If we have a square matrix thats invertible, do its row and column space coincide? If we have a square matrix thats invertible, do its row and column space coincide?
Regarding an nxn invertible matrix:
-The row space of the matrix is R^n
-The column space of the matrix is R^n
-The rank of the matrix is n
Is this a sufficient way of proving the question, or am I missing something?
 A: Yes. Suppose you have a matrix $ \mathbf{A} \in \mathbb{R}^{n \times n} $, and it is known to be full rank. Therefore, the number of basis vectors for the column space $ C(\mathbf{A}) $ and the row space $ C(\mathbf{A}^\text{T}) $ is the same, namely $ n $, and these (not necessarily identical) bases span $ \mathbb{R}^n $, so $ C(\mathbf{A}) = \mathbb{R}^n $, and $ C(\mathbf{A}^\text{T}) = \mathbb{R}^n $. It follows that $ C(\mathbf{A}) = C(\mathbf{A}^\text{T}) $. Note, however, that this is a special case that is true only when the matrix is invertible.
A: $\newcommand{\Reals}{\mathbf{R}}$The row space and column space of an $n \times n$ matrix are not generally equal, e.g.,
$$
A = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix},\quad
\text{row space} = \{0\} \times \Reals,\quad
\text{column space} = \Reals \times \{0\}.
$$
The row space and column space of an $n \times n$ matrix do always have the same dimension, however, and if this dimension is $n$, then each space is equal to $\Reals^{n}$.
