# Consider $A_{n\times n}$ and $B_{n\times n}$. If $AB = I_n$, are the columns of $A$ or $B$ linearly independent?

So we have that $AB = I_n$. This means that $A = B^{-1}$ and $B = A^{-1}$. In particular, we know that an inverse of both $A$ and $B$ exists. Thus, $rank(A) = rank(B) = n$. We know that the $im(A)$ or the $im(B)$ is spanned by the columns that correspond to pivot columns in $RREF(A)$ and $RREF(B)$, respectively. Since we know the ranks are both equal to $n$, we know that the images are spanned by all columns of $A$ and $B$, and that each has columns that are all linearly independent.

Is this correct reasoning? Is there another way of proving the same result?

Columns (and rows) of both are linearly independent. Indeed, $$AB = I \implies \det A \det B = 1 \implies \det A, \det B \neq 0$$ If the columns were linearly dependent, one of them would be a linear combination of the others, and so the determinant would be zero. (contrapositive)
• The determinant is a multilinear, alternating function: $$\det: \Bbb R^n \times \cdots \times \Bbb R^n = (\Bbb R^n)^n \to \Bbb R$$ such that it is linear in each variable and $$\det(v_1, \ldots, v_i, \ldots, v_j, \ldots, v_n) = - \det(v_1, \ldots, v_j, \ldots, v_i, \ldots, v_n)$$ I think most linear algebra book must talk about this a bit, but no specific book comes to mind now. Using only what I told you, you can prove that if one of the $v_i$ is a linear combination of the others, then the determinant is zero. Hint: start proving that $$\det(v_1, \ldots, w, \ldots, w, \ldots, v_n) = 0.$$ – Ivo Terek Oct 13 '14 at 10:23
First, suppose that the columns of $B$ are linearly dependent: this means that there is a nonzero $v$ such that $Bv=0$. But then $$0=A0=A(Bv)=(AB)v=Iv=v$$ contradicting $v\neq 0$. So the columns of $B$ are independent.
Second, suppose that the columns of $A$ are linearly dependent: there is a nonzero $w$ such that $Aw=0$. Because $B$ is $n\times n$, its (linearly independent) columns form a basis in $\mathbb{R}^n$ so there is $u\neq 0$ such that $Bu=w$. Then $$0=Aw=A(Bu)=(AB)u=Iu=u$$ contradicting $u\neq 0$. Hence, the columns of $A$ are indeed independent.