Understanding the concept of direct sum

[JD_PM]

TL;DR Summary

I want to understand what I am missing in the generalization I made for the direct sum when $3$ subspaces are involved.

Given two subspaces $U_1, U_2$, I understand the concept of direct sum

$$ W= U_1 \oplus U_2 \iff W= U_1 + U_2, \quad U_1 \cap U_2 = \{ 0 \}$$

Where $W$ is a subspace of $V$.

I am trying to generalize it for more than $2$ subspaces, say $3$. I thought of the following.

$$ W= U_1 \oplus U_2 \oplus U_3 \iff U_1 \cap U_2 = \{ 0 \}, U_1 \cap U_3 = \{ 0 \}, U_2 \cap U_3 = \{ 0 \}, U_1 + U_2 + U_3 = W $$

It does not seem to have the same structure that for the statement with $k$ subspaces

\begin{align*}
W= U_1 \oplus U_2 \oplus ... \oplus U_k \iff& U_i \cap \left(U_1 + ... + U_{i-1} + U_{i+1} + ... + U_k\right) = \{ 0 \} \\
&U_1 + U_2 + ... + U_k = W
\end{align*}

In particular, the issue lies on the intersection statement. Might you explain why my thought is faulty? I should be able to find a counterexample once I see it :)

Thanks!

 

[Office_Shredder]

Let $W=\mathbb{R}^2$, $U_1=span( (1,0)),U_2=span(1,1),U_3=span(0,1)$ be three one dimensional subspaces Under your attempt, we would have $W=U_1\oplus U_2 \oplus U_3$ but that's not desirable for the definition of direct sum.

 

[JD_PM]

Let $W=\mathbb{R}^2$, $U_1=span( (1,0)),U_2=span(1,1),U_3=span(0,1)$

But, if I am not mistaken, given

$$U_1 = \{ (x, 0) | x \in \Bbb R\}, \quad U_2 = \{ (x, y) | x, y \in \Bbb R\}, \quad U_3 = \{ (0, y) | y \in \Bbb R\}$$

We could find $\Bbb R^2= U_1 + U_2 + U_3$ in more than one way. Another would be say $U_1=span( (1,0)),U_2=span(1,2),U_3=span(0,1)$. So I do not see it as a counterexample as the sum is not unique.

Am I missing something?EDIT: Oops I realized you meant

$$U_1 = \{ (x, 0) | x \in \Bbb R\}, \quad U_2 = \{ (y, y) | y \in \Bbb R\}, \quad U_3 = \{ (0, y) | y \in \Bbb R\}$$

So indeed, we can write $\Bbb R^2= U_1 + U_2 + U_3$ in a unique way (up to a scalar factor). EDIT 2: " we can write $\Bbb R^2= U_1 + U_2 + U_3$ in a unique way (up to a scalar factor)" is not true, see #5.

 

[JD_PM]

I have been thinking that the key might be in finding a counterexample that satisfies $W= U_1 \oplus U_2 \oplus U_3$ but fails to meet $\left(U_1+U_2\right)\cap U_3 = \{ 0 \}$.

However I do not seem to see it... Might you give me a hint?

 

[JD_PM]

Let's start all over again.

I found a more convincing definition of direct sum (Axler, page 21)

$U_1 + U_2 + \ ... \ + U_k$ is a direct sum if $x \in U_1 + U_2 + \ ... \ + U_k$ can be written in a unique way as $x = u_1 + u_2 + \ ... \ + u_k$, where $u_i \in U_i$

$$U_1 = \{ (x, 0) | x \in \Bbb R\}, \quad U_2 = \{ (y, y) | y \in \Bbb R\}, \quad U_3 = \{ (0, y) | y \in \Bbb R\}$$

$\Bbb R^2= U_1 \oplus U_2 \oplus U_3$ does not hold in this case because there are (at least) two ways of obtaining the zero vector

$$(0, 0) = (1, 0) + (-1, -1) + (0, 1)$$

And the trivial case

$$(0, 0) = (0, 0) + (0, 0) + (0, 0)$$

So, if I am not mistaken, that is the reason why that is a valid counterexample to the given definition of the direct sum

 

[nuuskur]

There are several equivalent formulations in finite dimensions. Equivalent are:

1. $\sum X_i$ is a direct sum
2. $0$ decomposes uniquely
3. $X_j\cap \sum _{i\neq j} X_i = \{0\}$ for every $i=1,\ldots, n$
4. $\dim \sum X_i = \sum \dim X_i$

 

[FactChecker]

So, if I am not mistaken, that is the reason why that is a valid counterexample to the given definition of the direct sum

You got it, but allow me to be a little nitpicky about your words. It is a counterexample to your definition attempt being a correct definition of the direct sum.

 

[mathwonk]

The first step is to master the correct definition of a direct sum, from category theory. A direct sum of three spaces A,B,C is a space X with 3 subspaces A,B,C such that for any target space Y, any three maps A-->Y, B-->Y, C-->Y, always extends uniquely to a map X-->Y. It follows immediately that the union of the three spaces generate X, and that they intersect pairwise in zero. You, may then try to check if the converse holds. I doubt it, by visualizing three lines in the plane, all through zero. Youm might want to add that the sum of any two meets the third in zero.