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.

You have $n$ column vectors and if you can show that they are linearly independent then they form a basis for $R^n.$ All you have to do show the columns $a_1, a_2, \cdots, a_n  $ are linearly independent. Suppose
$$x_1a_1 + x_2a_2 + \cdots + x_n a_n = 0  \tag 1 $$ We need to show that $x_1 = 0,x_2 = 0, \cdots, x_n = 0$, i.e. that the vector equation $(1)$ is equivalent to $Ax = 0$. Since $A$ is invertible, we can multiply on the left by $A^{-1}$ and get $x = 0$ which is what we needed.
