Let \[ B=\begin{pmatrix} a&0&0\\ 0&b&0\\ 0&0&c \end{pmatrix}, \] where \(a,b,c\in\mathbb R\) and \(|a|,|b|,|c|\) are distinct. Show that there are exactly four symmetric matrices of the form \(BQ\), where \(Q\) is a real orthogonal matrix with determinant \(1\).
Solution
Let \[ A=BQ, \] where \(Q\in SO(3)\), and suppose that \(A\) is symmetric.
1. One has\[ A^2=B^2. \]
Since \(A\) is symmetric, \[ A^2=AA^T. \] On the other hand, \[ AA^T = BQ(BQ)^T = BQQ^TB = B^2, \] because \(Q\) is orthogonal and \(B^T=B\).
2. The matrix \(A\) commutes with \(B^2\).
By step 1, \[ AB^2 = AA^2 = A^3 = A^2A = B^2A. \]
3. The matrix \(A\) is diagonal.
Write \[ B^2 = \operatorname{diag}(a^2,b^2,c^2). \] The numbers \(a^2,b^2,c^2\) are pairwise distinct because \(\abs a,\abs b,\abs c\) are pairwise distinct. If \(A=(A_{ij})\), the equality \[ AB^2=B^2A \] from step 2 gives, for \(i\neq j\), \[ A_{ij}(d_j-d_i)=0, \] where \((d_1,d_2,d_3)=(a^2,b^2,c^2)\). Since \(d_i\neq d_j\), every off-diagonal entry \(A_{ij}\) is zero.
4. If \(a,b,c\) are all nonzero, there are exactly four possiblematrices \(A\).
By steps 1 and 3, \[ A = \operatorname{diag}(\epsilon_1a,\epsilon_2b,\epsilon_3c), \qquad \epsilon_i\in\{1,-1\}. \] Since \(B\) is invertible, \[ Q=B^{-1}A = \operatorname{diag}(\epsilon_1,\epsilon_2,\epsilon_3). \] Thus \[ \det Q=\epsilon_1\epsilon_2\epsilon_3. \] The condition \(\det Q=1\) is therefore \[ \epsilon_1\epsilon_2\epsilon_3=1. \] Exactly four of the eight sign triples satisfy this condition, and each gives a distinct symmetric matrix \(A=BQ\).
5. If one of \(a,b,c\) is zero, there are exactly four possiblematrices \(A\).
At most one of \(a,b,c\) can vanish because their absolute values are pairwise distinct. After relabeling coordinates, suppose \(a=0\). Steps 1 and 3 then force \[ A = \operatorname{diag}(0,\epsilon_2b,\epsilon_3c), \qquad \epsilon_2,\epsilon_3\in\{1,-1\}. \] Hence there are at most four possibilities.
Conversely, for every choice of \(\epsilon_2,\epsilon_3\), set \[ \epsilon_1=\epsilon_2\epsilon_3 \] and \[ Q = \operatorname{diag}(\epsilon_1,\epsilon_2,\epsilon_3). \] Then \(Q\) is orthogonal and \[ \det Q = \epsilon_1\epsilon_2\epsilon_3 = 1, \] while \[ BQ = \operatorname{diag}(0,\epsilon_2b,\epsilon_3c). \] Thus all four possibilities occur.
6. There are exactly four symmetric matrices of the required form.
Either none of \(a,b,c\) is zero, in which case step 4 applies, or exactly one is zero, in which case step 5 applies.
7. Q.E.D.
Step 6 is the required conclusion.