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let $V = M_{2×2}(\Bbb{R})$.

If $x = \begin{pmatrix}x_1 & x_2\\ x_3 & x_4 \end{pmatrix}$ and $y = \begin{pmatrix}y_1 & y_2\\ y_3 & y_4 \end{pmatrix}$ ; define $x ⊕ y = \begin{pmatrix}x_1 & y_2\\ y_3 & x_4 \end{pmatrix}$

and $λ \otimes \begin{pmatrix}x_1 & x_2\\ x_3 & x_4 \end{pmatrix} = \begin{pmatrix}\lambda x_1 & \lambda x_2\\ \lambda x_3 & \lambda x_4 \end{pmatrix}$, for any $x$, $y \in V$ and $λ \in \Bbb{R}$.

I don't really understand how to show that this defines a vector space using the axioms?

Thanks for your help

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    $\begingroup$ Why not just verify the axioms? $\endgroup$ – anomaly Oct 17 '17 at 18:52
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It is not a vector space because $V$ with ⊕ operation is not a commutative group.

Indeed operation $\oplus$ is not commutative :

$y ⊕ x = \begin{pmatrix}y_1 & x_2\\ x_3 & y_4 \end{pmatrix}$ is different from $x ⊕ y$ in general.

One can check that the ⊕ operation is associative:

$\left(\begin{pmatrix}a & c\\ b & d \end{pmatrix}⊕\begin{pmatrix}e & g\\ f & h \end{pmatrix}\right)⊕\begin{pmatrix}i & k\\ j & l \end{pmatrix} \ = \ \begin{pmatrix}a & c\\ b & d \end{pmatrix}⊕\left(\begin{pmatrix}e & g\\ f & h \end{pmatrix}⊕\begin{pmatrix}i & k\\ j & l \end{pmatrix}\right)$

with a common value which is $\begin{pmatrix}a & k\\ j & d \end{pmatrix}$.

But, as remarked by @jabo, there is no neutral element.

Thus $V$ with $\oplus$ operation is not even a group.

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  • $\begingroup$ Yes. Neither is there a neutral element of $(V,\oplus)$. $\endgroup$ – Jan Bohr Oct 17 '17 at 17:07

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