# Showing that V is a direct sum of two subspaces

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?

## Answers and Replies

jbunniii
Science Advisor
Homework Helper
Gold Member
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.

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.

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?

jbunniii
Science Advisor
Homework Helper
Gold Member
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?
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:
jbunniii
Science Advisor
Homework Helper
Gold Member
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:
Wow thanks for your detailed explanation. I just want to make sure for this last time that I understand this correctly.

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##.

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)

jbunniii
Science Advisor
Homework Helper
Gold Member
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)
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##.