Let $A$ a matrix $n\times n$ over $\mathbb R$. I'm trying to prove that $A$ is invertible $\iff\det A\neq 0$.
If $A$ is invertible, there is $B$ s.t. $AB=I$, and thus $$\det(A)\det(B)=\det(AB)=\det(I)=1,$$
and thus $\det A\neq 0$.
I have problem to prove the converse. Let $A$ s.t. $\det(A)\neq 0$. I would like to say that $$\det(A)\det(A)^{-1}=1.$$
I know that if $A^{-1}$ exist then $\det(A)^{-1}=\det(A^{-1})$, but since I have to prove that $A^{-1}$ exist, I can't use this formula… so how can I conclude ?
Best Answer
To continue the proof using your approach:
Use the definition of $\det( )$ as the unique alternating multilinear function acting on columns of the matrix such that $\det(I) = 1$ (as opposed to the definition that gives an algebraic expression for $\det()$ in terms of permuations).
$\det(A) \ne 0 \implies $ columns of $A$ are linearly independent $ \implies A$ is full rank.
To get the first arrow, you can do proof by contradiction. So assume the columns of $A$ are dependent and $A_1 = \sum_{i=2}^n c_iA_i$ for some $c_i$. Then $\det(A_1, …, A_n) = \det(A_1 - \sum_{i=2}^n c_iA_i, A_2 + c_2A_2, …, A_n + c_nA_n) = \det(0, A_2 + c_2A_2, …, A_n + c_nA_n) = 0$.
This is the contradiction.
To get the second arrow, it's just the definition of rank.
So $A$ is surjective. By rank-nullity theorem, it is also injective. Therefore it is bijective, proving the existence of $A^{-1}$ as a function. Then it is a small step to prove that $A^{-1}$ linear and then you can complete the proof.