Give an example of two self-adjoint operators $ \ T_1, T_2 \ ∈ L(F^4) \ $

Question

Give an example of two self-adjoint operators $ \ T_1, T_2 \ ∈ L(F^4) \ $ such that the eigenvalues of both operators are $ \ 2, 5, 7 \ $ but there does not exist an Isometry $ \ S ∈ L(F^4) \ $ such that $ \ T_1 = S^* T_2 S \ $.

Be sure to explain why there is no isometry with the required property.

where $ \ F \ $ is any field.

Answer:

Since $ \ T_1 \ T_2 \ $ has eigen values $ \ 2,5,7 \ $ and since $ \ T_1, \ T_2 \in L(F^4) $ , we conclude that

at least one eigen values is repeated .

Thus both of $ \ T_1 , \ T_2 \ $ are not diagonalisable.

So there does not exists any invertible operator $ \ T' \ $ such that

$ \large T_1=(T^{'})^{-1} TT' \ $

But I am unable conclude whether why no Isometry is there .

Help me out

Answer 1

It it not true that both operators being non-diagonalisable implies they are not similar. Furthermore, repeated eigenvalues does not imply non-diagonalisability either.

In fact, consider

$$T_1=\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 7 \\ \end{pmatrix}$$

and

$$T_2=\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 7 \\ \end{pmatrix}$$

For $T_1$, the eigenvalues are $2,2,5,7$ (counting multiplicity), while for $T_2$ they are $2,5,5,7$.

Both these matrices are in Jordan normal form. If there existed any operator $S$ with $T1=S^{-1}T_2S$, then $T_1$ and $T_2$ would be similar, hence would have the same Jordan normal form - but they don't.

For an isometry, it happens that $S^*=S^{-1}$.

Answer 2

Here is a reasoning that the operators introduced by @lisyarus indeed work without using Jordan forms. $$T_1=\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 7 \\ \end{pmatrix}, T_2=\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 7 \\ \end{pmatrix}$$ Define operators $T_1, T_2$ according to the matrices above with respect to some orthonormal basis. Then since these matrices equal their conjugate transpose their respective operators are self-adjoint. We will prove there is not any matrix $P$ such that $T_1 = P^{-1}T_2P$ then since for isometries $S^* = S^{-1}$ holds, the conclusion follows. Assume that there is such $P$. Then we would have $$\operatorname{trace}(T_1)=\operatorname{trace}(P^{-1}T_2P)= \operatorname{trace}((P^{-1}T_2)P) = \operatorname{trace}(P(P^{-1}T_2)) = \operatorname{trace}(T_2).$$ Where we used $\operatorname{trace}(AB)=\operatorname{trace}(BA)$ and matrix associativity. The last equality is a contradiction assuming $\mathbb{F}\in\{\mathbb{R}, \mathbb{C}\}$.