IMC 2006 · Day 2 · P12

We note that the problem is trivial if $A_j = \lambda I$ for some $j$, so suppose this is not the case. Consider then first the situation where some $A_j$, say $A_3$, has two distinct real eigenvalues. We may assume that $A_3 = B_3 = \begin{pmatrix} \lambda & \\ & \mu \end{pmatrix}$ by conjugating both sides. Let $A_2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $B_2 = \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}$. Then $$\begin{align*} a + d &= \operatorname{Tr} A_2 = \operatorname{Tr} B_2 = a' + d' \\ a \lambda + d \mu &= \operatorname{Tr}(A_2 A_3) = \operatorname{Tr} A_1^{-1} = \operatorname{Tr} B_1^{-1} = \operatorname{Tr}(B_2 B_3) = a' \lambda + d' \mu. \end{align*}$$ Hence $a = a'$ and $d = d'$ and so also $bc = b'c'$. Now we cannot have $c = 0$ or $b = 0$, for then $(1, 0)^\top$ or $(0, 1)^\top$ would be a common eigenvector of all $A_j$. The matrix $S = \begin{pmatrix} \frac{c'}{c} & \\ & 1 \end{pmatrix}$ conjugates $A_2 = S^{-1} B_2 S$, and as $S$ commutes with $A_3 = B_3$, it follows that $A_j = S^{-1} B_j S$ for all $j$.

If the distinct eigenvalues of $A_3 = B_3$ are not real, we know from above that $A_j = S^{-1} B_j S$ for some $S \in GL_2 \mathbb{C}$ unless all $A_j$ have a common eigenvector over $\mathbb{C}$. Even if they do, say $A_j v = \lambda_j v$, by taking the conjugate square root it follows that $A_j$'s can be simultaneously diagonalized. If $A_2 = \begin{pmatrix} a & \\ & d \end{pmatrix}$ and $B_2 = \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}$, it follows as above that $a = a'$, $d = d'$ and so $b' c' = 0$. Now $B_2$ and $B_3$ (and hence $B_1$ too) have a common eigenvector over $\mathbb{C}$ so they too can be simultaneously diagonalized. And so $S A_j = B_j S$ for some $S \in GL_2 \mathbb{C}$ in either case. Let $S_0 = \operatorname{Re} S$ and $S_1 = \operatorname{Im} S$. By separating the real and imaginary components, we are done if either $S_0$ or $S_1$ is invertible. If not, $S_0$ may be conjugated to some $T^{-1} S_0 T = \begin{pmatrix} x & 0 \\ y & 0 \end{pmatrix}$, with $(x, y)^\top \ne (0, 0)^\top$, and it follows that all $A_j$ have a common eigenvector $T (0, 1)^\top$, a contradiction.

We are left with the case when no $A_j$ has distinct eigenvalues; then these eigenvalues by necessity are real. By conjugation and division by scalars we may assume that $A_3 = \begin{pmatrix} 1 & b \\ & 1 \end{pmatrix}$ and $b \ne 0$. By further conjugation by upper-triangular matrices (which preserves the shape of $A_3$ up to the value of $b$) we can also assume that $A_2 = \begin{pmatrix} 0 & u \\ 1 & v \end{pmatrix}$. Here $v^2 = \operatorname{Tr}^2 A_2 = 4 \det A_2 = -4u$. Now $A_1 = A_3^{-1} A_2^{-1} = \begin{pmatrix} -(b+v)/u & 1 \\ 1/u & 0 \end{pmatrix}$, and hence $(b+v)^2/u^2 = \operatorname{Tr}^2 A_1 = 4 \det A_1 = -4/u$. Comparing these two it follows that $b = -2v$.

What we have done is simultaneously reduced all $A_j$ to matrices whose all entries depend on $u$ and $v$ ($= -\det A_2$ and $\operatorname{Tr} A_2$, respectively) only, but these themselves are invariant under similarity. So $B_j$'s can be simultaneously reduced to the very same matrices.