Let $H$ and $E$ be $n \times n$ matrices satisfying the relation
\[HE-EH=2E.\]
Let $\lambda$ be an eigenvalue of the matrix $H$ such that the real part of $\lambda$ is the largest among the eigenvalues of $H$.
Let $\mathbf{x}$ be an eigenvector corresponding to $\lambda$. Then prove that
\[E\mathbf{x}=\mathbf{0}.\]
Add to solve later

Proof.

Using the given relation we have
\begin{align*}
HE\mathbf{x}-EH\mathbf{x}&=2E\mathbf{x}\\
\Rightarrow
HE\mathbf{x}-\lambda E\mathbf{x}&=2E\mathbf{x}\\
\Rightarrow
HE\mathbf{x}&=(\lambda+2)E\mathbf{x}\tag{*}.
\end{align*}

Let $\mathbf{v}=E\mathbf{x}$ and assume that $\mathbf{v}\neq \mathbf{0}$. Then we have from (*)
\[H\mathbf{v}=(\lambda+2)\mathbf{v} \text{ for the nonzero vector } \mathbf{v}.\]
This means that $\lambda+2$ is an eigenvalue of the matrix $H$ and $\mathbf{v}$ is a corresponding eigenvector.

However, the real part of the eigenvalue $\lambda+2$ is greater than that of $\lambda$. This contradicts the choice of $\lambda$.
Therefore the vector $\mathbf{v}=E\mathbf{x}$ must be zero.

Comment.

You might wonder why we consider the relation $HE-EH=2E$.
In fact, this relation is a part of the relations of the Lie algebra $\mathfrak{sl}(2)$.

Although you don’t have to know anything about Lie algebra to solve this problem,
it might be good to know that these computations are actually used in a more advanced mathematic.

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