4.4 Sums and direct sum

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$Schilling.jpg$
Contributed by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling
Professor (Mathematics) at University of California, Davis

Throughout this section, $V$ is a vector space over $\mathbb{F}$, and $U_1 , U_2 \subset V$ denote subspaces.

Definition 4.4.1: (subspace) sum

Let $U_1 , U_2 \subset V$ be subspaces of $V$ . Deﬁne the (subspace) sum of $U_1$

Figure 4.4.1: The union $U \cup U^\prime$ of two subspaces is not necessarily a subspace.

and $U_2$ to be the set

\[ U_1 + U_2 = \{u_1 + u_2 | u_1 \in U_1 , u_2 \in U_2 \}. \tag{4.4.1}\]

Check as an exercise that $U_1 + U_2$ is a subspace of $V$ . In fact, $U_1 + U_2$ is the smallest subspace of $V$ that contains both $U_1$ and $U_2$ .

Example 4.4.2. Let

\[ U_1 = \{(x, 0, 0) \in \mathbb{F}^3 | x \in \mathbb{F}\}, \\ U_2 = \{(0, y, 0) \in \mathbb{F}^3 | y \in \mathbb{F}\}.\]

Then

\[U_1 + U_2 = \{(x, y, 0) \in \mathbb{F}^3 | x, y \in \mathbb{F}\}. \tag{4.4.2}\]

If, alternatively, $U_2 = \{(y, y, 0) \in \mathbb{F}^3 | y \in \mathbb{F}\}$, then Equation (4.4.2) still holds.

If $U = U_1 +U_2$ , then, for any $u \in U$, there exist $u_1 \in U_1$ and $u_2 \in U_2$ such that $u = u_1 +u_2.$

If it so happens that $u$ can be uniquely written as $u_1 + u_2$ , then $U$ is called the direct sum of $U_1$ and $U_2.$

Definition 4.4.3: Direct Sum

Suppose every $u \in U$ can be uniquely written as $u = u_1 + u_2$ for $ u_1 \in U_1$ and $u_2 \in U_2$ . Then we use

\[U = U_1 \oplus U_2\]

to denote the direct sum of $U_1$ and $U_2.$

Example 4.4.4. Let
\[U_1 = \{(x, y, 0) \in \mathbb{R}^3 | x, y \in \mathbb{R}\}, \\
U_2 = \{(0, 0, z) \in \mathbb{R}^3 | z \in \mathbb{R}\}.\]
Then $\mathbb{R}^3 = U_1 \oplus U_2$. However, if instead
\[U_2 = \{(0, w, z) | w, z \in \mathbb{R}\},\]
then $\mathbb{R}^3 = U_1 + U_2$ but is not the direct sum of $U_1$ and $U_2$ .
Example 4.4.5. Let
\[U_1 = \{p \in \mathbb{F}[z] | p(z) = a_0 + a_2 z^2 + \cdots + a_{2m}z^{2m} \}, \\
U_2 = \{p \in \mathbb{F}[z] | p(z) = a_1z + a_3z^3 + \cdots + a_{2m+1}z^{2m+1} \}.\]
Then $\mathbb{F}[z] = U_1 \oplus U_2.$

Proposition 4.4.6. Let $U_1 , U_2 \subset V$ be subspaces. Then $V = U_1 \oplus U_2$ if and only if the following two conditions hold:

$V = U_1 + U_2;$
If $0 = u_1 + u_2$ with $u_1 \in U_1$ and $u_2 \in U_2$ , then $u_1 = u_2 = 0.$

Proof.
$(“\Rightarrow”)$ Suppose $V = U_1 \oplus U_2$. Then Condition 1 holds by deﬁnition. Certainly $0 = 0 + 0$, and, since by uniqueness this is the only way to write $0 \in V$ , we have $u_1 = u_2 = 0$.

$(“\Leftarrow”)$ Suppose Conditions 1 and 2 hold. By Condition 1, we have that, for all $v \in V$ , there exist $u_1 \in U_1$ and $u_2 \in U_2$ such that $v = u_1 + u_2$ . Suppose $v = w_1 + w_2$ with $w_1 \in U_1$ and $w_2 \in U_2$ . Subtracting the two equations, we obtain

\[0 = (u_1 − w_1 ) + (u_2 − w_2 ),\]

where $u_1 − w_1 \in U_1$ and $u_2 − w_2 \in U_2$. By Condition 2, this implies $u_1 − w_1 = 0$ and $u_2 − w_2 = 0$, or equivalently $u_1 = w_1$ and $u_2 = w_2$ , as desired.

Proposition 4.4.7. Let $U_1 , U_2 \subset V$ be subspaces. Then $V = U_1 \oplus U_2$ if and only if the following two conditions hold:

$V = U_1 + U_2;$
$U_1 \cap U_2 = \{0\}.$

Proof.
$(“\Rightarrow”)$ Suppose $V = U_1 \oplus U_2$. Then Condition 1 holds by deﬁnition. If $u \in U_1 \cap U_2$ , then $0 = u + (−u)$ with $u \in U_1$ and $−u \in U_2$ (why?). By Proposition 4.4.6, we have $u = 0$ and $−u = 0$ so that $U_1 \cap U_2 = \{0\}.$

$(“\Leftarrow”)$ Suppose Conditions 1 and 2 hold. To prove that $V = U_1 \oplus U_2$ holds, suppose that

\[0 = u_1 + u_2, \rm{~where~} u_1 \in U_1 \rm{~and~} u_2 \in U_2. \tag{4.3}\]

By Proposition 4.4.6, it suﬃces to show that $u_1 = u_2 = 0$. Equation (4.3) implies that $u_1 = −u_2 \in U_2$. Hence $u_1 \in U_1 \cap U_2$, which in turn implies that $u_1 = 0$. It then follows that $u_2 = 0$ as well.

Everything in this section can be generalized to m subspaces $U_1 , U_2 , \ldots U_m,$ with the notable exception of Proposition 4.4.7. To see, this consider the following example.

Example 4.4.8. Let

\[U_1 = \{(x, y, 0) \in \mathbb{F}^3 | x, y \in \mathbb{F}\}, \\
U_2 = \{(0, 0, z) \in \mathbb{F}^3 | z \in \mathbb{F}\}, \\
U_3 = \{(0, y, y) \in \mathbb{F}^3 | y \in \mathbb{F}\}.\]

Then certainly $\mathbb{F}^3 = U_1 + U_2 + U_3$ , but $\mathbb{F}^3 \neq U_1 \oplus U_2 \oplus U_3$ since, for example,

\[(0, 0, 0) = (0, 1, 0) + (0, 0, 1) + (0, −1, −1).\]

But $U_1 \cap U_2 = U_1 \cap U_3 = U_2 \cap U_3 = \{0\}$ so that the analog of Proposition 4.4.7 does not hold.

Contributors

Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.