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Let $T$ be a linear transformation from $\mathbb R^3$ to $\mathbb R^3$. Determine whether or not $T$ is onto.

  1. Suppose $T(4,-2,-2)= u$, $T(0,2,2)= v$, $T(4,1,0) = u + v$. Is $T$ onto?
    I really do not know how to show if (1) is onto. I tried to to set $u$, $v$, $w = u+v$ as column vectors and from there determine if the columns span. But I'm not so sure if that's correct.

Or if $T$ is a one-to-one function then is also onto?

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Ordinarily it might help to know what $u$ and $v$ are, but in this case it doesn't affect the answer. Are the vectors $u$, $v$, and $u+v$ linearly independent? Do they span $\mathbb R^3$? – Jonas Meyer Dec 7 '12 at 7:21

Note that $(4,-2,-2), (0,2,2), (4,1,0)$ are linearly independent, so they span $\mathbb{R}^3$. Hence $T$ is completely defined by its action on these vectors.

You have the rank-nullity theorem that states $\dim {\cal R}(T) + \dim \ker T = 3$. $T$ will be onto iff $\dim {\cal R}(T) = 3$.

You have $T(4,-2,-2) = u$ and $T((4,1,0)-(0,2,2)) = u$. What does that say about $\dim \ker T$? (In fact, you can compute $\ker T$ explicitly.)

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