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Direct sum of subspaces examples


The philosophy behind the direct sum of subspaces is the decomposition of vector spaces as a sum of disjoint spaces. In fact, this is very important for defining the projections; so restricting the work only on the subspaces instead of working on the enter vector space.

Definition of the direct sum of subspaces

Let $E$ be a vector space and let $F$ and $H$ be subspaces of $E$. We define the sum of the two spaces $F$ and $H$ by \begin{align*}F+H:=\{x+y:x\in F,\;y\in H\}.\end{align*} We say that this sum is direct if $F\cap H=\{0\}$. In this case, we write $F\oplus H$.

On the other hand, we say that $F$ and $H$ supplementary vector spaces in $E$ is the sun $F+H$ is direct and $E=F+H$. We then write $E=F\oplus H$.

An example of supplementary vector spaces

Let $\mathcal{E}$ the vector space of all functions from $[-1,1]$ to $\mathbb{R}$. On this space, we consider three subspaces \begin{align*}F_1&:=\{f\in \mathcal{E}: f\;\text{is constant}\},\cr F_2&:=\{f\in \mathcal{E}: f_{|[-1,0]}=0\},\cr F_3&:=\{f\in \mathcal{E}: f_{|[0,1]}=0\}.\end{align*}Prove the following direct sum \begin{align*}\mathcal{E}=F_1\oplus F_2\oplus F_3.\end{align*}

First of all let us prove that each function in $\mathcal{E}$ can be decomposed as the sum of three functions belonging to $F_1,F_2$ and $F_3$. In fact, let $f\in \mathcal{E}$. Let define the following functions\begin{align*}f_1(t):=f(0),\qquad \forall t\in [-1,1],\cr f_2(t):=\begin{cases} 0,& t\in [-1,0],\cr f(t)-f(0),& t\in (0,1],\end{cases}\cr f_3(t):=\begin{cases} f(t)-f(0),& t\in [-1,0),\cr 0,& t\in [0,1].\end{cases}\end{align*}Denote $g=f_1+f_2+f_3$ and prove that $g=f$. As $f_2(0)=f_3(0)=0,$ then $g(0)=f_(0)=f(0)$. On the other hand, let $t\in [-1,0),$ by definition of $f_2$ and $f_2$ we have $f_2(t)=0$ and $f_3(t)=f(t)-f(0)$. Hence for any $t\in [-1,0),$\begin{align*}g(t)=f(0)+0+(f(t)-f(0))=f(t).\end{align*}Now let $t\in (0,1]$. We know that $f_2(t)=f(t)-f(0)$ and $f_3(t)=0$. hence, for any $t\in (0,1],$\begin{align*}g(t)=f(0)+(f(t)-f(0))+0=f(t).\end{align*}Finally, $g(t)=f(t)$ for any $t\in [-1,1]$, so that $\mathcal{E}=F_1+F_2+F_3$.

Let us now $h_1\in F_1,\;h_2\in F_2$ and $h_3\in F_3$ such that $h_1+h_2+h_3=0$. We shall prove that all these functions are null. In fact, as $h_2(0)=h_3(0)=0$, then for all $t\in [-1,1]$ we have \begin{align*}0=h_1(0)+h_2(0)+h_3(0)=h_1(0).\end{align*}As $h_1$ is the constant function $[-1,0]$ then for any $t\in [-1,1],$ $h_1(t)=h_1(0)=0$. So that\begin{align*}f_2+f_3=0\quad\text{on}\; [-1,0].\end{align*}As $f_2\in F_2,$ then $f_2=$ on $[-1,0]$. This implies that $f_3=0$ on $[-1,0]$. But we know that $f_3=0$ on $[0,1]$. Then $f_3=0$ on $[-1,1]$. Hence $f_2=0$ on $[-1,1]$. this ends the proof.

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