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Let $T\colon \mathbb{R}^2\to\mathbb{R}^3$ be a linear transformation defined by

$$T(x,y) = (10x+2y, -10x+10y, -8x-6y)$$

Find a vector $w$ that is not in the image of $T$.

I just really have no clue on where to start with this. Any ideas?

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3 Answers 3

up vote 4 down vote accepted

$T(x,y)=(10x+2y,−10x+10y,−8x−6y)=x(10,-10,-8)+y(2,10,-6)$. So the image of $T$ is the plane spanned by $(10,-10,-8)$ and $(2,10,-6)$. Any vector not in this plane will not be in the image, for example the cross product of the spanning vectors $(10,-10,-8)\times(2,10,-6)$.

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Well for a vector $(a,b,c)$ to lie in the image of $T$, there must be values $x$ and $y$ such that:

$10x+2y = a$

$-10x+10y = b$

$-8x-6y = c$

Notice in particular that we cannot have $a,b,c$ equal and non-zero since then the equations would have no solutions (check this).

So any vector such as $(1,1,1)$ will do.

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Hint: First find a basis for the image of $T$

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@Linax: What are the basis for $\mathbb R^2$? –  Babak S. Sep 13 '12 at 13:56

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