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$L_1, L_2$ are two linear maps with $L_1 : \mathbb{R}^3 \to \mathbb{R}^2,\; L_2 :\mathbb{R}^2 \to \mathbb{R}^3$ and be $L_2 \circ L_1 : \mathbb{R}^3\to \mathbb{R}^3$ their concatenation.

Can $L_2\circ L_1$ be injective?
Can $L_2\circ L_1$ be surjective?

I found out that $L_2\circ L_1$ can be injective and surjective if and only if $L_2,\;L_2$ is bijective. That means, that
$\forall y\in\mathbb{R}^2$ exists only one $x\in\mathbb{R}^3:L_1(x)=y$ and
$\forall y'\in\mathbb{R}^3$ exists only one $x'\in\mathbb{R}^2:L_2(x')=y'$


Let $L_2(y)=x',\;L_1(x)=y$ with $x,x'\in\mathbb{R}^3,\;y\in \mathbb{R}^2$ and $L_2\circ L_1=L_2(L_1(x))$
Then $L_2\circ L_1=L_2(L_1(x))=L_2(y)=x'$ and because $L_1$ bijective, it reaches all $y\in \mathbb{R}^2$ and because $L_2$ bijective, we can reach all $x'$. Therefore, $L_2 \circ L_1$ is bijective.

Is this proof correct or am I missing something?


EDIT: $L_2\circ L_1$: First, we need to prove, that $L_1$ injective:

In order to be injective, $\dim(\operatorname{ker}(L_1))=1$ must hold.

$\dim(\mathbb{R}^3)=\dim(\operatorname{im}(L_1))+\dim(\operatorname{ker}(L_1))$.

Because $\dim(\mathbb{R}^2)=2 \implies \dim(\operatorname{im}(L_1))\leq2\implies \dim(\operatorname{ker}(L_1))\geq1$ $\implies \exists x_1,x_2\in L_1:L_1(x_1)=L_1(x_2)$ with $ x_1\neq x_2$ $\implies$ $L_1$ not injective and $L_2\circ L_1$ not injective

Second $L_2\circ L_1$ surjective? Need to prove, that $L_2$ surjective:

$\dim(\mathbb{R}^3)=\dim(\operatorname{im}(L_2))+\dim(\operatorname{ker}(L_2))$

$\dim(R^2)=2$ and $\dim(R^3)=3 \implies \dim(im(L_2))\leq 3$ Therefore, $\dim(ker(L_2))\geq -1$

What means, that $\dim(ker(L_2))=-1$?

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    $\begingroup$ Yes, missing more things. $L_i$ are linear.. Watch the dimensions.. Injective and surjective are posed as 2 separate questions. $\endgroup$ – Berci May 29 '19 at 7:07
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    $\begingroup$ Neither of the maps can be bijective and because of the answers supposing the composition is injective or surjective leads to a contradiction $\endgroup$ – user515599 May 29 '19 at 7:48
  • $\begingroup$ I found out, that $\dim(ker(L_2))≥−1$ What does this mean for surjectivity? $\endgroup$ – Doesbaddel May 29 '19 at 7:57
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    $\begingroup$ Empty statement. Dimensions are always nonnegative $\endgroup$ – user515599 May 29 '19 at 8:06
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    $\begingroup$ what? $L_2$ is not surjective because the image of a linear map has dimension atmost of its domain. The image is a plane in 3D. It obviously cannot cover all of the space $\endgroup$ – user515599 May 29 '19 at 8:20
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A linear map $$T:\mathbb{R}^3 \to \mathbb{R}^2,$$

can never be injective. Best case scenario it's a projection into a the plane.

A linear map

$$T:\mathbb{R}^2 \to \mathbb{R}^3,$$

can never be surjective. Best case: you're embedding a plane (through the origin) into $3$-D space.

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  • $\begingroup$ Thank you for pointing that out. I made an edit. $\endgroup$ – Doesbaddel May 29 '19 at 7:42
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Hints.

  • If $L_2\circ L_1$ is injective, then $L_1$ is injective: this is a general property of functions and does not depend on linearity. In your case, is it possible for $L_1$ to be injective?

  • If $L_2\circ L_1$ is surjective, then $L_2$ is surjective: again this does not depend on linearity. In your case, is it possible for $L_2$ to be surjective?

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  • $\begingroup$ Thank you for pointing that out. $L_1$ is obviously not injective. $\endgroup$ – Doesbaddel May 29 '19 at 7:43
  • $\begingroup$ I found out, that $\dim(ker(L_2))\geq-1$ What does this mean for surjectivity? $\endgroup$ – Doesbaddel May 29 '19 at 7:55
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    $\begingroup$ Since you already know that the dimension of a vector space has to be greater than or equal to zero, $\dim(\ker(L_2))\ge-1$ means nothing. $\endgroup$ – David May 29 '19 at 23:56

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