Characterization of External Direct Sum - Cooperstein, pages 359 - 360

[Math Amateur]

I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

In Section 10.2 Cooperstein writes the following, essentially about external direct sums ... ...
View attachment 5518
View attachment 5519
Cooperstein asserts that properties (a) and (b) above "characterize the space \(\displaystyle V\) as the direct sum of the spaces \(\displaystyle V_1, \ ... \ ... \ , V_n\)."

Can someone please explain how/why properties (a) and (b) above characterize the space \(\displaystyle V\) as the direct sum of the spaces \(\displaystyle V_1, \ ... \ ... \ , V_n\)?Help will be appreciated ...

Peter

 

[Deveno]

Let's investigate just two factors (extending this to arbitrarily finitely many isn't hard, but it takes more dots and more "index hell").

We want to show that given $V$, and two other vectors spaces $V_1,V_2$ along with four LINEAR maps:

$\pi_1:V \to V_1$
$\pi_2:V \to V_2$
$\epsilon_1:V_1 \to V$
$\epsilon_2:V_2 \to V$

such that:

$\pi_1\circ\epsilon_1 = 1_{V_1}$
$\pi_2\circ\epsilon_2 = 1_{V_2}$

and:

$\epsilon_1\circ \pi_1 + \epsilon_2\circ\pi_2 = 1_V$

that $V \cong V_1 \oplus V_2$.

So, one way we can do this is to exhibit an isomorphism:

$\phi: V \to V_1 \oplus V_2$ (Note I haven't said anything about whether or not the $V_i$ are even subspaces of $V$).

So here is how we will define $\phi$:

$\phi(v) = (\pi_1(v),\pi_2(v))$. Note this has the proper co-domain.

First, we'll show that $\phi$ is linear:

$\phi(v+v') = (\pi_1(v+v'),\pi_2(v+v')) = (\pi_1(v) + \pi_1(v'),\pi_2(v) + \pi_2(v'))$

$= (\pi_1(v),\pi_2(v)) + (\pi_1(v'),\pi_2(v')) = \phi(v) + \phi(v')$.

$\phi(av) = (\pi_1(av),\pi_2(av)) = (a\pi_1(v),a\pi_2(v)) = a(\pi_1(v),\pi_2(v)) = a\phi(v)$.

Now suppose that $\phi(v) = (0,0)$. We want to show $v = 0$.

Now $\phi(v) = (\pi_1(v),\pi_2(v))$, so if $\phi(v) = (0,0)$, we have (for this $v$):

$\pi_1(v) = 0$
$\pi_2(v) = 0$.

Since the $\epsilon_i$ are linear, we have: $\epsilon_1(0) = 0$, and $\epsilon_2(0) = 0$.

Thus $v = 1_V(v) = (\epsilon_1\circ \pi_1 + \epsilon_2\circ\pi_2)(v) = (\epsilon_1\circ \pi_1)(v) + (\epsilon_2\circ\pi_2)(v)$

$= \epsilon_1(\pi_1(v)) + \epsilon_2(\pi_2(v)) = \epsilon_1(0) + \epsilon_2(0) = 0 + 0 = 0$.

Thus $\phi$ is injective.

Before we tackle surjectivity, we prove a preliminary result:

$\pi_1 \circ \epsilon_2 = 0$
$\pi_2 \circ \epsilon_1 = 0$.

To prove this, we note that the $\pi_i$ are surjective, I will show this for $\pi_1$ (the proof is just the same for $\pi_2$):

Let $v_1 \in V_1$. Then we have $\pi_1(\epsilon_1(v_1)) = v_1$, so $v_1$ has the pre-image under $\pi_1$ of $\epsilon_1(v_1) \in V$.

Now for any $v \in V$, we have:

$v = \epsilon_1(\pi_1(v)) + \epsilon_2(\pi_2(v))$, so taking $\pi_2$ of both sides, we have:

$\pi_2(v) = \pi_2(\epsilon_1(\pi_1(v)) + \epsilon_2(\pi_2(v))) = \pi_2(\epsilon_1(\pi_1(v))) + \pi_2(\epsilon_2(\pi_2(v)))$

$= \pi_2(\epsilon_1(\pi_1(v))) + \pi_2(v)$.

Subtracting $\pi_2(v)$ from both sides, we have:

$0 = \pi_2(\epsilon_1(\pi_1(v)))$.

Now since $\pi_1$ is surjective, we can write ANY $v_1 \in V_1$ as $\pi_1(v)$ for SOME $v \in V$, so the above shows that (for such a $v$ so that $v_1 = \pi(v)$):

$0 = \pi_2(\epsilon_1(v_1))$, and thus this holds for any $v_1 \in V_1$.

The other condition is proved similarly.

Finally, let $(v_1,v_2)$ be any element of $V_1 \oplus V_2$.

Consider $\epsilon_1(v_1) + \epsilon_2(v_2) \in V$.

We have $\phi(\epsilon_1(v_1) + \epsilon_2(v_2)) = \phi(\epsilon_1(v_1)) + \phi(\epsilon_2(v_2))$

$= (\pi_1(\epsilon_1(v_1)),\pi_2(\epsilon_1(v_1))) + (\pi_1(\epsilon_2(v_2)),\pi_2(\epsilon_2(v_2)))$

$= (v_1,0) + (0,v_2) = (v_1,v_2)$, and $\phi$ is surjective.