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Consider the group homomorphism $\phi:M_2(\mathbb{R}) \to \mathbb{R}$ given by $\phi(A) = \text{trace}(A)$. The kernel of $\phi$ is isomorphic to which of the following groups?

a) $M_2(\mathbb{R})/ \{A \in M_2(\mathbb{R}):\phi(A)=0\}$

b) $\mathbb{R}^2$

c) $\mathbb{R}^3$

d) $GL_2(\mathbb{R})$

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    Try and write out a generic matrix of trace zero. – 2017-01-01
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    I think option a is correct? – 2017-01-01
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    No, please think again. That is the quotient over the kernel. – 2017-01-01
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    Kernel of ϕ is the collection of those matrices whose trace are zero, – 2017-01-01
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    I recommend consulting the help menu for advice on formatting math on this site. – 2017-01-01
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    i can not understand...how to solve the problem..plz help me – 2017-01-01
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    Find a nice formula that parametrizes all traceless matrices. – 2017-01-01
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    If $\begin{bmatrix} a&b\\c&d\end{bmatrix} $ has trace zero, what's the relation between $a$ and $d$? How many parameters are left? – 2017-01-01
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    a+d=0 and there are 2 parameters left – 2017-01-02
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    $a+d=0$ eliminates only one parameter. – 2017-01-02
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    yes..a+d=0 eliminates one parameter then there are total 3 parameter – 2017-01-02

1 Answers 1

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The map is linear and surjective. The rank-nullity theorem says $$ \dim M_2(\mathbb{R})=\dim\ker\phi+1 $$ because the image of $\phi$ has dimension $1$.

Therefore , one of the choices is obvious. However, if you are really dealing with group homomorphisms, then only one choice is excluded, namely the only group that has non trivial torsion elements. (Thanks to Omnomnomnom who drew more attention to the question's formulation.)

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    Note that with the axiom of choice, there is a *group isomorphism* between $\Bbb R^3$ and $\Bbb R$, since both are free $\Bbb Z$ modules with a basis of the same uncountable cardinality. – 2017-01-02
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    @Omnomnomnom Added, thanks. But they're free $\mathbb{Q}$-modules with a basis of the same uncountable cardinality. – 2017-01-02