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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.

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  • $\begingroup$ It is just an incomplete argument. To complete it you would need, for example, to prove that equivalence to the identity is a condition that is equivalent (logically) to the determinant not being zero. $\endgroup$ – user647486 Mar 16 at 22:13
  • $\begingroup$ Also, the second "then" in your argument is not explained. If I were evaluating you I would not assume that you know why that is true. You would need to explain it. $\endgroup$ – user647486 Mar 16 at 22:17
  • $\begingroup$ The quantity $A\vec x$ is a vector. What, then is $\det(A\vec x)$? $\endgroup$ – amd Mar 16 at 22:17
  • $\begingroup$ Gaps in the argument notwithstanding, you’ve only shown one direction of the equivalence. $\endgroup$ – amd Mar 16 at 22:18
  • $\begingroup$ By the way, the argument in your book, is much more than just the proof of that theorem. It is an algorithm that is the single most important tool that you will have in an introduction to linear algebra. You could perfectly forget about that theorem. But you should never skip that proof. $\endgroup$ – user647486 Mar 16 at 22:22
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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.

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