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.
 A: For this statement, you don't even need anything about dimension or linear independence.
The statement that the columns of a matrix $A$ form a basis of $\Bbb R$ means that every vector $v\in\Bbb R^n$ can be uniquely written as a linear combination of them, in other words that the equation $Ax=v$ has a unique solution$~x$.
The square matrix $A$ being invertible means the existence of a matrix $B$ such that $AB=I=BA$. Now on one hand $A(Bv)=(AB)v=Iv=v$ shows that the equation has a solution $x=Bv$. On the other hand if $Ax=v$, then $x=Ix=(BA)x=B(Ax)=Bv$ which shows that the solution is unique.
A: 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 $\mathbb{R}^n$.
A: 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.
