%m211og86.tex: The direct sum
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\begin{center}
{\rmH\tcr{The Direct Sum of Subspaces}}
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\huge

\begin{description}
\item[\tcr{Definition:}] 
\tcr{(a)} If \mc{\vec{v}= (v_1,\dots,v_n)^t \in \C^{n}} and 
\mc{\vec{w} = (w_1,\dots,w_m)^t \in \C^m}, then the \tcg{vector}
$$
\tcb{\vec{v} \oplus \vec{w}:= (v_1,\dots,v_n,w_1,\dots,w_m)^t \in \C^{n+m}}
$$
is called the \tcr{\em direct sum} of \mc{\vec{v}} and \mc{\vec{w}}.

\tcr{(b)} If \mc{V \subset \C^n} and \mc{W \subset \C^m} are subspaces, then the 
\tcr{\em direct sum} of \mc{V} and \mc{W} is the \tcg{subspace} of \mc{\C^{n+m}} 
defined by
$$
\tcb{V \oplus W=\{\vec{v} \oplus \vec{w} \in \C^{n+m}:\vec{v} \in V, \vec{w} \in W\}.}
$$

\item[\tcr{Remarks:}] \tcr{1)} If \mc{\vec{v}_1,\ldots, \vec{v}_r} is a \tcg{basis} of \mc{V \subset \C^n} 
and \mc{\vec{w}_1, \ldots, \vec{w}_s} is one of \mc{W \subset \C^m}, then 
$$
\tcb{\vec{v}_1\oplus\vec{0}_m,\,  \ldots, \,\vec{v}_r\oplus\vec{0}_m,\,
\vec{0}_n\oplus\vec{w}_1,\, \ldots,\, \vec{0}_n\oplus\vec{w}_s}
$$
is a \tcr{basis} of \mc{V\oplus W}. 
(Here, \mc{\vec{0}_m = (\underbrace{0,\ldots,0}_m)^t}.\vspace{-10pt}) Thus
\begin{equation}
\tcb{\dim(V \oplus W) = \dim V + \dim W.} \label{6.21a}
\end{equation}

\tcr{2)} If \mc{A} is an \mc{a\times m} matrix and \mc{B} is a \mc{b\times n} 
matrix, then for every \mc{\vec{v} \in \C^m} and \mc{\vec{w} \in \C^n} we have 
\begin{equation}
\tcb{\mbox{diag}(A,B)(\vec{v} \oplus \vec{w}) =
(A \vec{v}) \oplus (B \vec{w}).}\label{6.21b}
\end{equation}


\iffalse
\begin{ex} \label{E6.22a}
Let $V = \{c_1(1,2)^t + c_2(3,4)^t: c_1, c_2 \in \C\}$ and 
$W = \{c'_1(1,2,1)^t + c'_2(3,4,1)^: c'_1, c'_2 \in \C\}$. 
Verify the addition rule \sref{6.21a} for $V$ and $W$.
\spm

\noi{\bf Solution.} If $\vec{v} \in V$, then 
$\vec{v} = c_1(1,2)^t + c_2(3,4)^t = (c_1, 2c_1)^t + (3c_2, 4c_2)^t = (c_1+3c_2, 2c_1+4c_2)^t$,
and similarly each $\vec{w} \in W$ has the form 
$\vec{w} = (c'_1+3c'_2, 2c'_1+4c'_2,c'_1+c'_2 )^t$. Thus
\begin{eqnarray*}
\vec{v}\oplus\vec{w} 
&=& (c_1+3c_2, 2c_1+4c_2)^t\oplus(c'_1+3c'_2, 2c'_1+4c'_2,c'_1+c'_2 )^t \\
&=& (c_1+3c_2, 2c_1+4c_2, c'_1+3c'_2, 2c'_1+4c'_2,c'_1+c'_2 )^t \\ 
&=& c_1(1,2,0,0,0)^t + c_2(3,4,0,0,0)^t + c_1'(0,0,1,2,1)^t + c_2'(0,0,3,4,1)^t,
\end{eqnarray*}
and so $V\oplus W = \{c_1(1,2,0,0,0)^t + c_2(3,4,0,0,0)^t + c_1'(0,0,1,2,1)^t + c_2'(0,0,3,4,1)^t : c_1, c_2, c_1', c_2' \in \C\}$.
Therefore, $\dim(V\oplus W) = 4 = \dim V + \dim W$.
\end{ex}\spm

\begin{ex}\label{E6.22b} 
Verify the multiplication rule \sref{6.21b} 
for $A = \smallmat{1}{2}{3}{4}$ and $B=\smallmat{5}{2}{6}{1}$. 
\spm

\noi{\bf Solution.} Write $\vec{v}=(v_1,v_2)$ and $\vec{w} = (w_1,w_2)$. Then
\begin{eqnarray*}
\Diag(A,B)(\vec{v}\oplus\vec{w})
& = &
\bmat{cc|cc}
1&2&0&0\\
3&4&0&0\\\hline
0&0&5&2\\
0&0&6&1
\emat
\bmat{c} v_1\\v_2\\ \hline w_1\\w_2 \emat
=
\bmat{c} 
v_1+2v_2 \\
3v_1+4v_2\\
5w_1+2w_2\\
6w_1+w_2
\emat \\
& = &
\exprow{1.1}
\bmat{c} 
v_1+2v_2 \\
3v_1+4v_2
\emat\oplus
\bmat{c}
5w_1+2w_2\\
6w_1+w_2
\emat 
\exprow{1.0}\\
& = &
\bmat{cc}
1&2\\
3&4
\emat
\bmat{c} v_1\\v_2\emat
\oplus
\bmat{cc}
5&2\\
6&1
\emat
\bmat{c} w_1\\w_2 \emat
= A\vec{v}\oplus B\vec{w}.
\end{eqnarray*}
\end{ex}
\spm\sps
\fi
\end{description}
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