Let $H$ be a Hilbert space and $B(H)$ is the space of all bounded linear operators on $H$.
An operator $A \in B(H)$ is said to be hyponormal if $A^*A-AA^* \ge 0 > $, where $A^*$ is the adjoint of $A$.
An operator $T \in B(H)$ is said to be $M$- hyponormal if there exists a positive constant $M$ such that
$$\|\left( T - \lambda\right)^*x \| \le M \| \left( T - \lambda x\right)\|,$$ for all $\lambda \in \mathbb{C}$ and $x \in H$.
Actually, if $H$ is a finite dimensional space then any hyponormal operator is normal.
How about the case of $M$- hyponormal operator? Does it become normal or hyponormal when $\dim {H} < \infty$?
I tried to find a counter example but still not get it.
Could you please help me to find counter example or the way to prove?
Let $H=\mathbb C^n$.
Since the inequality is invariant by unitary conjugation, we may assume (via the Schur Decomposition) that $T$ is upper triangular.
Consider first $x=e_1$, $\lambda=T_{11}$. Then $(T-\lambda I)x=0$, so $$ \|(T-\lambda I)^*x\|\leq M\|(T-\lambda I)x\|=0. $$ Thus $$ 0=T^*e_1-\overline{T_{11}}\,e_1=\sum_{k=2}^n\overline{T_{1k}}e_k. $$ So $T_{1k}=0$ for all $k\geq2$. Now let $x=e_2$, $\lambda=T_{22}$. Then, since $(T-\lambda I)x=0$, $$ 0=(T^*-\overline{T_{22}} I)e_2=\sum_{k=3}^n\overline{T_{2k}}e_k, $$ and it follows that $T_{23}=T_{24}=\cdots=T_{2n}=0$. By repeating this argument inductively, we obtain that $T$ is diagonal. That is, $T$ is normal.