Let $A$ be a square matrix. Prove that $A$ ~ $I_n$ if and only if $A\vec{x} = \vec{0}$ has only the trivial solution.

Question

I'm studying Linear Algebra for the second time, using Hoffmann & Kunze. Currently I'm trying to prove the following theorem:

Theorem 7. If $A$ is an $n \times n$ matrix, then $A$ is row-equivalent to the $n \times n$ identity matrix if and only if the system of equations $A\vec{x} = \vec{0}$ has only the trivial solution $\vec{x} = \vec{0}$.

The given proof in the textbook is, I find, obscure, and I don't understand it at all, so I came up with my own proof. However, it depends on $det(A) \neq 0$ and I'm not sure if this assumption can be made.

I rewrote the theorem as:

Let $A$ be a square matrix. Prove that $A$ ~ $I_n$ if and only if $A\vec{x} = \vec{0}$ has only the trivial solution.

My proof:

Let $det(A) \neq 0$. Then $rref(A) = I_n$. If $A\vec{x} = \vec{0}$, then $det(A\vec{x}) = 0 \rightarrow \vec{x} = \vec{0}$, so $\vec{0}$ is indeed the only solution.

The problem with my proof is, even if it's correct, the authors only introduce determinants around page $200$ and this theorem is from chapter $1$. So really I'm looking for a proof that does not rely upon determinants.

Answer 1

Recall that the elementary row operations correspond to left multiplication by elementary row operation matrices. (To figure out what these matrices are, apply the given row operation to the identity matrix!)

Therefore, if $A$ is row-equivalent to $I$, then there exist elementary matrices $E_1, \ldots, E_k$ such that

$$E_1 E_2 \ldots E_k A = I.$$

Hence,

\begin{align*} Ax = 0 &\implies E_1 E_2 \ldots E_k A x = E_1 E_2 \ldots E_k 0 \\ &\implies x = I x = E_1 E_2 \ldots E_k 0 = 0. \end{align*}

Combine this with the obvious fact $A0 = 0$, and we get that $x = 0$ is the unique solution to the matrix equation.