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I've been given the following problem as homework:

Prove that there exists a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ such that $T(1,1) = (1,0,2)$ and $T(2,3) = (1,-1,4)$.

Since it just says prove that one exists, I'm guessing I'm not supposed to actually identify the transformation.

One thing I tried is showing that it holds under addition/multiplication in the sense of:

1) $T(x + y) = T(x) + T(y)$

2) $T(cx) = cT(x)$

3) $T((1,1) + (2,3)) = T(1,1) + T(2,3)$? But that's not necessarily true.

I don't know how I'd accomplish this given my limited knowledge.

What's the approach to solving a problem like this?

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Could you give a rough idea of what you know? Do you know about bases, and perhaps representing linear transformations using bases that aren't the standard $\{(1, 0), (0, 1\})$ basis of $\mathbf R^2$? – Dylan Moreland Feb 9 '12 at 23:39
Well, since you know that $T$ must be linear, and since you know the images of $(1,1)$ and $(2,3)$ under $T$, first show that $(1,1)$ and $(2,3)$ form a basis of $\mathbb{R}^2$, and then extend $T$ linearly to all of $\mathbb{R}^2$. That is, every vector in $\mathbb{R}^2$ can be written as $a(1,1) + b(2,3)$, where $a, b$ are real numbers. Now define $T(a(1,1) + b(2,3)) = aT(1,1) + bT(2,3)$. By the very construction $T$ is linear. – William Feb 9 '12 at 23:40
up vote 6 down vote accepted

Hint: Show that $\{(1,1), (2,3)\}$ is a basis of $\mathbb{R}^2$.

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How exactly does that demonstrate that a linear transformation MUST exist? – CodyBugstein Oct 5 '12 at 0:58
@Imray: They form a basis... – Aryabhata Oct 5 '12 at 1:38
How does that prove they are linear though? Perhaps there are still two vectors in $\mathbb{R}^2$ that produce the same result? – CodyBugstein Oct 5 '12 at 14:46
@Imray: I don't understand your objection. – Aryabhata Oct 5 '12 at 15:23

In this case, it actually isn't that hard to compute the actual transformation.

If you know that $T(2,3)=(1,-1,4)$ and $T(1,1)=(1,0,2)$, what must $T(1,2)$ be equal to? (Note that $(2,3)-(1,1)=(1,2)$.)

Now compute $T(0,1)$ using similar reasoning with $(1,2)$ and $(1,1)$.

Now compute $T(2,0)$ using similar reasoning with $(2,3)$ and $(0,1)$.

Now compute $T(1,0)$ (do you see how to do this one?)

Finally, if $T(1,0)=(a_1,a_1,a_2)$ and $T(0,1)=(b_1,b_2,b_3)$, then you can calculate any $T(x,y)$ by $$T(x,y)=T(x(1,0)+y(0,1))=xT(1,0)+yT(0,1)=$$$$ x(a_1,a_2,a_2)+y(b_1,b_2,b_3)=(a_1x+b_1y,a_2x+b_2y,a_3x+b_3y).$$

If you do all these steps, you'll have the values for $a_1,a_2,a_3,b_1,b_2,b_3$, and you can show that the map $T$ you've written down is linear and has the property that $T(2,3)=(1,-1,4)$ and $T(1,1)=(1,0,2)$.

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