Let $M\in \text{SO}(3,\mathbb{R})$, prove that $\det(M-I_3)=0$.

Question

Let $M\in \text{SO}(3,\mathbb{R})$, prove that $\det(M-I_3)=0$.

My attempt:

$$ \begin{align} \det(M-I_3)&=\det(M-M^TM)\\&=\det((I_3-M^T)M)\\&=\underbrace{\det(M)}_{=1}\det(I_3-M^T) \end{align} $$

Hence $$ \begin{align} \det(M-I_3)&=\det(I_3-M^T)\\&=\det((I_3-M)^T)\\&=\det(I_3-M)\\&=\det(-(M-I_3))\\&=\underbrace{(-1)^3}_{=-1}\det(M-I_3), \end{align} $$ and thus $\det(M-I_3)=0.$

Is this proof correct or did I miss out on something?

Answer 1

Your answer is certainly the most direct proof, assuming you know $M\in SO(3,\mathbb R)$ if and only if $M^TM=I_3$ and $\det M=1.$

We could, instead, go back to another definition of orthogonal matrix:

A real matrix $M$ is orthogonal if $|Mv|=|v|$ for all real vectors $v.$

Then we study the the roots of the characteristic polynomial. The characteristic polynomial $p_M(x)$ of a $3\times 3$ real matrix must have either $1$ real root or $3$ real roots.

But the real roots $\lambda_i$ are real eigenvalues, and thus must be $\pm 1.$ Otherwise, if $|\lambda_i|\neq 1,$ then let $v$ be an eigenvector so $$|Mv|=|\lambda_i||v|\neq |v|,$$ so $M$ is not orthogonal.

If there are three real roots to $p_M,$ then the $\det M$ is the product of the roots, so the real roots can’t all be $-1.$

If there is only one real root, then the other two roots come in a pair: $\lambda,\overline{\lambda},$ and $\lambda\overline{\lambda}=|\lambda|^2>0,$ so the real root cannot be negative.

This shows more generally:

If $M$ is a real $n\times n$ matrix, with $n$ odd and $\det M>0,$ then $M$ must have a positive real eigenvalue.

Answer 2

This argument checks out. For an alternative approach, start with the fact that $M$ corresponds to a specific rotation of 3D space. Thinking geometrically, it is clear that this rotation will have a unique axis of rotation (itself a 1D subspace of $\mathbb{R}^3$) and any vector $\textbf{u}$ lying on this axis will be fixed in place by the rotation, so we can write

$$M\textbf{u} = \textbf{u}$$

This shows that every vector on the axis of rotation will be an eigenvector of $M$ with a corresponding eigenvalue of $1$. It follows by the determinant characterization of eigenvalues that

$$\det(M-1\cdot I_3)=0$$

or

$$\det(M-I_3)=0$$