# not injective/not surjective linear maps

Let $S$ be the vector space of real sequences, and for $x=(x_1,x_2,\dots)$ define $\alpha(x)=(0,x_1,x_2,\dots)$ and $\beta(x)=(x_2,x_3,\dots)$. The problem was asking for few other things to do, but I got stuck at showing that the first is not injective, while the second is not surjective.

Now, I realize I need to find two distinct arguments (two different sequences), plug it in $\alpha(x)$ and get the same value, which would show that is not injective. But I can not think of any example. Could somebody guide me, how I should structure my search for such example?

As for second, could I take for example a sequence of one member $\{x_1\}$? Then it would be undefined for $\beta$ proving it's not surjective, right?

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You got it mixed up. The first one is not surjective, while the second one is not injective. –  Julian Kuelshammer Jun 22 '13 at 14:30
Are you sure that you have to prove that $\alpha$ is not injective and $\beta$ is not surjective and not the other way around. I guess that $\alpha$ is not surjective because the sequence $(x_1,x_2,...)$ with $x_1 \neq 0$ isn't part of the image of $\alpha$. –  mvcouwen Jun 22 '13 at 14:33
lol, yeah, now I see. the book mixed me, but wasn't able to spot it. But then, I was wondering, I couldn't take sequence of one membe $\{x_1\}$ and plug it in $\beta$ showing its not surjective? –  Sarunas Jun 22 '13 at 14:33
..and the first one is injective while the second is surjective. These are standard examples. –  DonAntonio Jun 22 '13 at 14:34

Maybe this is backwards? $\alpha(x)$ looks injective to me and $\beta(x)$ looks surjective.

$\alpha(x)$ is not surjective though, and I think you have the right idea. Basically, the realization is that $\alpha(x)$ has a $0$ as its first coordinate for each $x$, and so in particular, there is no sequence that maps to $(1, 1, 1,...)$.

$\beta(x)$ is not injective. Intuitively, this is because $\beta(x)$ does not give you enough information to reconstruct $x$. In particular, you have no idea what the first coordinate of $x$ is given $\beta(x)$. In fact, you can make this into a proof by picking $x, x'$ that differ only in the first coordinate and showing they map to the same thing.

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Hints:

$$\{1,1,...,1,\ldots\}\notin\text{Im}\,(\alpha)\,,\,\,\text{but}\;\;\{0,x_1,x_2,\ldots\}=\{0,y_1,\ldots\}\iff$$

$$\{x_1,x_2,\ldots\}=\{y_1,y_2,\ldots\}$$

$$\forall\;x:=\{x_1,x_2,...\}\;,\;\;\beta\{0,x_1,x_2,\ldots\}=x\;,\;\;\text{yet}$$

$$\beta\{1,2,3,\ldots\}=\beta\{0,2,3,\ldots\}$$

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