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Consider the tetrahedron with vertices $(1,1,1), (-1,-1,1), (1,-1,-1)$, and $(-1,1,-1)$. Give the standard matrices for each of the following symmetries of the tetrahedron. Check that each of the matrices is an orthogonal 3∗3 matrix.

a) $+/- 120$ degree rotations about the line joining $(0,0,0)$ and $(1,1,1)$ (vertex)

b) 180 degree rotation about the lines joining $(0,0,1)$ and $(0,0,-1)$

c) Reflection across the plane containing one edge and the midpoint of the other edge (not sure what this means)

I did not learn linear algebra from a geometric perspective and I don't even know where to start. Any advice or links to useful videos are appreciated.

user3642365
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1 Answers1

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Hint:

Let's say you have a linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$.

For convenience, denote $\mathbf{x}_i = T(\mathbf{e}_i)$, where $\mathbf{e}_i$ is the $i$th standard basis vector. So, for example, $\mathbf{e}_1 = \langle 1, 0, 0 \rangle$ in $\mathbb{R}^3$. Likewise, $\mathbf{e}_2 = \langle 0, 1, 0 \rangle$, etc.

Then $T$ is encoded by the $n \times n$ matrix $[\mathbf{x}_1, \mathbf{x}_2, \cdots, \mathbf{x}_n]$.

As a simple example, consider a $90^\circ$ counterclockwise rotation about the origin in $\mathbb{R}^2$. Note that $\langle 1, 0 \rangle \mapsto \langle 0, 1 \rangle$, and further $\langle 0, 1 \rangle \mapsto \langle -1, 0 \rangle$. So our linear transformation is encoded by the matrix $\left[ \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right]$.

Now applying this to solve your problem will be a bit of work, but if I'm not mistaken, the standard basis vectors lie on the midpoints of edges of this tetrahedron, and edges will map to edges. It'll be your task to figure out precisely where they are mapping to.

Another approach:

Let $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$, and $\mathbf{v}_4$ denote the vertices of your tetrahedron. Let $T$ be one of the given linear transformations. Then $T(\mathbf{v}_1) = \mathbf{v}_i$ for some $i$, and so forth. In this manner you can arrive at a system of equations whose solution yields the entries of the matrix encoding $T$.

Kaj Hansen
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  • Thanks for the response. If the standard basis for the tetrahedron is the midpoints of the edges, wouldn't there be four vectors in the basis and contradict that each standard matrix should be 3x3? – user3642365 Jan 03 '15 at 06:10
  • Oh no, I guess I was a little ambiguous. I'm saying that $\langle 1, 0, 0 \rangle$, $\langle 0, 1, 0 \rangle$, and $\langle 0, 0, 1 \rangle$. will themselves lie on edges of this tetrahedron. You'll want to figure out where they map to. – Kaj Hansen Jan 03 '15 at 06:14
  • Also, you'll want to doublecheck me on that! I'm not really near a pen/paper right now to check my work, so you'll want to make sure that actually is where the basis vectors are relative to your tetrahedron. – Kaj Hansen Jan 03 '15 at 06:16
  • That sounds right... I'm at the airport now so I can't really do math either but I'll definitely re-try this problem and update. So this problem sounds like just rotating the standard basis itself and then interpolating the new vertices from that. Correct me if I'm wrong. – user3642365 Jan 03 '15 at 06:19
  • Indeed: Figure out where those standard basis vectors go and you'll instantly have the columns of your matrix. By far the hardest part will be visualizing what's going on in $3$-space. I actually solved this exact problem a few semesters back and I was very dependent on being able to hold and manipulate an actual tetrahedron. – Kaj Hansen Jan 03 '15 at 06:22
  • I should note that there are other ways of going about this. Alternatively, you know that when a transformation $T$ acts on a vertex of the tetrahedron, then it sends it to another known vertex. So you'll have a big system of equations $T\mathbf{v_i} = \mathbf{v_j}$. – Kaj Hansen Jan 03 '15 at 06:24