Determinant map is homomorphism and surjective.

Question

I just came from a course of abstract algebra, and my teacher told us that the determinant map $\det : GL(n, \mathbb{R}) \to \mathbb{R}^\times$ is a surjective homomorphism.

Here, $GL(n, \mathbb{R}) = $ the set of $(n \times n)$ matrices $M$ such that $\det(M) \neq 0$

Why is $\det$ surjective?

score 9 · Accepted Answer · answered Sep 04 '15 at 14:44

9

Note that $$\det\begin{pmatrix}a&0&\cdots & 0\\ 0&1&\cdots & 0\\ \vdots&\vdots&\ddots&\vdots\\0 & 0 & \cdots &1\end{pmatrix} =a$$

answered Sep 04 '15 at 14:44

Hagen von Eitzen

374,180

2

Very elegant proof by construction. – Sep 04 '15 at 14:45
That'll do :) nice (+1) – snulty Sep 05 '15 at 02:08
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Why is this all that is needed? – MeditationOrBust Oct 30 '21 at 21:56

Determinant map is homomorphism and surjective.

1 Answers1