Rearrange order of eigenvalues of triangular matrices

Question

I have given two square matrices $T_1$, $T_2$ and a basis $B$ such that $B^{-1}T_i B$, $i=1,2$, has block upper triangular form, i.e. $$B^{-1}T_i B= \left[\begin{matrix} Q_i & \ast & \ast \\ 0 & 2 & \ast \\ 0 & 0 & 1 \end{matrix}\right], $$ where $Q_i$ are square matrices and $\ast$ stands for anything. I am trying to find a basis $\tilde{B}$ such that $\tilde{B}^{-1}T_i \tilde{B}$, $i=1,2$, has block upper triangular form with eigenvalues $1$ and $2$ interchanged, i.e. $$\tilde{B}^{-1}T_i \tilde{B} = \left[\begin{matrix} \tilde{Q}_i & \ast & \ast \\ 0 & 1 & \ast \\ 0 & 0 & 2 \end{matrix}\right] $$

I do not know how to construct the matrix $\tilde{B}$ from $B$.

Any help is appreciated.

Answer 1

Interestingly, this seems to be impossible unless the eigenvalues (1 and 2 in my case) are the same, as a quick Mathematica calculation shows.

Let T1, and T2 be the original matrices, with eigenvalues d1 and d2 and let B be a basis we are looking for.

T1  = {{d1, t1 },{0, d2}}; 
T1t = {{d2, t1t},{0, d1}};
T2  = {{d1, t2 },{0, d2}};
T2t = {{d2, t2t},{0, d1}};
B = {{b1, b2},{b3, b4}};

ass = b1*b4-b2*b3!=0 && t1!=0 && t2!=0 && t1!=t2;
FullSimplify[
  Reduce[Inverse[B].T1.B == T1t && Inverse[B].T2.B == T2t && ass,
  {b1, b2, b3, b4}]
, ass]

Returns:

t1 == (t1t t2)/t2t && d1 == d2 && b3 == 0 && b4 t1 == b1 t1t