# Can $L_2\circ L_1$ be injective and/or can $L_2\circ L_1$ be surjective?

$$L_1, L_2$$ 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 $$\dim(ker(L_2))=-1$$?

• Yes, missing more things. $L_i$ are linear.. Watch the dimensions.. Injective and surjective are posed as 2 separate questions. – Berci May 29 '19 at 7:07
• Neither of the maps can be bijective and because of the answers supposing the composition is injective or surjective leads to a contradiction – user515599 May 29 '19 at 7:48
• I found out, that $\dim(ker(L_2))≥−1$ What does this mean for surjectivity? – Doesbaddel May 29 '19 at 7:57
• Empty statement. Dimensions are always nonnegative – user515599 May 29 '19 at 8:06
• 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 – user515599 May 29 '19 at 8:20

## 2 Answers

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.

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

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?

• Thank you for pointing that out. $L_1$ is obviously not injective. – Doesbaddel May 29 '19 at 7:43
• I found out, that $\dim(ker(L_2))\geq-1$ What does this mean for surjectivity? – Doesbaddel May 29 '19 at 7:55
• 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. – David May 29 '19 at 23:56