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Let $f : GL_n (\mathbb{R}) \rightarrow (\mathbb{R^*}, \cdot)$ be a group homomorphism such that $\forall a \in G$

$$ f(a) = det(a) $$

(a) Describe $Ker(f)$

(b) Describe $Im(f)$

(a) Describe $Ker(f)$

The kernel of $f$ is the set of all matrices whose determinants are $1$ so that $Ker(f) = SL_n(\mathbb{R})$.

(b) The image of $f$ is all of $\mathbb{R^*}$ since for any non-zero real $r \in \mathbb{R^*}$, we have some matrix $a \in GL_n(\mathbb{R})$ s.t. $det(a) = r$.

Is there anything big I'm missing here?

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Can you fully prove (b)? Other than this, the answer's correct. – DonAntonio Nov 15 '12 at 23:25
up vote 3 down vote accepted

Your answer for (a) is correct. For (b), taking the diagonal matrix $\operatorname{diag}(r,1,\dots,1)$, we can see that $\operatorname{Im}(f)=\Bbb R^*$.

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