Ok, I was asked this strange question that I can't seem to grasp the concept of..
Let $T$ be a linear transformation such that: $$T \langle1,-1\rangle = \langle 0,3\rangle \\ T \langle2, 3\rangle = \langle 5,1\rangle $$ Find $T$.
Is there suppose to be a function out of this? A matrix of some kind? Maybe both? If so, what is it?
$\endgroup$ 14 Answers
$\begingroup$Yes, $T$ is a matrix of some kind. You can tell that it is $2\times 2$ since both its inputs and outputs are vectors of length $2$. My recommendation for solving this is the following. Suppose $T$ has matrix representation
$$ T \;\; =\;\; \left [ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right ]. $$
Start with the first equation which you can write as
$$ \left [ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right ] \left [ \begin{array}{c} 1 \\ -1\\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{c} 0 \\ 3\\ \end{array} \right ]. $$
You can find the values of $a,b,c,d$ by working through these equations.
$\endgroup$ $\begingroup$"Given T find T" makes no sense! But if you are given "$T$" (standard font) and asked to find "$\mathbf{T}$" (bold face) where the bold face has already been defined to be the "matrix associated with the linear transformation, then, yes.
You should also understand that the matrix representing a given linear transformation depends upon what basis you are using for the vector space. Here, nothing is said about a basis but it is in $\mathbb R^2$ so I think we are to assume the "standard" basis, $\langle 1, 0\rangle$ and $\langle 0, 1\rangle$.
Since this is $\mathbb R^2$ to $\mathbb R^2$ any such matrix is of the form $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$. We are told that this linear transformation maps $\langle1, -1\rangle$ to $\langle0, 3\rangle$ so we must have $$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ -1 \end{bmatrix}= \begin{bmatrix}a- b \\ c- d\end{bmatrix}= \begin{bmatrix}0 \\ 3\end{bmatrix}.$$
We are also told that this linear transformation maps $\langle2, 3\rangle$ to $\langle5, 1\rangle$ so we must have $$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}2 \\ 3 \end{bmatrix}= \begin{bmatrix}2a+ 3b \\ 2c+ 3d\end{bmatrix}= \begin{bmatrix}5 \\ 1\end{bmatrix}.$$
Solve the two equations $a- b= 0$ and $2a+ 3b= 5$ for $a$ and $b$. Solve the two equations $c- d= 3$ and $2c+ 3d= 1$ for $c$ and $d$.
$\endgroup$ $\begingroup$Technically the $T$ you are asked to find is a function and not a matrix. Though you can find the matrix representation of $T$ as others have noted in their answers. I think that the point of this question is to make sure you understand how bases and linear transformations (as functions) work. The question isn't worded very precisely, but I'm sure that its asking you find what $T(x,y)$ equals for an arbitrary $(x, y)$.
First notice that, from the values given, $T: F^2 \rightarrow F^2$, where $F$ is the scalar field. Next observe that $\{(1, -1), (2, 3)\}$ forms a basis for $F^2$. I imagine this question came from the same section as some form of the following theorem (taken from Friedberg, Insel and Spence's Linear Algebra 4th edition).
$\textbf{Theorem (2.6):}$ Let $V$ and $W$ be vector spaces over $F$, and suppose that $\{v_1, v_2, ..., v_n\}$ forms a basis for $V$. For $w_1, w_2, ..., w_n$ in $W$, there exists a unique linear transformation $T: V \rightarrow W$ such that $T(v_i) = w_i$ for $i = 1, 2, ..., n$.
Since the linear transformation $T$ given in your question defines what $T$ does on a basis, this theorem shows that there can be only one such linear transformation. Thus we should be able to figure out what $T$ does to any arbitrary $(x, y) \in F^2$. Since $\{(1, -1), (2, 3)\}$ is a basis for $F^2$, there exists $a, b \in F$ such that
$\begin{equation*} (x, y) = a(1, -1) + b(2, 3) \end{equation*}$
for any $(x, y) \in F^2$. We can solve for $a$ and $b$ to find that $a = \frac{3}{5}x - \frac{2}{5}y$ and $b = \frac{1}{5}x + \frac{1}{5}y$. Since we know $T$ is linear we have
$\begin{equation*} T(x, y) = T(a(1, -1) + b(2, 3)) = aT(1, -1) + bT(2, 3)\end{equation*}$.
But since we know what $T$ does to each element of the basis, we can conclude that
$\begin{equation*} T(x, y) = a(0, 3) + b(5, 1) = (5b, 3a + b). \end{equation*}$
Plugging in the $a$ and $b$ we found before we obtain an explicit formula for $T$.
$\endgroup$ 1 $\begingroup$Assuming $T:\mathbb{K}^2 \rightarrow \mathbb{K}^2$ some vector field $\mathbb{K}$, the required transformation can be expressed as a matrix. To do this recall $T<1,0>$ gives the first column and $T<0,1>$ gives the second.
Then $T<1,0> = \frac{1}{5}(3<0,3>+<5,1>)=<3,2>$ and $T<0,1> = \frac{1}{5}(-2<0,3>+<5,1>)=<1,-1>$.
