What's the "dimensionality" of a matrix (not dimension!)? I know that a $m\times n$ matrix has dimension... well, $m\times n$.

However, in a paper (not a mathematics paper!) I encountered the term dimensionality of a matrix, whose definition is nowhere to be found. 

Apparently, the dimensionality has to be a scalar, as it says e.g.:

"For all real, skew-symmetric matrices, the rank is 2 times the dimensionality, as eigenvalues come in conjugate imaginary pairs."

So, can anyone tell me what I have to deal with here?

Thanks!
Trying to find information on the 'dimensionality' of a matrix is a bit difficult for me! But I think I figured it out in laymens terms:

The dimensionality of a vector space is the number of unique ways a set of vectors 'point' within the vector space. 

Consider a 2x2 vector space  $\vec{V} = \begin{bmatrix}
   a       &amp; b \\
    c       &amp; d \\
\end{bmatrix}$

The dimensionality of $\vec{V}$ is $4$, because it has four independant components that 'point' in different directions and $\vec{V}$ can be represented by a set of four basis vectors:
$A\begin{bmatrix} 1 &amp; 0\\0 &amp; 0\end{bmatrix}  + B\begin{bmatrix} 0 &amp; 1\\0&amp;0\end{bmatrix}+ C\begin{bmatrix} 0 &amp; 0\\1&amp;0\end{bmatrix} +D\begin{bmatrix} 0 &amp; 0\\0&amp;1\end{bmatrix}$. The basis vectors are arbitrary and all kinds of sets can satisfy the basis. But, you get what I mean.

Also consider the vector space $\vec{A}=A_x\hat{x} + A_y\hat{y} + 0\hat{z}$. It has dimensionality $2$ because the components only exist in the $x$-$y$ plane. 

Lastly, consider the vector space $\vec{A}=k\hat{x} + k\hat{y} + k\hat{z}$. This one is tricky, because $\vec{A}$ appears to point in three different directions. But since the components are all the same, the vector space is just a one dimensional line from $($-$\infty,$ -$\infty,$ -$\infty)$ to $(\infty, \infty, \infty)$. So the dimensionality is $1$.

Also, I am still confused a bit about vector space dimensionality, so if anyone cares to right my wrongs, please do! I'm really unclear about this kind of stuff and any help is appreciated. Thanks!
Trying to find information on the 'dimensionality' of a matrix is a bit difficult for me! But I think I figured it out in laymens terms:

The dimensionality of a vector space is the number of unique ways a set of vectors 'point' within the vector space. 

Consider a 2x2 vector space  $\vec{V} = \begin{bmatrix}
   a       &amp; b \\
    c       &amp; d \\
\end{bmatrix}$

The dimensionality of $\vec{V}$ is $4$, because it has four independant components that 'point' in different directions and $\vec{V}$ can be represented by a set of four basis vectors:
$A\begin{bmatrix} 1 &amp; 0\\0 &amp; 0\end{bmatrix}  + B\begin{bmatrix} 0 &amp; 1\\0&amp;0\end{bmatrix}+ C\begin{bmatrix} 0 &amp; 0\\1&amp;0\end{bmatrix} +D\begin{bmatrix} 0 &amp; 0\\0&amp;1\end{bmatrix}$. The basis vectors are arbitrary and all kinds of sets can satisfy the basis. But, you get what I mean.

Also consider the vector space $\vec{A}=A_x\hat{x} + A_y\hat{y} + 0\hat{z}$. It has dimensionality $2$ because the components only exist in the $x$-$y$ plane. 

Lastly, consider the vector space $\vec{A}=k\hat{x} + k\hat{y} + k\hat{z}$. This one is tricky, because $\vec{A}$ appears to point in three different directions. But since the components are all the same, the vector space is just a one dimensional line from $($-$\infty,$ -$\infty,$ -$\infty)$ to $(\infty, \infty, \infty)$. So the dimensionality is $1$.

Also, I am still confused a bit about vector space dimensionality, so if anyone cares to right my wrongs, please do! I'm really unclear about this kind of stuff and any help is appreciated. Thanks!