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I'm working on a homework that is asking me to show a few things about SO(3) matrices:

1) Show that the columns of $R\in SO(3)$ form a right-handed orthonormal set of basis vectors.

I have no trouble showing that the columns are orthonormal to each other (by setting $R=(\mathbf{v}_1 \; \mathbf{v}_2 \; \mathbf{v}_3)$ and considering that $R^T R = I$, but how do I use that $det(R)=1$ to show that this is a right-handed basis set? Ie, how do I show that $\mathbf{v}_i\times \mathbf{v}_j = \epsilon_{ijk}\mathbf{v}_k$?

2) $R$ has at least one eigenvector with eigenvalue $1$.

I'm more stuck on this one. I'm guessing that I need to use $det(R-I)=0$, but expanding this out in terms of the elements of $R$ doesn't seem to yield anything that's easy to prove... is this the correct approach?

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up vote 1 down vote accepted

1) What is the relationship between the triple product $(v_i \times v_j) \cdot v_k$ and $\det R$, and what would this triple product tell you about $v_i \times v_j$?

2) Be careful you know that $\det(R-I) = 0$ for a correct reason, e.g., $$\det(R-I) = \det(R - R^T R) = \det(I-R^T)\det R = (-1)^3 \det(R - I) \det R = -\det(R-I).$$ Now, what do you know, a priori, about the nullity of a non-invertible square matrix? Can a non-invertible square matrix have nullity $0$, or is this impossible?

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Great, I think I know what to do now. Thanks! – alexvas Feb 13 '13 at 9:31

For 1), you have $\langle x, a \times b \rangle = \det \begin{bmatrix} x & a & b \end{bmatrix}$. Since $\det \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} = 1$, it follows that $\langle v_i, v_j \times v_k \rangle = \det \begin{bmatrix} v_i & v_j & v_k \end{bmatrix} = \epsilon_{ijk}$. Since the $v_l$ are orthonormal, it follows that $\epsilon_{ijk} \, v_i = v_j \times v_k$.

For 2), if $\lambda_k$ are the eigenvalues of $R$, then since $\|Rx\| = \|x\|$ for all $x$, it follows that $|\lambda_k| = 1$. Also, $\lambda_1 \lambda_2 \lambda_3 = 1$, since $R \in SO(3)$. If all eigenvalues are real, then the number of negative eigenvalues must be even, hence at least one has the value $1$. If an eigenvalue is complex, it must occur in a conjugate pair whose product is positive, hence the remaining eigenvalue must be $1$.

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I was hoping for hints rather than solutions... – alexvas Feb 13 '13 at 9:30
You should ask for what you want. For 1), you wrote 'how do I show that...'. For 2) I didn't answer your question, I just showed that an eigenvalue of $1$ exists. – copper.hat Feb 13 '13 at 9:55

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