# Matrices Satisfying $HF-FH=-2F$

## Problem 69

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 $F^N\mathbf{v}=\mathbf{0}$.

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## Hint.

- For (a), take the trace of the both sides of the given relation.
- 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 $\mathfrak{sl}(2)$.

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a simpler approach for B

$HF − FH= −2F$

If F is singular, then there must be at least one eigenvalue = 0, and we are done. (i.e. we are satisfied with natural number N = 1.)

Now suppose F is non-singular, and right multiply the equation by $F^{-1}$. This gives us

$H − FHF^{-1}= −2I$

now take the trace

$trace(H) − trace(FHF^{-1})= −2trace(I)$

$trace(H) − trace(F^{-1}FH)= −2n$

$trace(H) − trace(H)= −2n$

$0= −2n$

a contradiction. Hence F is singular and we’re done.

Dear D,

Your argument is correct if $\lambda=0$. But $\lambda$ is any eigenvalue of $A$, and even if $A$ is singular, there could be a nonzero eigenvalue of $A$.