a matrix that maps the standard basis to the desired output is simply $$ \begin{bmatrix} 1&14&4\\10&32&13 \end{bmatrix}, $$ but we need it to map the given basis to this output instead. We can do that by multiplying by the change-of-basis matrix that maps from our basis to the standard one, which is just the inverse of the matrix that has those basis vectors as its columns, giving the same expression as (*).
Note that in both cases, the order in which you take the vectors and their images doesn’t matter as long as it’s consistent in both of the matrices.