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(Matrix)

$$I_2\text{ and } J = \begin{pmatrix} 0 & 1\\ -1 &0 \end{pmatrix} ∈ M_2(\mathbb{R})$$

Show that $C = [aI_2 + bJ | a,b ∈\mathbb{R}]$ is a field.

I can't solve this problem, could anyone help me? Thank you for your help:)

Bcpicao
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noany
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  • Are you defining the matrices as $diag(a_1,...a_n)$? – Bcpicao May 25 '20 at 12:31
  • No but I couldn't find the right format for the matrices. – noany May 25 '20 at 12:36
  • 2 x 2 matrices. – noany May 25 '20 at 12:37
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    I edited it. It is simply the field of complex numbers in disguise. I think this question gives you all the resources: https://math.stackexchange.com/questions/886872/history-of-the-matrix-representation-of-complex-numbers – Bcpicao May 25 '20 at 12:51
  • @Bcpicao: Not necessarily "in disguise". E T Copson, in his Introduction to the Theory of Functions of a Complex Variable, defines this set to be the complex numbers. He does this to avoid "the undefined symbol" $\sqrt{-1}$. – ancient mathematician May 25 '20 at 16:50
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    @ancientmathematician Indeed, although I maintain my formulation for two reasons: firstly, historically, that representation came afterwards. Secondly, the OP is clearly more familiar with the ad hoc representation of the complex numbers. I admit who's in disguise depends on your reference frame. – Bcpicao May 25 '20 at 17:27
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    Why can’t you solve it? What exactly are you getting stuck on? Do you know the properties that a field must have? Are there any particular properties that you’re having trouble demonstrating? – amd May 25 '20 at 19:39

1 Answers1

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It is $\mathbb{C}$, the field of complex numbers. In fact, you can see the matrix $I$ as $1$, and $J$ as $i$, since it has the same properties (show it in the proof). Then you need to verify that the product is commutative and every matrix (i.e. every complex number) has a multiplicative inverse. The calculations are simple, since you have $2\times 2$ matrices.

Nennee
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