Jordan canonical form for $m\times n $ matrices?

Question

Is the Jordan canonical form only for a $n\times n $ matrix?

Can we generalize it somehow to $m\times n $ matrices? Here $m\neq n$.

Answer 1

The Jordan normal form is kind of the "simplest" matrix that describes a linear map $F\colon V\to V$. It is crucial that the space $V$ is the same since we perform the same change of basis, that is searching for the simplest among $S^{-1}AS$ with the same $S$. If the linear map is $F\colon V\to W$, then it is unclear what the "same change of basis" is. If we change the bases of $V$, $W$ independently (looking for the simplest matrix among $S^{-1}AT$), we can get a diagonal matrix with some identities and some zeros on the diagonal. If we restrict the bases to orthonormal ones ($S$, $T$ are unitary), we get SVD.

P.S. For matrices over a ring (no division is available) then the $S^{-1}AT$ kind of the "simplest" (with special $S$, $T$ to avoid troubles with no division) is known as the Smith normal form.

Answer 2

If you view the Jordan canonical form as a way to express a linear operator in a simple way by choosing a suitable basis, then the analogue for linear transformations between two different spaces is the singular value decomposition. It says that by choosing suitable bases for the two spaces, the transformation can be expressed by a diagonal matrix. The bases are very nice: orthogonal bases.

The Jordan canonical form exists for matrices over any algebraically closed field, typically the complex numbers. The singular value decomposition only exists for matrices over the reals or complex numbers.