# 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 '13 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 $n=2$.

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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. –  jesterII Mar 10 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. –  Gerry Myerson Mar 10 at 4:29
@jesterII We have $$\pmatrix{2&4\\ -1&-2}=\pmatrix{-2&-1\\ 1&0}\pmatrix{0&1\\ 0&0}\pmatrix{-2&-1\\ 1&0}^{-1}.$$ –  user1551 Mar 10 at 5:45

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

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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 $P = (\vec{e}_i^T)$ and $Q = P^{-1}$. Clearly, both $P$ (as well as $Q$) is invertible since its column vectors $\{e_1,\ldots,e_n\}$ are linearly independent. 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 $\mathcal{N}(A) = \mathcal{R}(A)$.

$Proof.$ 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 $\mathcal{N}(A) = \mathcal{N}(A^T)$.

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