# Find a linear transformation such that 2 vectors are mapped to these two values

Find a linear transformation $T\colon \mathbb{R}^2 \to P_2(\mathbb{R})$ such that $T(1,2) = 3 + 4x^2$ and $T(3,5) = 1 - 2x + 3x^2$.

I was thinking of approaching the problem the following way:

$$T \left( \begin{matrix} 1 \\ 2 \end{matrix} \right) = \left( \begin{matrix} 3 \\ 0 \\ 4 \end{matrix} \right)$$

$$T \left( \begin{matrix} 3 \\ 5 \end{matrix} \right) = \left( \begin{matrix} 1 \\ -2 \\ 3 \end{matrix} \right)$$

$$\text{Let } T = \left( \begin{matrix}A & B \\ C & D \\ E & F \end{matrix} \right) \text{ then }$$

$$\begin{align*} A + 2B = 3\\ C + 2B = 0\\ E + 2F = 4\\ 3A = 5B = 1\\ 3C + 5D = -2\\ 3E + 5F = 3 \end{align*}$$

But converting this system of equations into a matrix and performing row reduction gave me

$$T = \left( \begin{matrix} -13 & 8\\-16 & 46/5 \\-14&9 \end{matrix}\right)$$

which when checked against the two given untransformed vectors do not give me the right result.

In addition, the question asks to prove that this linear transformation T is unique. How do I show that a transformation is indeed unique?

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For future reference, questions should be asked as such.  Using the imperative mode implies that you are giving orders. – JavaMan Mar 29 '11 at 2:15
The second equation you derived for $T$ is incorrect: it should be $C+2D=0$, not $C+2B=0$, and your solution does not satisfy this. – Arturo Magidin Mar 29 '11 at 4:20
I'm sorry if my question sounded like a command, that was most certainly not my intention. Thank you Arturo for pointing this out, it appears like I was too tired to notice this stupid mistake. – Millianz Mar 29 '11 at 15:37

## 1 Answer

Note that $(1,2)$ and $(3,5)$ form a basis of $\mathbb{R}^2$. A linear transformation is completely determined by its action on a basis, so that gives uniqueness.

More explicitly: Since $(0,1) = 3(1,2) - (3,5)$ and $(1,0)=-5(1,2)+2(3,5)$, then knowing that $T$ does to $(1,2)$ and to $(3,5)$, and knowing that $T$ is linear, tells you what $T$ does to $(1,0)$ and to $(0,1)$, which tells you what $T$ does to everything, and gives you an easy way to write down what $T$ does to any vector $(a,b)$ (in terms of what it does to $(1,0)$ and to $(0,1)$).

Since everything will be forced by the values of $T(1,2)$ and of $T(3,5)$, if $U$ were any linear transformation that has the same values at $(1,2)$ and at $(3,5)$ as $T$ does, then it would also agree with $T$ on $(1,0)$ and on $(0,1)$, and therefore $T$ and $U$ would agree everywhere, that is, $T=U$.

(This is the "standard" way of proving uniqueness: assume you have two objects [in this case, linear transformations] that have the properties you want, and show that this implies that they are equal.)

The moral here is: if you know what a linear transformation does to a basis, you know what the linear transformation does to everything. )And you can define a linear transformation by specifying what it does to a basis.)

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