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Let $$\begin{array}{ccc} M : & \mathbb{R}^3 & \longrightarrow & M_3(\mathbb{R}) \\ & (a, b, c) & \longmapsto & {\begin{pmatrix} a & b & c \\ b & a+c & b \\ c & b & a\end{pmatrix}} \end{array}$$

Let $E = \left\{M(a, b, c), (a, b, c) \in \mathbb{R}^3\right\}$

1) Show that $(E, +, \cdot)$ is a vector space.

2) Show that $(E, +, \times)$ is a ring.

1) Clearly : $$E= \left\{a\cdot I_3 + b \cdot J + c \cdot K, (a, b, c) \in \mathbb{R}^3\right\}$$ Where : $$I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$ $$J = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}$$ $$K = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{pmatrix}$$

So $E = \mathrm{Vect}\left(I_3, J, K\right) \iff (I_3, J, K)$ basis of $E$ as linearly independent.

Hence : $E$ is a vector space, and $\dim(E)=3$.

2) $E$ is a vector space, so we just have to show :

  1. $I_3$ and $-I_3$ are in $E$:

True because $I_3 = M(1, 0, 0)$, $-I_3 = M(-1, 0, 0)$

  1. $\times$ is stable for $E$

A quite tedious demonstration here, but I managed to show it..

$\to$ We can also show that $E$ is commutative...


Now my question is : Is there any more elegant way to answer these two questions ?

I was thinking maybe using homorphism and the fact that $M_3(\mathbb{K})$ is both a vector space and a ring (non-commutative, but that was not part of the questions), but I don't know how to do it (also, if you got an even more elegant way, I would appreciate too !)

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Hint: Show that $M:\mathbb R^3\to M_3(\mathbb R)$ is a $\mathbb R$-linear map and hence $E$ being the image set $M(\mathbb R^3)$ of $\mathbb R^3$, $E$ is a subspace of $M_3(\mathbb R)$. Hence $(E+,\cdot)$ is a vector space over $\mathbb R$.

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  • $\begingroup$ Ah indeed this one is quite elegant. Can I do the same for ring ? (Is there an equivalent of linear map for rings ?) $\endgroup$ – servabat Feb 14 '15 at 18:04
  • $\begingroup$ I think no. It seems to me that $M$ would not be a ring homomorphism. I am trying. I will try to fix this. $\endgroup$ – user149418 Feb 14 '15 at 18:06
  • $\begingroup$ Since $E$ already has an additive structure, it is better to verify the rest of the properties for $E$ to have a ring structure than finding any ring homomorphism. $\endgroup$ – user149418 Feb 14 '15 at 21:30
  • $\begingroup$ I guess you are right ! Anyway that method is yet way more elegant. Thanks ! By the way, because it is the image of $\matbb{R}^3$ means it is of the same dimension that it ? Or I still need to apply my method to find the dimension ? $\endgroup$ – servabat Feb 14 '15 at 21:32

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