Fundamental Theorem of Linear Algebra

Given a $m\times n$ matrix $A$, there are four fundamental subspaces:

• $Ran(A)$
• $Ker(A)$
• $Ran(A^T)$
• $Ker(A^T)$

Proposition

$A$ and $A^T$ have the same rank.

Proof

Suppose $A$ has rank r and the rank normal form of $A$ is $RAC=\left[\begin{array}{cc}{I}_{r\times r}& O\\ O& O\end{array}\right]$ for invertible matrices $R,C$, then $C^T{A}^T{R}^T=\left[\begin{array}{cc}{I}_{r\times r}& O\\ O& O\end{array}\right]$, and $C^T,R^T$ are still invertible. Therefore, the rank normal form of $A^T$ is $\left[\begin{array}{cc}{I}_{r\times r}& O\\ O& O\end{array}\right]$, and hence its rank is also $r$.