# Can a matrix have a null space that is equal to its column space?

I had a question in a recent assignment that asked if a $3\times 3$ matrix could have a null space equal to its column space... clearly no, by the rank+nullity theorem... but I have a hard time wrapping my head around the concept of such a matrix, no matter what size, even existing. How could this be possible, and does anybody have an example of such a $m\times n$ matrix?

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The question is: what does it mean the rank is zero? –  Mhenni Benghorbal Oct 3 at 12:22
The nullspace lies inside the domain, while the column space lies inside the codomain. Therefore, if the nullspace is equal to the column space, you must have $m=n$. Also, by the rank-nullity theorem, $n$ must be an even number. It follows that if $n=2k$, the nullspace must be $k$-dimensional. Denote by $\{e_1,\ldots,e_n\}$ the canonical basis of $\mathbb{F}^n$. By a change of basis, we may assume that the nullspace is spanned by $e_1, \ldots, e_k$. Therefore, if the nullspace and column space of $A$ coincide, $A$ must be similar to a matrix of the form $$A=\pmatrix{0&B_{k\times k}\\ 0&0},$$ where $B$ is invertible. For instance, consider $A=\pmatrix{0&1\\ 0&0}$ when $n=2$.
$$\pmatrix{2&4\cr-1&-2\cr}$$