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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?

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    $\begingroup$ Looks good to me. $\endgroup$
    – Thomas Andrews
    Commented Sep 18, 2021 at 16:26
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    $\begingroup$ The $M$ on the second line should be on the right. It doesn't make a difference, but is more rigorous. $\endgroup$
    – Trebor
    Commented Sep 18, 2021 at 16:31
  • $\begingroup$ You're right! Thanks for pointing it out, @Trebor! $\endgroup$
    – Analysis
    Commented Sep 18, 2021 at 17:22
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    $\begingroup$ It is a neat answer and works for odd sized matrices over any ring ( without $2$ torsion). $\endgroup$
    – orangeskid
    Commented Sep 18, 2021 at 21:50

2 Answers 2

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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.

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  • $\begingroup$ That's funny. My definition of orthogonal matrix $A$ has always been $A^\top A = I$. $\endgroup$
    – Ted Shifrin
    Commented Sep 18, 2021 at 21:59
  • $\begingroup$ It’s not hard to go from one to the other. @TedShifrin As long as you know $x\cdot (Ay)=(A^Tx)\cdot y.$ I prefer the distance-preserving definition because it is more geometric. But the $A^TA=I$ version generalizes to all fields. $\endgroup$
    – Thomas Andrews
    Commented Sep 18, 2021 at 22:22
  • $\begingroup$ Yes, as a geometer and linear algebra textbook author, I'm fully aware of this. I'm just saying that your definition is perhaps not the usual one. I don't disagree with your view on the geometry. $\endgroup$
    – Ted Shifrin
    Commented Sep 18, 2021 at 22:28
  • $\begingroup$ @TedShifrin btw, is that the definition even over $\mathbb C?$ I’d have thought it would be the conjugate transpose, but my linear algebra is certainly not good. $\endgroup$
    – Thomas Andrews
    Commented Sep 19, 2021 at 2:24
  • $\begingroup$ No, the orthogonal group has no conjugate; it’s an algebraic group. The unitary group $U(n)$ preserves the hermitian inner product and has the conjugate. :) $\endgroup$
    – Ted Shifrin
    Commented Sep 19, 2021 at 2:26
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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

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

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  • $\begingroup$ “Must have an axis of rotation.” Why? That seems like an assertion. It is true, but it isn’t true for $2\times 2$ matrices or $4\times 4$ matrices, so the assertion seems not at all obvious. $\endgroup$
    – Thomas Andrews
    Commented Sep 18, 2021 at 17:02
  • $\begingroup$ I understand the intuition, but geometry requires more than intuition. This is an intuitive argument, not a replacement for a proof. $\endgroup$
    – Thomas Andrews
    Commented Sep 18, 2021 at 17:49
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    $\begingroup$ @Thomas Andrews Although this argument isn't totally rigorous it nevertheless demonstrates the invaluable property that every matrix belonging to $SO_3(\mathbb{R}^3)$ is a proper rotation in some two dimensional subspace in $\mathbb{R}^3$. This argument is one that someone who barely knows linear algebra can understand and, in my opinion, is more reflective than an argument consisting of mere symbol pushing. I'm all for it. $\endgroup$
    – user721971
    Commented Sep 18, 2021 at 20:49
  • $\begingroup$ But that brings up another point. $SO$ is not usually defined as rotation matrices, because to do so, you’d have to define a rotation matrix, which is tedious in general dimensions. So the fact that $SO$ is rotation matrices, while true, is already a fairly high-powered statement. It requires, at the very least, a definition of “rotation matrix,” and then a proof. So applying a fairly high-powered theorem along with a mere intuition seems to make this “proof” all the more dubious. $\endgroup$
    – Thomas Andrews
    Commented Sep 18, 2021 at 21:00
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    $\begingroup$ The only way I know how to justify/prove your assertion is to use the result the OP is asking to prove. $\endgroup$
    – Ted Shifrin
    Commented Sep 18, 2021 at 22:00

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