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Been working on this assignment for hours now and I got to a point where I have a couple of questions left and I just can't get anywhere with them... One of the questions is as follows:

Let $U, V, W$ be three finite-dimensional $K$-Vectorspaces with $\dim W \ge 1$. Let $\alpha \in L(U, V)$ be a specifically chosen linear transformation. Using this we define the Pullback (What does Pullback even mean? Never had this anywhere in my lecture) of linear transformations:

$$\alpha^* \colon L(V, W) \to L(U, W); \quad f \mapsto f \circ \alpha$$

1) Show that, $\alpha^*$ is a linear transformation.

2) Show that, $\alpha^*$ is injective, when $\alpha$ is surjective.

3) Show that, $\alpha$ is injective, when $\alpha$ is surjective.

My first problem is that I'm unable to understand $\alpha^*$. What does the $f$ mean here? It's no where defined. I know what a linear transformation is and I know that in order to prove it, I have to show that:

(i) $\alpha^*(x + y) = \alpha^*(x) + \alpha^*(y)$

(ii) $\alpha^*(cx) = c\alpha^*(x)$ for all $x, y \in V$ and $c \in K$.

Any help would be appreciated. It's already 1:30 AM and I'm not getting any further without understanding $\alpha^*$...

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  • $\begingroup$ You're used to linear transformations $(x,y) \mapsto 2x+y$. What does $(x,y)$ mean here? In this case, it's a stand-in for an arbitrary vector (of two entries). For this question, the linear transformation is $f \mapsto f \circ \alpha$. What is $f$ here? In this case, its a stand-in for an arbitrary linear transformation from $U$ to $V$. $\endgroup$ – Omnomnomnom Jan 16 '17 at 0:35
  • $\begingroup$ Also, of course you haven't heard of a pullback: they're using this question to tell you what a pullback is. Hence the phrase "we define the pullback...". It's called a pullback because it takes linear transformations $f:V \to W$ and pulls them back to produce a transformation from $U$ to $V$ (i.e. it pulls the domain back, against the arrows, from $V$ to $U$). $\endgroup$ – Omnomnomnom Jan 16 '17 at 0:37
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You've written down the definition of $\alpha^*$, so it is not true that $\alpha^*$ is "nowhere defined". Granted, the definition is a bit more "abstract" than you're probably used to, but the key is to remember that all the same tools and rules apply.

For 1, we want to show that for any two functions $f,g \in L(V,W)$ and any scalar $k$, we have $$ \alpha^*(f + kg) = \alpha^*(f) + k\alpha^*(g) $$ note that on each side of the equation, we have an element of $L(U,W)$. How do we show that two linear transformations are the same? We show that for any input $u \in U$, they produce the same output. That is, your goal is to show that for any $u \in U$, we have $$ [\alpha^*(f + kg)](x) = [\alpha^*(f)](x) + k[\alpha^*(g)](x) $$ From here, all you have to do is unpack the definitions.


Part 2: Suppose that $\alpha$ is surjective. We want to show that for any $f,g$: we will only have $\alpha^*(f) = \alpha^*(g)$ if $f = g$.

So, suppose that $\alpha^*(f) = \alpha^*(g)$, which is to say that $f \circ \alpha = g \circ \alpha$. Then for all $u \in U$, we have $$ f(\alpha(u)) = g(\alpha(u)) $$ now, for every $v \in V$: there exists a $u$ such that $\alpha(u) = v$. Thus, we deduce from the above that for every $v \in V$, $$ f(v) = g(v) $$ which is to say that $f = g$.


Part 3: Suppose that $\alpha$ is injective. We want to show that for every $g \in L(U,W)$, there is an $f \in L(V,W)$ with $g = \alpha^*(f)$.

Let $u_1,\dots,u_n$ be a basis of $U$. By injectivity, $\alpha(u_1),\dots,\alpha(u_n)$ form a linearly independent set in $V$. Extend this set into a basis $v_1,\dots,v_n,v_{n+1},\dots,v_m$ of $V$, were $v_i = \alpha(u_i)$ for $i = 1,\dots,n$.

Now, consider any $g: U \to W$. In order to define map $f$ for which $f \circ \alpha = g$, it suffices to say $$ f(v_i) = w_i \qquad i = 1,\dots,n\\ f(v_i) = 0 \qquad i > n $$ and extend to a linear map by linearity.

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  • $\begingroup$ Don't you mean $\alpha^*$ rather than $\alpha_*$? $\endgroup$ – ಠ_ಠ Jan 16 '17 at 1:03
  • $\begingroup$ I did the following for i: Let f, g be elements of L(V,W) then: a*(f+g) = (f+g)(a(x)) = f(a(x)) + g(a(x)) = a*(f) + a*(g) and a*(kf) = k * f(a(x)) = k * a* -------------- =a* ~ I hope it's correct or close to being correct... This is just so abstract I can't get in my head... You gotta understand that I've only had 2 lectures about vector spaces so far, and this is my first semester studying Computer Science... So excuse me for being a bit slow here... This is all new territory for me. $\endgroup$ – Rudy Ailabouni Jan 16 '17 at 1:18
  • $\begingroup$ I did the following for i: Let f, g be elements of L(V,W) then: a*(f+g) = (f+g)(a(x)) = f(a(x)) + g(a(x)) = a*(f) + a*(g) and a*(kf) = k * f(a(x)) = k * a* \n I hope it's correct or close to being correct... This is just so abstract I can't get in my head... You gotta understand that I've only had 2 lectures about vector spaces so far, and this is my first semester studying Computer Science... So excuse me for being a bit slow here... This is all new territory for me. $\endgroup$ – Rudy Ailabouni Jan 16 '17 at 1:36
  • $\begingroup$ @RudyAilabouni looks right to me. The "part 1" should feel like you're not doing much at all. It's parts 2 and 3 that are a bit more challenging. Still, it should all come down to carefully putting the definitions together. $\endgroup$ – Omnomnomnom Jan 16 '17 at 1:45
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    $\begingroup$ Thank you! Much appreciated! $\endgroup$ – Rudy Ailabouni Jan 17 '17 at 0:36
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You are given that $\alpha$ is a linear transformation from $U$ to $V$. Then, you are defining a function $\alpha^*$ which is a function of functions. In particular, the definition says that if you take any linear map $f$ from $V$ to $W$, you can define one from $U$ to $W$ by $f\circ \alpha$. That is, $x\mapsto f(\alpha(x))$ is a linear transform from $U$ to $W$. Then, you are just giving this the name $\alpha^*f$. Treating the set of functions $L(V,W)$ and $L(U,W)$ as vector spaces under pointwise addition, it makes sense to ask whether a map between them is linear.

What you need to show is that if $f_1$ and $f_2$ are linear functions from $V$ to $W$, then $\alpha^*(c(f_1+f_2))=c\alpha^*f_1 + c\alpha^*f_2$, which is just plugging $\alpha^*$ into your definition of linearity.

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