Let $M= \lbrace g\lambda g^{-1} , $ $ g \in SU(3)\rbrace $ such that $\lambda = diag(i\lambda_1, i\lambda_2, i\lambda_3) $, $\lambda_i \in \mathbb{R}$ and $\lambda_1 > \lambda_2 > \lambda_3 .$
Let $T= \lbrace diag (t,t,t') , t,t' \in U(1)\rbrace$ be the subgroup of SU(3) of diagonal matrices such that the first and the second entries are equal. Consider the action of T on M by conjugation.
What is the set $M^T=\lbrace m \in M , t.m=m \forall t \in T\rbrace $ of fixed point in $ M $ under the action of T on M ?
I greatly appreciate any help in this.
Hint: We have $A \cdot B = B$ if and only if $AB = BA$, where $AB$ denotes the matrix product of the matrices $A$ and $B$. Now, consider the specific matrix $$ A_0 = \pmatrix{-1 \\ & -1\\ && 1} \in T. $$ If $B$ is any matrix such that $A_0B = BA_0$, what can we say about the matrix $B$? Show that if $B$ is of this required form, then we have $AB = BA$ for every matrix $A \in T$.
Then, note that the set $M$ can be nicely characterized with the help of the spectral theorem.