Let \(A:\mathbb R^3\to\mathbb R^3\) have a matrix, in the standard basis, that is both symmetric and orthogonal. Prove that \(A\) is one of the following:
- \(I\) or \(-I\);
- a rotation through \(180^\circ\) about an axis;
- a reflection across a two-dimensional subspace.
Solution
1. One has\[ A^2=I, \]so every eigenvalue of \(A\) is either \(1\) or \(-1\).
Orthogonality gives \[ A^{\mathsf T}A=I, \] while symmetry gives \[ A^{\mathsf T}=A. \] Hence \(A^2=I\). Therefore every eigenvalue \(\lambda\) satisfies \[ \lambda^2=1, \] so \(\lambda=\pm1\).
2. There is an orthogonal direct-sum decomposition\[ \mathbb R^3=V_+\oplus V_-, \]where\[ V_+=\ker(A-I), \qquad V_-=\ker(A+I), \]and \(A\) acts as \(I\) on \(V_+\) and as \(-I\) on \(V_-\).
Since \(A\) is real symmetric, the spectral theorem gives an orthonormal basis of eigenvectors. Step 1 shows that all corresponding eigenvalues are \(1\) or \(-1\). Thus the spans of the two kinds of eigenvectors are \(V_+\) and \(V_-\) and are orthogonal complements.
3. If\[ \dim V_+=3 \quad\text{or}\quad \dim V_+=0, \]then respectively\[ A=I \quad\text{or}\quad A=-I. \]
If \(V_+=\mathbb R^3\), step 2 says that \(A\) fixes every vector. If \(V_-=\mathbb R^3\), it negates every vector.
4. If \(\dim V_+=1\), then \(A\) is a rotation through\(180^\circ\) about the axis \(V_+\).
Let \(L=V_+\). By step 2, \[ V_-=L^\perp. \] Thus \(A\) fixes every vector on the axis \(L\) and sends every vector in the perpendicular plane \(L^\perp\) to its negative. On each plane perpendicular to \(L\), this is exactly rotation by angle \(\pi\). Hence \(A\) is rotation through \(180^\circ\) about \(L\).
5. If \(\dim V_+=2\), then \(A\) is reflection across the plane \(V_+\).
Let \(P=V_+\). Then \[ V_-=P^\perp \] is one-dimensional. Step 2 says that \(A\) fixes \(P\) pointwise and negates the normal direction \(P^\perp\). This is precisely reflection across \(P\).
6. These cases exhaust all possibilities.
The integer \(\dim V_+\) is one of \[ 0,1,2,3. \] Steps <1>3–<1>5 identify the transformation in each case.
7. Q.E.D.
Step 6 gives exactly the classification stated in the problem.