I've usually used that given a square matrix $A$ with determinant $\det(A)\neq0$, then its inverse $A^{-1}$ is the matrix that meets:




However, I've found in several sources, including Wikipedia that if $A$ is finite and square, then:

$$A^{-1}A=\mathbb{I} \Longrightarrow AA^{-1}=\mathbb{I}.$$

I'm struggling to find a proof. Does anybody know one?


If $f, \, g \; : \; \mathbb{R}^{n} \; \longrightarrow \; \mathbb{R}^{n}$ are two linear maps such that : $f \circ g = \mathrm{Id}$ then, it implies that :

  • $f$ is invertible
  • $g$ is invertible
  • $g \circ f = \mathrm{Id}$.

To prove that $f$ and $g$ are invertible, you only need to prove that $\mathrm{ker}(f) = \lbrace 0 \rbrace$ and $\mathrm{ker}(g)=\lbrace 0 \rbrace$. Let $x \in \mathrm{ker}(g)$. Then, $(f \circ g)(x) = x$ and $(f \circ g)(x) = f(0)=0$. So, $x=0$ and it follows that $\mathrm{ker}(g) = \lbrace 0 \rbrace$ and $g$ is invertible. You can prove as well that $\mathrm{ker}(f) = \lbrace 0 \rbrace$ (so $f$ is invertible).

Let $x \in \mathbb{R}^{n}$. Since $g$ is invertible, there exist $u \in \mathbb{R}^{n}$ such that $x = g(u)$. Therefore, $(g \circ f)(x) = g\big( (f \circ g)(u) \big) = g(u) = x$. As a consequence, $g \circ f = \mathrm{Id}$.

  • $\begingroup$ Wouldn't we also need to prove that $Im(f)=Im(g)=\mathbb{R}^n$? $\endgroup$ – Pablo S. Ocal Oct 6 '14 at 10:15
  • $\begingroup$ No, we do not need to. Since $f$ and $g$ are endomorphisms (of $\mathbb{R}^{n}$), it follows from the Rank-nullity theorem (see en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem). $\endgroup$ – pitchounet Oct 6 '14 at 10:20
  • $\begingroup$ Yes, thanks, I didn't realize the original space and the final space have the same dimension, and thus being injective means being surjective (for linear functions). $\endgroup$ – Pablo S. Ocal Oct 6 '14 at 10:23
  • $\begingroup$ Yes, absolutely. $\endgroup$ – pitchounet Oct 6 '14 at 10:25

If $A^{-1}A$ is the identity, this means that $A$ must be injective. Since we are in finite dimension and $A$ is square, this means that $A$ is surjective too. So it is invertible: let's call $B$ its inverse. Then, if you have $CA=Id$, it follows that $CAB=B$ and then $C=B$, which is what you wanted: $C$ is actually the inverse af $A$.

Let me remind you that this applies only in finite dimension. In infinite dimension you can very easily have linear operators that are injective but not surjective (or surjective, but not injective).

  • $\begingroup$ I don't understand why this was downvoted. It seems to be correct to me. Con the downvoter please explain? $\endgroup$ – Giovanni Mascellani Oct 11 '14 at 13:24

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