# Rearrange order of eigenvalues of triangular matrices

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.

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

• Just wondering: If the two eigenvalues are identical, then how do you even know that they’ve been swapped?
– amd
May 17, 2020 at 18:17