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If so, can you give an example of such a matrix? Otherwise, what is the proof of this condition (NulA=RowA) not being possible?

Jucinda Lynon
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  • Try looking up the fundamental theorem of linear algebra. In short, no. – Jacky Chong May 18 '20 at 00:11
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    What about $[\begin{smallmatrix} 1 & i \ 1 & i \end{smallmatrix}]$? – anon May 18 '20 at 00:16
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    @runway44 Good example! This illustrates the difference between real and complex matrices. Also, for complex matrices, $\operatorname{Row}(A) \neq \operatorname{Col}(A^\ast)$ where $A^\ast$ is the adjoint matrix. – Jacky Chong May 18 '20 at 01:06
  • Wouldn’t the row vector $[1, i ]$ correspond to the column vector $\begin{bmatrix} 1 \ -i \end{bmatrix}$, so that that vector is in the row space, but not the null space? – Joe May 18 '20 at 01:13

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If you’re looking at $2 \times 2$ real matrices, then this is not possible. Here is a simple proof:

Proof: Let $A = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ have real entries, and suppose that its null space equals its row space. Note that $A \neq 0$ since the null space of the zero matrix is $\mathbb{R}^2$ while the row space is $0$. So $A$ has some nonzero element. Note that $\begin{bmatrix}a \\ b \end{bmatrix}$ and $\begin{bmatrix}c \\ d\end{bmatrix}$ are in the row space, hence they are in the null space. In other words, $$\begin{bmatrix}0 \\ 0 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix}a \\ b \end{bmatrix} = \begin{bmatrix}a^2 + b^2 \\ ac + bd \end{bmatrix} \quad \text{and} \quad \begin{bmatrix}0 \\ 0 \end{bmatrix} = \begin{bmatrix}a&b\\ c&d \end{bmatrix}\begin{bmatrix}c \\ d\end{bmatrix} = \begin{bmatrix}ac + bd \\ c^2 + d^2 \end{bmatrix}.$$

Thus $a^2 + b^2 = 0$ and $c^2 + d^2 = 0$, contradicting the fact that at least one of $a, b, c$ or $d$ is nonzero. $\square$

If you're looking at $2 \times 2$ complex matrices, then it is possible; as runway pointed out in the comments, consider the matrix $$\begin{bmatrix}1 & i \\1 & i \end{bmatrix}.$$

Proof: To show that the null space of this matrix equals its row space, we show that an element is in the null space if and only if it is in the row space. It is easy to see that any element of the row space, i.e. any element of the form $\begin{bmatrix}z \\ zi \end{bmatrix}$ (where $z$ is a complex number), is in the null space (just compute the matrix multiplication). Hence the row space is contained in the null space. Conversely, suppose that $\begin{bmatrix}x \\ y\end{bmatrix}$ is in the null space; if we multiply out the matrices, this corresponds to saying that $x + iy = 0$. This implies that $x = -iy$, or $y = ix$. Hence $\begin{bmatrix}x \\ y \end{bmatrix}$ is in the row space. $\square$

twosigma
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    More generally, for real matrices, the null space is the orthogonal complement of the row space, and the intersection of a subspace with its orthogonal complement is $0$. – twosigma May 20 '20 at 06:05