Subsection
{\sc\large This Section is a Draft, Subject to Changes}
\bigskip
Our first decomposition applies only to diagonalizable (Definition
DZM) matrices, and yields a decomposition into a sum of very simple matrices.
Theorem ROD Rank One Decomposition
Suppose that $A$ is a diagonalizable matrix of size $n$ and rank $r$. Then there are $r$ square matrices $A_1,\,A_2,\,A_3,\,\dots,\,A_r$, each of size $n$ and rank $1$ such that
\begin{align*}
A&=A_1+A_2+A_3+\dots+A_r
\end{align*}
Furthermore, if $\scalarlist{\lambda}{r}$ are the nonzero eigenvalues of $A$, then there are two sets of $r$ linearly independent vectors from $\complex{n}$,
\begin{align*}
X&=\set{\vectorlist{x}{r}}
&
Y&=\set{\vectorlist{y}{r}}
\end{align*}
such that $A_k=\lambda_k\vect{x}_k\transpose{\vect{y}_k}$, $1\leq k\leq r$.
Proof
We record two observations that was not stated in our theorem above. First, the vectors in $X$, chosen as columns of $S$, are eigenvectors of $A$. Second, the product of two vectors from $X$ and $Y$ in the opposite order, by which we mean $\transpose{\vect{y}_i}\vect{x}_j$, is the entry in row $i$ and column $j$ of the matrix product $\inverse{S}S=I_n$ (Theorem
EMP). In particular,
\begin{align*}
\transpose{\vect{y}_i}\vect{x}_j
&=
\begin{cases}
1&\text{if $i=j$}\\
0&\text{if $i\neq j$}
\end{cases}
\end{align*}
We give two computational examples. One small, one a bit bigger.
Example ROD2 Rank one decomposition, size 2
Here's a slightly larger example, and the matrix does not have full rank.
Example ROD4 Rank one decomposition, size 4
