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 :
- $I_3$ and $-I_3$ are in $E$:
True because $I_3 = M(1, 0, 0)$, $-I_3 = M(-1, 0, 0)$
- $\times$ 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 !)