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?
Best Answer
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.