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So I have this problem here which I'm not exactly sure of. I need to find the formula for a linear transformation. Here's what it says:

Find the formula for $ T(x,y,z)$ for the linear transformation $ T:\mathbb{R}^3 \rightarrow \mathbb{R}^2$

such that:

$T(1,1,1) = (-1,0)$

$T(-1,1,1) = (-3,-2)$

$T(1,1,-1) = (3,0$)

I'm not sure what to do with that information, other than assuming that those three vectors within T are all basis.

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1 Answer

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Any linear map is known if we have its values in a basis. So the first thing to do is seeing if $$\{(1,1,1),(-1,1,1),(1,1,-1) \}$$ is a basis for $\Bbb R^3$. You don't need to assume it, since we can check it: $$\begin{vmatrix} 1 & 1 & 1 \\ -1 & 1 & 1 \\ 1 & 1 & -1\end{vmatrix} = -2-(0)+(-2) = -4 \neq 0,$$so we're good. Now you take $(x,y,z)$ and write it as: $$(x,y,z) = \alpha (1,1,1)+\beta(-1,1,1)+\gamma(1,1,-1),$$ with $\alpha,\beta,\gamma \in \Bbb R$. You can find $\alpha,\beta$ and $\gamma$ by solving: $$\begin{cases}x = \alpha - \beta + \gamma \\ y = \alpha + \beta + \gamma \\ z = \alpha + \beta -\gamma \end{cases}$$I trust that you can do this. Note that $\alpha,\beta$ and $\gamma$ will stay in terms of $x,y$ and $z$, though. Since $T$ is linear, we have: $$T(x,y,z) = \alpha T(1,1,1)+\beta T(-1,1,1)+\gamma T(1,1,-1),$$ but you know all of those values now. Go for it!

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