# Determinant defined using multilinear alternating maps, and invertibility of linear endomorphisms

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.

• 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. – Hoot Jun 1 '15 at 1:53
• He just shows how to construct a basis using a basis for $V$. Its Lemma 8.17 in his text. – ಠ_ಠ Jun 1 '15 at 11:06
• I'll try to find a copy. Anyway, does the first thing I said fit your requirements? – Hoot Jun 1 '15 at 15:51
• 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. – Hoot Jun 1 '15 at 16:11

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.