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Im not sure if my question is worded very well, but I'm having trouble understanding how to tackle this problem.

Let $T\colon\mathbb{R}^3\to\mathbb{R}^2$ be the linear transformation such that $T(1,-1,2)=(-3,1)$ and $T(3,-1,1) = (-1,2)$. Find $T(9,-1,10)$.


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Linear transformations are linear. So try to express $(9, -1, 10)$ as a linear combination of $(1, -1, 2)$ and $(3, -1, 1)$. –  Qiaochu Yuan Jun 18 '12 at 4:19

1 Answer 1

up vote 7 down vote accepted

Recall that a linear transformation is, well, linear. If you know the value at vectors $\mathbf{v}_1$ and $\mathbf{v}_2$, then you can compute the value at any linear combination of those two vectors, by using linearlity: $$T(\alpha\mathbf{v}_1+\beta\mathbf{v}_2) = \alpha T(\mathbf{v}_1) + \beta T(\mathbf{v}_2).$$

Here, for example, you know the value of $T$ at $(1,-1,2)$ and at $(3,-1,1)$. So, for instance, you can easily calculate the value of $T$ at $(5,-3,5) = 2(1,-1,2)+(3,-1,1)$: $$\begin{align*} T(5,-3,5) &= T\Bigl( 2(1,-1,2) + (3,-1,1)\Bigr)\\ &= 2T(1,-1,2) + T(3,-1,1)\\ &= 2(-3,1) + (-1,2)\\ &= (-7,4).\end{align*}$$

So... if you can find a way of writing $(9,-1,10)$ as a linear combination of $(1,-1,2)$ and $(3,-1,1)$, then you'll be set.

Added. Unfortunately, this cannot be done here. Namely, there are no scalars $a$ and $b$ such that $(9,-1,10) = a(1,-1,2) + b(3,-1,1)$. Note that you would need $a+3b=9$ and $-a-b=-1$. Adding these two you get $2b = 8$, so $b=4$; then $a=-3$, but then the last coordinate does not work out: $$-3(1,-1,2) + 4(3,-1,1) = (-3+12, 3-4, -6+4) = (9,-1,-2).$$ So the information you have does not determine the value of $T$ at $(9,-1,10)$. Given any vector $(r,s)$ in $\mathbb{R}^2$, you can find a linear transformation $U$ that agrees with $T$ on $(1,-1,2)$, on $(3,-1,1)$, and that sends $(9,-1,10)$ to $(r,s)$.

That suggests to me that you have either miscopied the problem, or else that whoever assigned the problem made a mistake. One possibility is that the first vector should have been $(1,-1,-2)$, as then you do get $(9,-1,10)$.

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Ok. Is it possible to find T for a vector that can't be represented by the other 2? Or does this not work? –  user979616 Jun 18 '12 at 4:26
@user979616: The information you have been provided only determines $T$ uniquely in the span of the two given vectors. Given any one vector $\mathbf{w}$ outside the span, if you tell me your favorite vector $\mathbf{z}$ of $\mathbb{R}^2$ I can find a linear transformation that agrees with $T$ on $(1,-1,3)$ and $(3,-1,1)$, but sends $\mathbf{w}$ to $\mathbf{z}$. Change your opinion on what your favorite vector of $\mathbb{R}^2$ is, I can give you a new linear transformation that sends $\mathbf{w}$ to your new favorite vector. So... no. You need the values at a basis to uniquely determine $T$. –  Arturo Magidin Jun 18 '12 at 4:28

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