Let $F$ and $H$ be an $n\times n$ matrices satisfying the relation
\[HF-FH=-2F.\]

(a) Find the trace of the matrix $F$.

(b) Let $\lambda$ be an eigenvalue of $H$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$. Show that there exists an positive integer $N$ such that $F^N\mathbf{v}=\mathbf{0}$.

Hint.

1. For (a), take the trace of the both sides of the given relation.
2. For (b), show that if $F^k\mathbf{v}\neq \mathbf{0}$ then there are infinitely many eigenvalues, hence a contradiction.

Proof.

(a) The trace of the matrix $F$

Using the given relation we compute the trace of $F$ as follows.
By taking the trace of both sides we have
\[\tr(-2F)=\tr(HF-FH).\]

The right hand side is $-2\tr(F)$ and the left hand side is
\begin{align*}
\tr(HF-FH)&=\tr(HF)-\tr(FH)\\
&=\tr(HF)-\tr(HF)=0.
\end{align*}
Therefore we have $\tr(F)=0$.

(b) There exists an positive integer $N$ such that $F^N\mathbf{v}=\mathbf{0}$.

Since $\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $\lambda$ of $H$, we have $H\mathbf{v}=\lambda \mathbf{v}$ or equivalently $(H-\lambda I)\mathbf{v}=\mathbf{0}$.

Now we compute
\begin{align*}
& F(H-\lambda I)=FH-\lambda F\\
&=(HF+2F)-\lambda F=(H-(\lambda-2)I)F.
\end{align*}

Therefore we have
\[F(H-\lambda I)=(H-(\lambda-2)I)F.\]

Evaluating at $\mathbf{v}$, we obtain
\[\mathbf{0}=F(H-\lambda I)\mathbf{v}=(H-(\lambda-2)I)F\mathbf{v}.\]

If $F\mathbf{v} \neq \mathbf{0}$, then this equality implies that $F\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $\lambda-2$ of $H$. In this case we further calculate
\begin{align*}
\mathbf{0}&=F(H-(\lambda-2)I)F\mathbf{v} \\
&=(FH-(\lambda-2)F)F=(HF+2F-(\lambda-2)F)F\mathbf{v}\\
&=(H-(\lambda-4))F^2\mathbf{v}.
\end{align*}

If the vector $F^2\mathbf{v}\neq \mathbf{0}$, then this equality implies that $F^2\mathbf{v}$ is an eigenvector corresponding to the eigenvalue $\lambda-4$ of $H$.
Repeating this procedure, we see that
\[\mathbf{0}=(H-(\lambda-2k))F^k\mathbf{v}\] for all $k$.

Therefore, if $F^k\mathbf{v}$ is nonzero vector for all $k$, then there are infinitely many eigenvalues $\lambda-2k$ but this is impossible since $H$ is an $n \times n$ matrix and hence $H$ has at most $n$ eigenvalues. Therefore there exists $N$ such that $F^N\mathbf{v}=\mathbf{0}$.

Related Question.

See the problem “Matrices satisfying the relation HE-EH=2E” for similar questions.

As noted there, the relation $HF-FH=-2F$ comes from the Lie algebra $\mathfrak{sl}(2)$.

The post Matrices Satisfying $HF-FH=-2F$ first appeared on Problems in Mathematics.

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
\[E\mathbf{x}=\mathbf{0}.\]

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.

Related Question.

See the problem Matrices satisfying HF−FH=−2F for a similar question.

Another relation comes from the Lie algebra $\mathfrak{sl}(2)$ is studied there.

The post Matrices Satisfying the Relation $HE-EH=2E$ first appeared on Problems in Mathematics.