Htykuut

Is it? I've been wondering if
$$O =Biggl(begin{matrix}0 & 0 & 0\ 0 & 0 & 0\ 0 & 0 & 0end{matrix}Biggl)$$
is a canonical form or not?(technically speaking)

Depends on what you define as a canonical form for a matrix. If you say that a matrix is in a canonical form if it's in Row Echelon Form or Reduced Row Echelon form, then truly, the null matrix $mathbf{O}$ is indeed in canonical form, as :

• all nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix),


• the leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it (some texts add the condition that the leading coefficient must be 1).

It is also in Reduced Row Echelon form, in a logical sense, as :

• All zero rows are at the bottom of the matrix.


• The leading entry of each nonzero row subsequently to the first is right of the leading entry of the preceding row.


• The leading entry in any nonzero row is a 1.


• All entries in the column above and below a leading 1 are zero.

For the sake of information, as Egreg mentioned in the comments, the null matrix $mathbf{O}$ is also a Jordan Canonical form. if you have been introduced to it, it is essentialy a diagonalization manipulation for not standarly diagonalizable matrices. More information can be found here.

These assumptions, though, do not lead to any "spectacular" results, as the null matrix is really trivial.