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I'm supposed to determine the matrix of the reflection of a vector $v \in \mathbb{R}^{3}$ around the plane $z = 0$, parallel to the line $x = y = z$. I think this means that, denoting the plane by $E$ and the line by $F$, we will have $\mathbb{R}^{3} = E \oplus F$ and thus for a vector $v$, we write $v = z + w$ where $z \in E$ and $w \in F$, and then we set $Rv = z + Rw$? Then I guess we'd have $Rw = -w$ making $Rv = z - w$. Here $R$ denotes the reflection. Is this the correct definition?

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  • $\begingroup$ That sounds like it would work to me. $\endgroup$ – Nathan Reed Jan 24 '13 at 1:28
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Technically $E \ \dot{+} \ F = \mathbb{R}^{3}$, or $E \oplus F \simeq \mathbb{R}^{3}$, but it is a matter of debating $\mathbb{R}^{2} \oplus \mathbb{R} = \mathbb{R}^{3}$ should be an equality. The left and right hand side also are homeomorphic to each other and they don't need any distinctions between them.

If you want to work with $E \oplus F$, your $v \in \mathbb{R}^{3}$ is thus (technically) has an ismorphic image, which you called $v$ in $E \oplus F$.

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  • $\begingroup$ But by $E$ we mean $E = \{(x, y, 0); x, y \in \mathbb{R}\}$ and by $F$ we mean $F = \{(x, x, x); x \in \mathbb{R}\}$. They are both subspaces of $\mathbb{R}^{3}$ this way, so I don't see how can't we write $\mathbb{R}^{3} = E \oplus F$. $\endgroup$ – Pedro Jan 24 '13 at 10:07
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The definition is exactly as stated in the question.

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