Suggest edit — direct sum

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---
title: "direct sum"
type: concept
related: [Additive Identity, Sum Of Subsets, Direct Sum, Inverses, Linear Independence]
source: https://www.jemoka.com/posts/kbhdirect_sum/
confidence: high
status: active
---

A direct sum is a sum of subspaces (not just subsets!!) where there&rsquo;s only one way to represent each element.
constituents subspaces of $V$ named $U_1, \dots, U_{m}$
requirements The sum of subsets of $U_1+\dots+U_{m}$ is called a direct sum IFF:
each element in $U_1+\dots +U_{m}$ can only be written in one way as a sum $u_1 +\dots +u_{m}$ (as in, they are linearly independent?)
We use $\oplus$ to represent direct sum.
additional information why is it called a direct sum? Something is not a direct sum if any of its components can be described using the others. Its kind of line linear independence but! on entire spaces.
a sum of subsets is a direct sum IFF there is only one way to write $0$ Given $U_1, \dots, U_{m}$ are subspaces of $V$, then $U_1+\dots +U_{m}$ is a direct sum IFF the only way to write $0$ as a sum $u_1 +\dots +u_{m}$ is by taking each element to $0$.
Proof:
if&mdash; If some $U_1 + \dots +U_{m}$ is a direct sum, definitionally there is only one way to write $0$. And you can always write $0$ by taking all the constituents to $0$ as they are subspaces, so the additive identity exists.
only if&mdash; We are given that there is only one way to write $0$, that:
\begin{equation} 0 = u_1+ u_2+ \dots+ u_{m}: u_j \in U_{j} \end{equation}
as $U_{j}$ are all subspaces, and the additive identity exists, we can say that $u_1=u_2=\dots =0$.
Assume for the sake of contradiction that $U_1 + \dots +U_{m}$ is not a direct sum. Therefore:
\begin{equation} \exists\ v_1 = u_1+u_2+\dots + u_{m}: u_{j} \in U_{j} \end{equation}
and
\begin{equation} \exists\ v_1 = w_1+w_2+\dots + w_{m}: w_{j} \in U_{j} \end{equation}
&ldquo;there are two unique representations of a vector given the sum of subsets&rdquo;
Subtracting these representations, then:
\begin{equation} (v_1-v_1) = (u_1-w_1) + \dots +(u_{m}-w_{m}): u_{j}, w_{j} \in U_{j} \end{equation}
Finally, then:
\begin{equation} 0 = (u_1-w_1) + \dots +(u_{m}-w_{m}): u_{j}, w_{j} \in U_{j} \end{equation}
We have established that each slot that makes up this particular sum $=0$. Therefore, $u_{i}-w_{i} = 0$. This means $ui=wi$&mdash;there are no two unique representations of $v_{1}$. Reaching contradiction. $\blacksquare$
a sum of subsets is only a direct sum IFF their intersection is set containing $0$ Take $U$ and $W$, two subspaces of $V$. $U+V$ is a direct sum IFF $U \cap W = \{0\}$.
Proof:
if&mdash; Suppose $U+V$ is a direct sum. $\forall v \in U \cap V$, as $v$ is equal to itself, we have that:
\begin{equation} 0 = v+(-v) \end{equation}
where, $v$ is in $U$ and $-v$ is in $V$ (as both $U$ and $V$ are vector spaces, both would contain $-1v=-v$ as we are given $v \in U \cap V$ and scalar multiplication is closed on both.)
By the unique representation in the definition of direct sums, you have only one way to construct this expression: namely, that $v=0$ as both are vector spaces so the additive identity exists on both.
Hence:
\begin{equation} \{0\} = U \cap V \end{equation}
only if&mdash; Suppose $U \cap W = \{0\}$. Take also $u \in U$ and $w \in W$; we can construct an expression:
\begin{equation} u + w = 0 \end{equation}
If we can show that there is only one unique combination of $u$ and $w$ to write $0$, we satisfy the previous proof and therefore $U+W$ is a direct sum.
The expression above implies that $w$ is the additive inverse of $u$; therefore; $u = -w$. As both $U$ and $W$ are vector spaces, their elements all have inverses. As $u$ is the inverse of $w$, and given the definition of sum of subsets that $u \in U$ and $w \in W$, $u$ and $w$ are both in both $U$ and $W$.
As the intersection of $U$ and $V$ is $0$, $u=w=0$. Therefore, there is only one unique representation of $0$, namely with $u=0,w=0$, making $U+W$ a direct sum. $\blacksquare$
direct sum proofs are not pairwise! Those two proofs above only deal with pairs of sum of subsets. If you have multiple subsets, they don&rsquo;t apply!
every subspace of $V$ is a part of a direct sum equaling to $V$ For every subspace $U$ of a finite-dimensional $V$, there is a subspace $W$ of $V$ for which $V = U \oplus W$.
Because $V$ is defined to be finite-dimensional, and the fact that a finite-dimensional subspace is finite-dimensional, $U$ is finite-dimensional.
Therefore, because every finite-dimensional vector space has a basis, $U$ has a basis $u_1, \dots u_{m}$.
Because bases are linearly independent, and $U \subset V$, $u_1, \dots u_{m}$ is a linearly independent list in $V$.
Because a linearly independent list expends to a basis, we can construct $u_1, \dots u_{m}, w_{1}, \dots w_{n}$ as the basis of $V$. We will construct a $W = span(w_1, \dots w_{n})$ &mdash; the space formed as the span of the &ldquo;extension&rdquo; vectors to make the basis in $V$.
Because the list $u_{j}\dots w_{k}$ we made is a basis in $V$, $U+W=V$.
You can see this because every element $v \in V$ can be constructed with a linear combination $u_1, \dots u_{m}, w_{1}, \dots w_{n}$ (again, because this list shown to be a basis of $V$ therefore it spans $V$.) Then, to show that $U+W=V$, we can collapse $a_{1}u_1\dots + a_{m}u_{m}=u \in U$, and $c_{1}w_1 \dots +c_{m}w_{m} = w \in W$. Hence, every element $v \in V$ can be constructed by some $u \in U + w \in W$, making $U+W=V$.
Now, we have to show that the combination is a direct sum. There is a few ways of going about this, the one presented by Axler is leveraging the fact that a sum of subsets is only a direct sum IFF their intersection is set containing $0$&mdash;that $U \cap W = \{0\}$.
Given some element $v$ that lives in the intersection between $U$ and $W$, it must be formed as a linear combination of two linearly independent lists (as $u_j, \dots w_{j}$ is a basis, they are linearly independent.)
Intuition: if an non-zero element lives in the intersection between two linearly independent lists which together is still linearly independent, it must be able to be written by a linear combination of other elements of that linearly independent list to live in the intersection of the two lists&mdash;which is absurd (violates the definition of linearly dependent). The only element for which this is an exception is $0$.
Actual proof:
suppose $v \in U \cap W$, so $v = a_1u_1\dots +a_{m}v_{m}$ as well as $v=b_1w_{1} + \dots b_{n}w_{n}$. Subtracting the two lists results in:
\begin{equation} 0 = a_1u_1+ \dots a_{m} u_{m} - b_1w_1+ \dots +b_{n}w_{n} \end{equation}
having already declared this list linearly independent, we see that each scalar $a_1, \dots -b_{n}$ must equal to $0$ for this expression. Therefore, the intersection $v$ must be $\{0\}$ as $0u_1 + \dots +0u_{m}=0$.