Rank Normal Form

Rank Normal Form

Rank Normal Form

Given a $m\times n$ matrix $A$ with rank $r$, we can find invertible $m\times m$ matrix $R$ and inveribel $n\times n$ matrix $C$ such that $RAC=\left[\begin{array}{cc}{I}_{r\times r}& O\\ O& O\end{array}\right]$.

It can be seen as a process of simplifying a matrix by changing the basis of its domain and codomain.

Full Rank Decomposition

Full Rank Decomposition

For any $m\times n$ matrix $A$ with rank $r$, we can find an $m\times r$ matrix $B$ with rank $r$ and an $r\times n$ matrix $C$ with rank $r$ such that $A=BC$.

Proof

The proof is straight forward, pick the right basis, then $A=\left[\begin{array}{cc}{I}_{r\times r}& O\\ O& O\end{array}\right]$, and

$$\left[\begin{array}{cc}{I}_{r\times r}& O\\ O& O\end{array}\right]=\left[\begin{array}{c}{I}_{r\times r}\\ O\end{array}\right]\left[\begin{array}{cc}{I}_{r\times r}& O\end{array}\right]$$

So set $B=\left[\begin{array}{c}{I}_{r\times r}\\ O\end{array}\right]$ and $C=\left[\begin{array}{cc}{I}_{r\times r}& O\end{array}\right]$.

So basically any linear transformation is a composition of a projection to a subspace and an inclusion to another space.

Rank-one Decomposition

Rank-one Decomposition

A matrix of rank $r$ is the sum of $r$ rank-one matrices.

Proof

Do the full rank decomposition, $A=BC$. Say $B=\left[\begin{array}{ccc}{b}_1& \dots & b_r\end{array}\right],C=\left[\begin{array}{c}{c}_1^T\\ \dots \\ c_r^T\end{array}\right]$, then

$$A=BC=\left[\begin{array}{ccc}{b}_1& \dots & b_r\end{array}\right]\left[\begin{array}{c}{c}_1^T\\ \dots \\ c_r^T\end{array}\right]=\sum _{i=1}^r{b}_i{c}_i^T$$

Each $b_i{c}_i^T$ is a rank-one matrix.