Section 2.1 Sums and Direct Sums
Let’s return to the example above in which \(T:V\to W\) is a linear map, and suppose \(\{v_1, \dots, v_k\}\) is a basis for \(U=\ker(T)\text{,}\) and we extend that basis to a basis \(\cB=\{v_1, \dots, v_n\}\) for \(V\text{.}\) If we put \(U' =
\Span(\{v_{k+1}, \dots, v_n\})\text{,}\) then every element \(v \in
V\) can be written as \(v=u+u'\) for unique vectors \(u\in
U\) and \(u' \in U'.\)
Taking this one step further, if \(v=u+u'\) as above we know that
\begin{equation*}
T(v) = T(u+u') = T(u) + T(u') = 0+T(u') = T(u'),
\end{equation*}
so that understanding the action of \(T\) on \(V\) has been reduced to understanding the action on the subspace \(U'.\) So effectively we have reduced the size of our problem.
The situation we described above is actually rather special, so let’s begin with a slightly more general notion.
Let \(U,W\) be subspaces of a vector space \(V\text{.}\) Denote by
\begin{equation*}
U+W := \{ u+w \mid u\in U,\ w\in W\}.
\end{equation*}
That is, \(U+W\) is the set of vectors \(v\in V\) which can be written as \(v = u+w\) for some \(u\in U\) and some \(w \in W.\) That seems very similar to what happened in the example above, except in that example, the vectors \(u,w\) were uniquely determined.
It is easy to check that \(U+W\) is a subspace of \(V,\) (indeed the smallest subspace of \(V\) containing \(U\) and \(W\)), but before going too far, we should make a few simple observations. First, it is immediate to check that \(U+W = W+U\) since addition in a vector space is commutative. What if we have more than two subspaces?
If we had three subspaces \(U_i\text{,}\) \(i=1,2,3,\) we could easily check (since we know how to add pairs of subspaces) that
\begin{equation*}
(U_1+U_2)+U_3 = U_1+(U_2+U_3),
\end{equation*}
so we can unambiguously define
\begin{equation*}
U_1+U_2+U_3 :=
(U_1+U_2)+U_3\text{,}
\end{equation*}
and inductively we define
\begin{equation*}
U_1+\cdots + U_n :=
(U_1+\cdots + U_{n-1}) + U_n.
\end{equation*}
But as with any new concept, some examples help us better understand it.
Example 2.1.1. A standard decomposition of \(F^n\).
Let \(\{e_1, \dots, e_n\}\) be the standard basis for \(F^n,\) and put \(U_i = \Span\{e_i\},\) the line through the origin in the direction of \(e_i.\) So when \(n=3,\) these subspaces are just the \(x,y,\) and \(z\) axes. Then we see that \(F^n = U_1+U_2+\cdots +
U_n.\) We also see that every element of \(F^n\) is the sum of uniquely determined elements from the \(U_i\text{.}\) As row vectors,
\begin{equation*}
(a_1, \dots, a_n) = (a_1,0,\dots,0) +
(0,a_2,0,\dots,0) + \cdots + (0,\dots,0,a_n).
\end{equation*}
Example 2.1.2. Decomposing \(V=F^3\).
Let \(\{e_1, e_2, e_3\}\) be the standard basis for \(V\text{,}\) and let \(U= \Span\{e_1, e_2\}\) and let \(W =
\Span\{e_3, e_1+e_2+e_3\}\text{.}\) It is straightforward to show that
\begin{equation*}
U+W = \Span\{e_1,e_2,e_3,e_1+e_2+e_3\} =
\Span\{e_1,e_2,e_3\} = V,
\end{equation*}
so every element of \(V\)can be written as the sum of vectors from \(U\) and \(W\text{,}\) but in this case not necessarily uniquely.
As a trivial example, let \(v=e_3.\) Then \(v\) can be written as \(v = u+w\) with \(u=0\) and \(w=e_3\text{,}\) or with \(u= -e_1 -e_2\) and \(w=e_1+e_2+e_3.\)
The source of this non-uniqueness is actually easy to discover. Suppose that \(V=U+W,\) and for some \(v\in V,\)
\begin{equation*}
v = u_1 + w_1 = u_2 + w_2.
\end{equation*}
Then of course \(r=u_1-u_2 = w_2 -w_1.\) For uniqueness, we would need \(u_1 = u_2\) and \(w_1=w_2.\) Said another way, we would need \(r=u_1-u_2 = w_2 -w_1=0.\) But \(u_1-u_2 \in U\) and \(w_2-w_1 \in W,\) so the only way to force uniqueness is if \(U\cap W = \{0\}.\)
We summarize this as
Proposition 2.1.3.
Let \(U,W\) be subspaces of a vector space \(V\text{,}\) and suppose that \(V=U+W.\) Then every element of \(V\) is representable as a sum of uniquely determined elements of \(U\) and \(W\) if and only if \(U\cap W = \{0\}.\)
In the case that \(V=U+W\text{,}\) and \(U\cap W = \{0\},\) we write
\begin{equation*}
V = U\oplus W
\end{equation*}
and call \(V\) the direct sum of the subspaces \(U\) and \(W.\)
Checkpoint 2.1.4.
Suppose that \(U_i\text{,}\) \(i=1,2,3\) are subspaces of a vector space \(V,\) and that \(V = U_1 + U_2+U_3.\) We want necessary and sufficient conditions so that every element of \(V\) can be represented as a unique sum of elements from the \(U_i.\) What about when \(V=U_1+\cdots + U_n\) for \(n\ge 3?\)
Hint.
To gain some insight, first find an example in \(\R^3\) where \(U_i\cap
U_j=\{0\}\) whenever \(i\ne j,\) but not every element of \(\R^3\) has a unique representation as a sum.
