Multiplicity of the Determinant.

Abstract.
This post shows a way to quickly show that the determinant is multiplicative without getting your hands dirty.

I just learned about a nice trick to show that $\det(A B) = \det A \cdot \det B$ for matrices $A, B \in K^{n \times n}$ from my colleague Francesco Sica, who attributed it to F. Catanese.

Assuming that it is known that there is, up to scale, only one alternating $n$ -linear form $K^{n \times n} \to K$ (i.e. that $\dim_K \bigwedge^n K^n = 1$ ), one can proceed as follows. Given $A \in K^{n \times n}$ , consider the map $f_A : K^{n \times n} \to K$ , $B \mapsto \det(A B)$ . This is clearly $n$ -linear and alternating, whence there exists some $\lambda \in K$ such that $f_A = \lambda \cdot \det$ . Evaluating $f_A$ at the identity matrix $I$ gives $\lambda = \lambda \det(I) = f_A(I) = \det(A I) = \det A$ . Evaluating $f_A$ at $B$ gives $\det(A B) = f_A(B) = \lambda \det B = \det A \cdot \det B$ .

Of course, using the trick similar to the first lemma here, it suffices to show this for $K = \C$ to obtain it for any unitary commutative ring, after showing that the determinant is in fact a polynomial equation with integer coefficients (for example, by showing the Leibniz formula).

Tags: determinant, multiplicative

This entry was posted on wednesday, november 10th, 2010 at 00:55 utc and is filed under Beautiful Proofs, Linear Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

two responses to “Multiplicity of the Determinant.”

Jens says:

19.09.2011 at 17:39 utc

Hi! This proof looked too familiar to me… and indeed: Quebbemann (WS 99/00) used the same proof in his Lineare Algebra. :)
Felix Fontein says:

19.09.2011 at 18:10 utc

And again I regret that I didn’t start studying in Oldenburg two years earlier… :)

Multiplicity of the Determinant.

Abstract.

two responses to “Multiplicity of the Determinant.”

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