# Linear independence, Spanning, Injectivity, and Surjectivity

We were told in class that when we proved that a function between two vector spaces is injective and surjective, essentially we were proving that a basis for the vector space is linearly independent and spanning, respectively. I was just wondering how this works exactly?

I understand the definitions of these concepts, I am just not sure how to use the definitions of linear independence and spanning to prove that a function is bijective.

Let $T\colon V\to W$ be a linear transformation of vector spaces. Choose a basis $\mathcal{B}$ of $V$. An obvious question is whether the image $T(\mathcal{B}):=\{T(\mathbf{e})\mid \mathbf{e}\in \mathcal{B}\}$ of this basis is a basis of $W$. There are two ways this can fail: (1) $T(\mathcal{B})$ is linearly dependent, and (2) $T(\mathcal{B})$ doesn't span $W$.

Here's what you should prove for yourself:

1. $T(\mathcal{B})$ is linearly independent if and only if $T$ is injective.
2. $T(\mathcal{B})$ spans $W$ if and only if $T$ is surjective.

For 1., you'll want to use the fact that $T$ is injective if and only if its kernel is $0$. If that fact is unfamiliar to you, prove it also!

This basically relies on the following propositions. Let $V, W$ be vector spaces.

Lemma: Given $E \in Hom(V, W)$, $E$ is mono $\iff Ker(E) = {0}$

Proof:

$\implies$ ) Suppose $Ker(E) \neq {0} \iff \exists x \in Ker(E) : x \neq 0$, but then we have that $E(0) = 0 \wedge E(x) = 0$, hence it is not injective.

$\impliedby$ ) Let $a, b \in V : E(a) = E(b)$, this happens iff $E(a) - E(b) = 0 \iff E(a - b) = 0$, but then $a - b = 0 \iff a = b$, since 0 is the only element mapped to 0.

Corollary: Given $E \in Hom(V, W)$ mono, $\{v_i : i \in I\}$ is linearly independent set $\iff$ $\{f(v_i) : i \in I\}$ is linearly independant

Lemma: Given $E \in Hom(V, W), E$ is epi, $\{v_i : i \in I\}$ spans $V \implies \{f(v_i) : i \in I\}$ spans $Im(f)$

Proof:

Take any $v \in V$, then $v = \sum_{i \in I} \alpha_i v_i$, hence $f(v) = \sum_{i \in I} \alpha_i f(v_i) \in \langle \{ f(v_i) : i \in I \} \rangle$ The other inclusion is clear, since $f(v_i) \in Im(f) \forall i \in I$, thus: $Im(f) = \langle \{ f(v_i) : i \in I \} \rangle$

Hence, suppose you prove that a linear transform $f$ is an isomorphism, and you have $\beta = \{v_i : i \in I\}$ a basis of $V$, the first corollary says that $f(\beta) = \{ f(v_i) : i \in I \}$ is linearly independant. Since $Im(f) = W$, we also have that $f(\beta)$ spans $W$, thus $f(\beta)$ is also a basis for $W$.

There's a few things that are unclear in your question.

First, by definition a basis of a vector space is a linearly independent spanning set. It's cardinality, ie the number of vectors in the basis gives us a measure of the size of the space - it's called the dimension.

Now given two different spaces with the same dimension, we can map bijectively the basis of one to the other; and then by linearity this extends to a linear bijection map between the two.

I believe your confusion comes from the sentence "we were proving that a basis for the vector space is linearly independent and spanning".

What it is that you're proving is that the image (the set of images) of a base for the first space is again a base for the second.

More precisely: let

$f:V \to W$

a linear map between vector spaces. Say $\mathscr{B}$ is a base for $V$. Then what you have to see is that the set $f(\mathscr{B}) = \{f(b)\in W \textrm{such that } b\in\mathscr{B}\}$ is actually a basis for $W$ iff $f$ is a (linear) bijection.

This is easy. Let's do one of the implications for surjectivity. So assume $f(\mathscr{B})$ is a basis for $W$, and I'd like to see that $f$ is surjective. So we take any $w\in W$ and see it has a preimage. Since $f(\mathscr{B})$ is a basis, we can write

$w = \sum a_i f(b_i)$

for some $a_i$ in the field, and some (finite) $b_i\in\mathscr{B}$. By linearity of $f$, we can write

$w = \sum a_i f(b_i) = \sum f(a_i b_i) = f(\sum a_i b_i)$

But $\sum a_i b_i$ is an element in $V$, since is a (finite) linear combination of elements in the basis. Thus $w$ has a preimage. You can work out the other implications to practice with the concepts!