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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?

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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)

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  • $\begingroup$ What is the relation between linear independence and the determinant of a matrix? Where can I learn about this? $\endgroup$ – Joel B Oct 13 '14 at 10:19
  • $\begingroup$ 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 $$\det(v_1, \ldots, w, \ldots, w, \ldots, v_n) = 0.$$ $\endgroup$ – Ivo Terek Oct 13 '14 at 10:23
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If the given vectors are LI ,when u write them as in rows of matrix . then det of matrix should be always a non zero number , if determinant is 0 then they are LD ..this is because when we write av + bu + cw = 0 where u,v,w are vectors and a b c are scalars . when plugging in vectors u get n homogenous equations if vectors u plugged has n components , then after gettin n homogenous equations . by cramers rule we if detA is not equal to zero then it is trivial solution which is a=b=c=0 . and hence LI

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A direct argument not relying on determinant:

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.

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