# Showing that V is a direct sum of two subspaces

1. Apr 8, 2014

### Seydlitz

Hi guys, I have this general question.

If we are asked to show that the direct sum of $U+W=V$where $U$ and $W$ are subspaces of $V=\mathbb{R}^{n}$, would it be possible for us to do so by showing that the generators of the $U$ and $W$ span $V$? Afterwards we show that their intersection is a zero-vector. For example:

$U$ is a subspace generated by $(0,1)$ and $W$ is a subspace generated by $(2,2)$. Clearly those generators span two dimensional $V$, and their intersection is $(0,0)$. Therefore the conclusion can be made that their direct sum is $V$. Is this kind of reasoning okay?

2. Apr 8, 2014

### jbunniii

Yes, this reasoning is valid. In general, $V$ is the direct sum of $U$ and $W$ if and only if $V = U+W$ and $U \cap W = \{0\}$. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.

However, be aware that this only works for two subspaces. If you have three or more subspaces, say $\{U_i\}_{i=1}^{N}$, then it's possible to have $V = U_1 + \ldots U_N$ and $U_i \cap U_j = \{0\}$ for all $i \neq j$, but the sum is not direct. It's a good exercise to construct an example where this occurs.

3. Apr 9, 2014

### Seydlitz

Ah ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? $V_1 - V_2 = V_1 + V_2$ if we define subtraction to be $\{ x-y | x \in V_1, y \in V_2 \}.$ This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?

4. Apr 9, 2014

### jbunniii

Since $y \in V_2$ if and only if $y \in -V_2$, it doesn't matter whether you add or subtract elements of $V_2$, you get the same result in both cases: $V_1 - V_2 = V_1 + V_2$.

Admittedly, the notation $V_1 - V_2$ can seem a bit misleading. Fortunately, since it is the same as $V_1 + V_2$, there's no reason to use $V_1 - V_2$.

Note that you could even define $aV_1 + bV_2 = \{a x + b y | x \in V_1, y \in V_2\}$, where $a$ and $b$ are nonzero scalars. Then $V_1 - V_2$ is just a special case with $a=1$ and $b=-1$.

Once again we have $aV_1 + bV_2 = V_1 + V_2$ for any nonzero $a,b$, because $x \in aV_1$ if and only if $x \in V_1$ and similarly, $y \in V_2$ if and only if $y \in bV_2$. So we can just stick with the notation $V_1 + V_2$ since all the other "linear combinations" of $V_1$ and $V_2$ give the same result.

Last edited: Apr 9, 2014
5. Apr 9, 2014

### jbunniii

By the way, here's a bit of extra info in case it is helpful. Suppose we start with a collection of subspaces $\{U_i\}_{i=1}^{N}$ of $V$. We may be interested in combining the $U_i$'s to form a larger subspace. The naive thing to do would be to form the set theoretic union $\cup_{i=1}^{N}U_i$. But this won't generally be a subspace.

So what we really want is the smallest subspace containing all of the $U_i$ (or equivalently, the smallest subspace containing $\cup_{i=1}^{N}U_i$). This turns out to be exactly $U_1 + \ldots + U_N$. Proof: certainly $U_1 + \ldots + U_N$ is a subspace containing each $U_i$. If $S$ is another such subspace then it must contain all elements of the form $u_1+\ldots+u_N$ with $u_i \in U_i$. Thus $U_1 + \ldots + U_N \subset S$ so $U_1 + \ldots + U_N$ is the smallest such subspace.

Now in general, the $U_i$'s may not be "linearly independent" of each other: maybe there is some nonzero element of $U_1$ which can be expressed as a linear combination of elements of the other $U_i$'s. But if they ARE linearly independent, then we say that the sum is a direct sum, and we write it as $U_1 \oplus \ldots \oplus U_N$ instead of $U_1 + \ldots + U_N$. We can obtain a basis for a direct sum $U_1 \oplus \ldots \oplus U_N$ by simply selecting a basis $B_i$ for each $U_i$ and taking the union: $B = \cup_{i=1}^{N} B_i$. Indeed, this union is disjoint because there are no common elements among the $B_i$'s. In the finite-dimensional case, this immediately tells us that
$$\text{dim}(U_1 \oplus \ldots \oplus U_N) = \sum_{i=1}^{N}\text{dim}(U_i)$$
This is not true if the sum is not direct. In that case, all we can say is that
$$\text{dim}(U_1 + \ldots + U_N) \leq \sum_{i=1}^{N}\text{dim}(U_i)$$

Last edited: Apr 9, 2014
6. Apr 10, 2014

### Seydlitz

Wow thanks for your detailed explanation. I just want to make sure for this last time that I understand this correctly.

Can I also intuitively think that $V_2$ = $-V_2$ if $V$ is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)

7. Apr 10, 2014

### jbunniii

Sure, the fact that $V_2$ is a subspace and therefore contains its additive inverses is exactly the reason why $y \in V_2$ if and only if $y \in -V_2$, and this latter statement is the definition of set equality: we conclude that $V_2 = -V_2$.

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