# Prove that $T(v) = v,$ for some $v \in \Bbb R^3 \setminus \{0\}.$

Let $$T \in \text {SO} (3) : = \left \{T \in M_3 (\Bbb R)\ |\ TT^t = T^tT = I,\ \det (T) = 1 \right \}.$$ Prove that there exists $$v \in \Bbb R^3 \setminus \{0\}$$ such that $$T(v) = v.$$ Hence conclude that the elements of $$\text {SO} (3)$$ are rotations by some angle about some uniquely determined axis.

The first part is easier. To prove that $$T(v) = v,$$ for some $$0 \neq v \in \Bbb R^3$$ it is enough to show that $$\det (T - I) = 0.$$ Now \begin{align*} \det (T - I) & = 1 \cdot \det (T - I) \\ & = \det (T) \det (T - I) \\ & = \det (T^t) \det (T - I) \\ & = \det (T^t T - T^t) \\ & = \det (I - T^t) \\ & = \det (I^t - T^t) \\ & = \det ((I - T)^t) \\ & = \det (I - T) \\ & = (-1)^3 \det (T - I) \\ & = - \det (T - I) \end{align*} This shows that $$2 \det (T - I) = 0 \implies \det (T - I) = 0,$$ as required.

How to prove the second conclusion? Any help in this regard will be appreciated.

• ${\rm span}(v)$ and $v^\perp$ are both invariant subspaces of $T$ and $T|_{v^\perp}\in SO(2)$.. – Berci Dec 12 '20 at 19:12

Take $$v\ne0$$ such that $$T(v)=v$$. Let $$w\in\Bbb R^3$$ be a vector orthogonal to $$v$$. Then\begin{align}\langle T(w),v\rangle&=\langle T(w),T(v)\rangle\\&=\langle w,T^tT(v)\rangle\\&=\langle w,v\rangle\\&=0.\end{align}Now let $$u=v\times w$$. Then $$u$$ and $$v$$ are orthogonal and therefore $$T(u)$$ and $$v$$ are orthogonal too. And, since $$w$$ and $$u$$ are orthogonal, then so are $$T(w)$$ and $$T(u)$$. The restriction of $$T$$ to $$\operatorname{span}(\{w,u\})$$ is a rotation (since it is an isometry and its determinant is $$1$$). So, $$T$$ is a rotation around $$v$$.

• Since $T$ preserves distances and angles, then so does its restriction to $\operatorname{span}(\{w,u\})$. So, $(0,a,c)$ and $(0,b,d)$ must have norm $1$ and be orthogonal. So, for some $\theta\in\Bbb R$,$$\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\quad\text{or}\quad\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}.$$But only the first of these matrices has determinant $1$. – José Carlos Santos Dec 12 '20 at 19:45
• Sir what I can observe is that $\text {span} \left (\{w,u\} \right )$ is invariant under $T$ and the matrix of $T$ relative to the ordered basis $\{v,w,u \}$ is of the form $\begin{pmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & c & d \end{pmatrix}$ where $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = 1.$ Since $T$ is orthogonal it follows that $(0,a,c)$ and $(0,b,d)$ are orthonormal and hence $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text {SO} (2).$ How does it imply that $T$ is a rotation around $T\$? – math maniac. Dec 12 '20 at 19:46
• What does that mean? What I claimed is that $T$ is a rotation around $v$. – José Carlos Santos Dec 12 '20 at 19:47

As you showed that there is $$v_1$$ such that $$Tv_1=v_1$$. Let $$P$$ be the plane whose normal vector is $$v_1$$. Firstly we should show $$Tx \in P$$ for all $$x \in P$$. This equivalent to show that $$Tx$$ and $$v_1$$ are orthogonal.

$$==x^TT^TTv_1=x^Tv_1=0 \hspace{0.2cm} \text{since x and v_1 are orthogonal.}$$

Secondly, we must show that the induced map $$T|_P$$ is rotation. Let this be denoted by $$G$$. So $$G$$ can be represented by $$2 \times 2$$ matrix with eigenvalues $$e,f$$.

Actually $$e$$ and $$f$$ are eigenvalues of $$T$$, too. So they are $$e=e^{i \theta}$$ and $$f=e^{-i \theta}$$.

Since the eigenvalues are distinct, we can directly say that $$G$$ is similar to the matrix $$\begin{bmatrix} cos \theta & -sin \theta \\ sin \theta & cos \theta \end{bmatrix}$$. We know that similar matrices represent same linear transformation, we can directly to say that $$G$$ is a rotation.

You have already proven that there exists $$u$$ such that $$T(u) = u$$. Complete $$u$$ to an orthogonal basis $$u, v, w$$. The matrix of $$T$$ in this basis is

$$M(T) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & c & d \end{pmatrix}$$

where the form of the first column follows from $$T(u) = u$$ and the form of the first row follows from the fact that rows and columns of an orthogonal matrix have unit 2-norm. Applying this property to the second column and second row, we also see that

$$a^2 + b^2 = 1 \\ a^2 + c^2 = 1$$

so $$|b| = |c|$$. Similarly, $$|a| = |d|$$. Therefore we can write

$$M(T) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & \pm b & \pm a \end{pmatrix}$$

where so far all four possibilities for the signs in the bottom row are possible. However, since the columns are orthogonal to each other we see that either

$$M(T) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & b & -a \end{pmatrix}$$

or

$$M(T) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & a & b \\ 0 & -b & a \end{pmatrix}\tag1.$$

However, the first case is ruled out because then $$\det(M(T)) = -a^2 - b^2 = -1$$.

Now, notice that $$a \in [-1, 1]$$, so there exists an angle $$\beta \in [0, \pi]$$ such that $$a = \cos\beta$$. Also, comparing $$a^2 + b^2 = 1$$ to $$\cos^2\beta + \sin^2\beta = 1$$ we see that $$b = \sin\beta$$ or $$b = -\sin\beta$$. In the letter case, we can absorb the minus sign in the angle by defining $$\theta = 2\pi - \beta$$. In the first case, we set $$\theta = \beta$$. Substituting this into $$(1)$$, we get

$$M(T) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & \sin\theta \\ 0 & -\sin\theta & \cos\theta \end{pmatrix}$$

which we recognize as the rotation matrix around the $$u$$ axis by angle $$\theta$$.

• Not along the $x$-axis but along the $u$-axis. Here the ordered basis is $\{u,v,w\}$ not $\{e_1,e_2,e_3\}.$ So we have found that any non-trivial element of $\text {SO} (3)$ is a rotation by an angle around a uniquely determined axis. Can we generalize this result for $\text {SO} (n)\$? – math maniac. Dec 12 '20 at 20:34
• Thanks, @mathmaniac. Fixed. The structure of matrices in $SO(n)$ is somewhat more complicated, but yes, you can generalize certain aspects of SO(3). In particular, if $n$ is odd, then $1$ is always among the eigenvalues. However, there are also differences. For example, the "rotoreflection" blocks (with negative cosine on diagonal) whose presence we ruled out above can appear in pairs for $n\geq4$. – Adam Zalcman Dec 12 '20 at 20:43
• Thank you so much @Adam Zalcman for your remark. Can you please suggest me some book where I can find such things? – math maniac. Dec 13 '20 at 5:00