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L a linear map of $\mathbb{R}^m$ in $\mathbb{R}^n$.
1)Does a linear map of two linear dependent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear independent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?

2)a)Does a linear map of two linear independent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear dependent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?
b)And if it is what are the consequences for $ker(L)$?

1)No, because if two vectors are dependent:
$\exists k \in \mathbb{R}$,
so that: $k\underline{x}=\underline{y}$
and a linear map to two is linear dependent vectors $\in \mathbb{R}^n$is only possible by two vectors $\in \mathbb{R}^m$.

2)a)Yes, f.ex.: $f:\mathbb{R}^3 -> \mathbb{R}^2$,
with $f(x,y,z)^T = (x, y-z)^T$ for $(0,1,0)^T$ and $(0,0,1)^T$.
b)with the rank-nullity theorem:
$ dim V - dim (im T)= dim (ker T)$
$3 - 2 = 1 = dim (ker T)$ in the example

My questions are:
Is my explication for 1) & 2) sufficient?and if not how can I improve it?
Is 2)b) I don't really know how to interprete that...

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I'd like to suggest the following improvements for your MathJax code: use \to or \rightarrow for an arrow $\to$, and use \dim for $\dim$, \ker for $\ker$, and \operatorname{im} for $\operatorname{im}$ (these expressions should be upright). Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. –  Zev Chonoles May 28 '13 at 22:25
Your questions (1) and (2)(a) look identical... –  DonAntonio May 28 '13 at 22:26
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2 Answers

up vote 0 down vote accepted


I'd explain (1) as follows, applying explicitly a linear map:

$$x,y\in\Bbb R^m\;,\;\;a\in\Bbb R\;,\;\;T\;\text{a linear map}\;,\; x=ay\implies Tx= T(ay)=aTy$$

You don't need to worry about (2)-(b) since (1)=(2)-(a) cannot be.

Edit: After the correction to (2)-(a): the zero map is a trivial example of this possibility, and in general, if

$$x,y\in\Bbb R^m\;,\;a\in\Bbb R\;\wedge\;Ty=aTx=T(ax)\implies T(y-ax)=0\implies \ker T\neq 0$$

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I made a mistake in 2)a) it should be the linear map of 2 independent vectors to 2 dependent vectors... –  Phil May 28 '13 at 22:38
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Well, it seems you get the idea.

To improve, introduce a letter for the linear map we are talking about, say, $T$. Then you can write like $k\,T(x)=T(kx)=T(y)$.

For 2b, if $k\,T(x)=T(y)$ then $kx-y\in\ker T$.

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