# 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?

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 $n=2$.

• The matrix given by @GerryMyerson in the answer below does not appear to correspond to the form you suggested, although the nullspace and column space are indeed equivalent.
– jII
Commented Mar 10, 2014 at 3:56
• @jesterII, it says $A$ must be similar to a matrix of a specific form. That means my matrix is $P^{-1}AP$ for some invertible matrix $P$, with $A$ being the matrix in 1551's answer. Commented Mar 10, 2014 at 4:29
• @jesterII We have $$\pmatrix{2&4\\ -1&-2}=\pmatrix{-2&-1\\ 1&0}\pmatrix{0&1\\ 0&0}\pmatrix{-2&-1\\ 1&0}^{-1}.$$ Commented Mar 10, 2014 at 5:45

$$\pmatrix{2&4\cr-1&-2\cr}$$

Let us study this question in a general setting of $$n \times n$$ matrices.
For a given $$n \times n$$ matrix $$A$$, we denote its nullspace by $$\mathcal{N}(A)$$, and its column space by $$\mathcal{C}(A)$$.

Recall that the orthogonal complement of a vector subspace $$V$$ is $$V^\perp := \{\vec{x} : \forall \vec{v} \in V.~(\vec{x}^T \vec{v} = 0) \}.$$ It is well known that for any $$n\times n$$ matrix $$A$$ it holds that $$(\mathcal{N}(A))^\perp = \mathcal{R}(A)$$, where $$\mathcal{R}(A)$$ is the row space of $$A$$. In particular, $$\mathcal{C}(A) = \mathcal{R}(A^T)$$ implies that $$(\mathcal{C}(A))^\perp = \mathcal{N}(A^T)$$.

Now let $$A$$ be any $$n \times n$$ matrix with $$\mathcal{N}(A) = \mathcal{C}(A)$$. From our preceding discussion, it is thus necessary and sufficient that $$\mathcal{N}(A) = \mathcal{N}(A^T)$$. We shall first show that such an $$n \times n$$ matrix $$A$$ must be of the form $$Q \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right) Q^T,$$ where $$X$$ is an $$r \times r$$ matrix whose rank, $$r$$, is equal to that of $$A$$, and some bordering $$0$$'s may be absent if $$A$$ is of full rank $$n$$.

$$Proof$$. Extend any given basis $$\{e_{r+1},\ldots,e_n\}$$ for $$\mathcal{N}(A)$$ to a basis $$\{e_1,\ldots,e_r,e_{r+1},\ldots,e_n\}$$ for $$\mathbb{R}^n$$. Let $$P = (e_{i}^T)$$ and $$Q = P^{-1}$$. Clearly, $$P$$ (as well as $$Q$$) is an invertible and orthogonal matrix since its column vectors $$\{e_1,\ldots,e_n\}$$ are linearly independent and orthonormal. Since $$\mathcal{N}(A) = \mathcal{N}(A^T)$$, it follows that $$PAP^T = \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right),$$ where $$r(X) = r(PAP^T) = r(A) = r$$. Hence $$A$$ is of the desired form. The proof is thus complete.

Now we establish that any $$n \times n$$ matrix $$A$$ of the above form has to satisfy the equation $$\mathcal{N}(A) = \mathcal{N}(A^T),$$ and hence the condition that $$\mathcal{N}(A) = \mathcal{C}(A)$$.

$$Proof.$$ Suppose $$A = Q \left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right) Q^T$$, where $$X$$ is an invertible matrix with $$r(X) = r(A)$$. Writing the matrix $$\left(\begin{array}{cc} X & 0 \\ 0 & 0 \end{array}\right)$$ as $$Y$$, we have that $$\vec{x} \in \mathcal{N}(A)$$ iff $$QYQ^T \vec{x} = 0$$ iff $$YQ^T \vec{x} = 0$$. Now since $$X$$ is an invertible matrix of size $$r$$, the last condition is equivalent to $$Y^TQ^T \vec{x} = 0$$, which in turn is equivalent to $$QY^TQ^T \vec{x} = 0$$ iff $$x \in \mathcal{N}(A^T)$$. This shows that $$\mathcal{N}(A) = \mathcal{N}(A^T)$$.

1. if the nullspace is to be equal the column space the matrix should be square that is m=n and n=2k (should be even, k is any positive integer). so a 3x3 matrix cannot have a nullspace that is the same as the column space

reason : if A is an mxn matrix- from the fundamental theorem of linear algebra we know the dimensions of the fundamental spaces are;

dim[C(A)]= dim[R(A)]= r - the rank of the matrix

dim[N(A)]=n-r

nullspace is a subspace of n dimensional space where as the column space is a subspace of m dimensional space in order for C(A) & N(A) to be the same it is necessary that m=n. dimensions should match as well that is,

                               n-r = r
r = n/2  therefore n= 2k is a must.

1. AA = 0 when C(A) = N(A)