Do the columns of an invertible $n \times n$ matrix form a basis for $\mathbb R^n$? The columns of an invertible $n \times n$ matrix form a basis for $\mathbb R^n$.
I follow the definition from the text book, then I guess because the matrix is invertible, each vector in the matrix is linearly independent, thus the basis of column space is span in $\mathbb R^n$.
However, I am still confused.  Could someone tell me why or if I've missed something?
Thanks.

If you have an $n \times n$ invertible matrix, then the columns (of which there are $n$) must be independent. How do we know that these vectors -- the columns -- span $\mathbb{R}^n$? We know this because $\mathbb{R}^n$ is an $n$-dimensional space, so any independent set of $n$ or more vectors will span the space. Hence, the columns are both independent and span $\mathbb{R}^n$, so they form a basis for $\mathbb{R}^n$.
