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Let $T$: $\mathbb{R}^3 \to \mathbb{R}$ be a linear transformation.

Show that either $T$ is surjective, or $T$ is the zero linear transformation.

My approach:

First we start off by supposing T is not surjective and we want to show that $\forall \overrightarrow v \in \mathbb{R}^3, T(\overrightarrow v)=\overrightarrow 0$.

$T$ not surjective implies that $Image(T) \not = \mathbb{R}$ and that $\exists \overrightarrow y \in \mathbb{R}$ such that $$ \begin{bmatrix} a & b & c \end{bmatrix} . \overrightarrow v \not = \overrightarrow y$$ with a,b,c $\in \mathbb{R}$

$\Rightarrow$ $ax_{1}+bx_{2}+cx_{3} \not = y_{1}$

From here, I want to show that a,b,c are equal to zero. Should I say that an equation of a plane to a real number should have infinite solutions unless the coefficients are zero and conclude that a,b,c are zero or is there another way?

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  • $\begingroup$ How many subspaces does $\mathbb{R}$ have? What happens if a subspace of $\mathbb{R}$ contains a nonzero element? $\endgroup$ – egreg Mar 6 '16 at 20:39
  • $\begingroup$ Try to prove that $T$ is surjective if it is not zero. For this, choose some $x_0\in\mathbb R^3$ such that $Tx_0\neq 0$. Now, let $y\in\mathbb R$. Use the linearity of $T$ to prove that there is indeed some $x\in\mathbb R^3$ such that $Tx = y$. You can find this $x$ in the linear span of $x_0$. $\endgroup$ – Friedrich Philipp Mar 6 '16 at 20:40
  • $\begingroup$ @FriedrichPhilipp, Do I only have to prove it one way to answer the question? $\endgroup$ – Omrane Mar 6 '16 at 20:48
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A simple approach is to say that the image of $T$ is a subspace of $\Bbb R$ so it would be $\Bbb R$ or $\{0\}$. The former case is when $T$ is surjective and the last is when $T=0$.

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  • $\begingroup$ Damn can't believe I didn't see that one. Thanks a bunch! Should I also prove it the other way i.e: suppose T is not the zero linear transformation, show T is surjective or is one way enough to answer the question? $\endgroup$ – Omrane Mar 6 '16 at 20:43
  • $\begingroup$ You want to prove $T$ is either surjective or $T=0$ and it was proved. So one way is enough. $\endgroup$ – user296113 Mar 6 '16 at 20:50
  • $\begingroup$ Alright, I see. Thanks. $\endgroup$ – Omrane Mar 6 '16 at 20:51
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Assume that $T$ is not the zero transformation (this means that there is a $v$ such that $T(v) = r \neq 0$), and deduce that it is surjective (by using $v$ and linearity of $T$).

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