In Jeffrey Lee's differential geometry text on page 353 he defines the determinant in an interesting way using multilinear alternating maps:

Suppose $V$ is an $n$-dimensional $k$-vector space over some field $k$, and let $L^p_{\text{alt}}(V,k)$ denote the set of $p$-multilinear alternating maps from $V$ to $k$. Any $f \in \text{hom}(V,V)$ induces the pullback map $f^*: L^p_{\text{alt}}(V,k) \to L^p_{\text{alt}}(V,k)$ in the obvious way. In the case $p=n$ the dimension of $L^n_{\text{alt}}(V,k)$ is $1$, so $f^*$ acts by scalar multiplication, and we call this scalar the determinant of $f$. That is, $$f^* \omega = \det (f) \omega$$ for any $\omega \in L^n_{\text{alt}}(V,k)$.

Without using a basis, it is easy to see from this definition that $\det: \text{hom}(V,V) \to k$ is a monoid homomorphism with respect to composition in $\text{hom}(V,V)$ and multiplication in $k$. So it follows that $\det(f^{-1})=\det(f)^{-1}$ whenever $f$ is invertible.

But now my question is, how do we show without using a basis that $\det(f) \neq 0$ implies that $f^{-1}$ exists? I'm a bit rusty with my linear algebra so perhaps I'm missing something obvious.

  • $\begingroup$ I guess it depends on what you mean by "basis-free", but if $f$ didn't have full rank then it wouldn't be hard to show that $f^*\omega$ was always zero, right? You'd be shoving in $n$ arguments that had to be linearly dependent into an alternating map. $\endgroup$ – Hoot Jun 1 '15 at 1:53
  • $\begingroup$ He just shows how to construct a basis using a basis for $V$. Its Lemma 8.17 in his text. $\endgroup$ – ಠ_ಠ Jun 1 '15 at 11:06
  • $\begingroup$ I'll try to find a copy. Anyway, does the first thing I said fit your requirements? $\endgroup$ – Hoot Jun 1 '15 at 15:51
  • $\begingroup$ I see. So he does, as I expected, use the determinant in the proof of 8.17. It's still interesting to characterize the determinant in the way you describe, but the logic isn't "pure" in some sense. I guess that's my only point. $\endgroup$ – Hoot Jun 1 '15 at 16:11

I should probably stop trying to answer things in comments, so:

Suppose $f$ isn't invertible. Then I claim that for any $n$-form $\omega$ I have $f^*\omega = 0$. Indeed, the image of $f$ is a proper subspace of $V$, so if $v_1, \dots, v_n \in V$ then $f(v_1), \dots, f(v_n)$ are linearly dependent, so if you feed them into an alternating map you get zero.


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