Can't understand this operation: $\alpha^∗ \colon L(V, W) \to L(U, W); f \mapsto f \circ \alpha$

Question

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 $\alpha^*$...

Answer 1

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.

Answer 2

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.