The rank of the below matrix confused me. Since it has $3$ independent rows, so the dimension of rows space should be $3$. However, it also has $7$ independent columns, then the dimension of columns space is $7$ also?
\begin{pmatrix} 1& -2& 0& 19& -6& 0& -37\\ 0& 0& 1& -6& 2& 0& 6\\ 0& 0& 0& 0& 0& 1& 3\\ 0& 0& 0& 0& 0& 0& 0\\ \end{pmatrix}
On
It actually doesn't have $7$ independent columns (for example the second column is $-2$ times the first)
On
As mentioned David Sir, ""Dimension of a matrix" makes no sense"-it is true fact. I think it may be "Rank of a matrix". However in detailed solution of the problem we see that only $1st$ $3rd$ and $6th$ columns are linearly independent, remaining all rows can be expressed as a linear combinations of these three columns. So $dim (\text{column space})=3$. Also $dim (\text{row space})=3$, since it has three independent rows also. Moreover for any $m\times n$ matrix $A$
$$dim (\text{row space of A})=dim (\text{column space of A})= rank(A).$$
"Dimension of a matrix" makes no sense. Dimension is a property of vector spaces, not of individual vectors or matrices.
Your matrix has three independent rows (and no more), so the dimension of its row space is $3$.
If a matrix has seven independent columns (in fact, yours doesn't) and no more, then the dimension of its column space will be $7$.