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$D_4$ is generated by a rotation $\alpha$ of order 4 and a reflection $\beta$. Its elements $e$, $\alpha$, $\alpha^2$, $\alpha^3$, $\beta$, $\alpha\beta$, $\alpha^2\beta$, $\alpha^3\beta$ give an ordered basis for the algebra $\mathbb{R}[D_4]$.

I have to express the left action of $\alpha$ and $\beta$ as matrices with respect to this basis. I then have to find the invariant one dimensional subspaces for each of these actions.

I think the second part should be straightforward. If $A$ is the matrix representing the left action of $\alpha$, I would just write $Av=v$ and find all such vectors $v$, right? Then $v$ would be invariant. I'm not sure how to do the first part, however, in expressing $\alpha$ and $\beta$ as matrices. I think I should be achieving two $8\times 8$ matrices but I don't know how.

Any help is appreciated!

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  • $\begingroup$ Well, call the basis $e_1, \dots, e_8$. Then $Ae_1=\alpha e=\alpha = e_2$ -- and so on. And for ex $Be_2=\beta\alpha=\alpha^3 \beta=e_8$. OK? $\endgroup$
    – ancient mathematician
    Mar 10, 2017 at 16:40
  • $\begingroup$ You define a group action as group elements being mapped to functions on a set. I suppose action you want to find are working on vectors in $\mathbb{R}^2$, no? $\endgroup$
    – mathreadler
    Mar 10, 2017 at 18:32

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The action of $\alpha$ is calculated as follows. Consider the left multiplication map $L_{\alpha}:D_4 \rightarrow D_4$ defined by $ L_{\alpha} : x \mapsto \alpha x$. This is a permutation and $\alpha$ acts on $\mathbb{R}[D_4]$ by permuting it's basis elements by $L_{\alpha}$. So taking the order of the basis as given we get the matrix:

$$A := \begin{pmatrix} 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ \end{pmatrix}$$

Now let us find the invariant 1-dimensional subspaces of $A$. If $\langle v \rangle$ is such a subspace then $v$ satisfies $Av = \lambda v$ for some $\lambda \in \mathbb{R}$. Solving explicitly will give you the subspaces which are spans of the following vectors,

$$v_1 = \begin{pmatrix} 1 \\ 1 \\1 \\1 \\0 \\0 \\0 \\0\end{pmatrix}, v_2 = \begin{pmatrix} 0 \\ 0 \\0 \\0 \\1 \\1 \\1 \\1\end{pmatrix} v_3 = \begin{pmatrix} 1 \\ 1 \\1 \\1 \\1 \\1 \\1 \\1\end{pmatrix}$$

And similarly for $\beta$.

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