POTW Prove Hermitian Matrices Satisfy ##H^2 = H^\dagger H##

[Euge]

Show that an $n\times n$-matrix $H$ is hermitian if and only if $H^2 = H^\dagger H$.

 

[julian]

Spoiler

Only if part:

Assume $H$ is Hermitian. Then

\begin{align*}
H^\dagger = H
\end{align*}

implies

\begin{align*}
H^\dagger H = H^2 .
\end{align*}

Meaning if $H^\dagger H \not= H^2$ then $H$ couldn't be Hermitian.

If part:

Assume $H^2 = H^\dagger H$. As $H^2 = H^\dagger H$ is Hermitian its eigenvectors form a basis for the vector space $V = \mathbb{C}^n$. We can define an inner product. If $\{ e_i \}_{i =1,2 \dots , n}$ is an orthonormal basis for $V$, $u = \sum_{i=1}^n a_i e_i$ and $v = \sum_{j=1}^n b_j e_j$, then we define the inner product as:

\begin{align*}
<u,v> := \sum_{i=1}^n a_i^* b_i
\end{align*}

Obviously, $H u = 0$ implies $H^2 u = 0$. We have

\begin{align*}
\| H u \|^2 = < H u , H u > = <u , H^\dagger H u > = <u , H^2 u>
\end{align*}

from which we have $H^2 u = 0$ implies $H u = 0$. Therefore,

\begin{align*}
\ker (H) = \ker (H^2)
\end{align*}

Label the eigenvectors of $H^2$ with non-zero eigenvalues by $e_\alpha$ with $\alpha = 1,2, \dots, N$ ($N \leq n$). They form a basis for $\ker (H)^\perp$. Denote the eigenvalue of $e_\alpha$ by $\lambda_\alpha$.

Note that as $H^2 e_\alpha = \lambda_\alpha e_\alpha$, we have that $<u , H^2 e_\alpha> = 0$ for all $\alpha$ if and only if $u \in \ker H$. Note $H u \in \ker H$ if and only if $Hu =0$. We have

\begin{align*}
<u , H^\dagger H^2 e_\alpha > = <u , H^3 e_\alpha> ,
\end{align*}

from which we have

\begin{align*}
<H u , H^2 e_\alpha > = <H^\dagger u , H^2 e_\alpha > .
\end{align*}

Thus, we have $H u=0$ if and only if $H^\dagger u \in \ker H$. So if $H u =0$ then $H H^\dagger u=0$. If $H u \not=0$ then $H H^\dagger u \not=0$. Thus

\begin{align*}
\ker (H) = \ker (H H^\dagger)
\end{align*}

Note $(H^\dagger)^2 = H^\dagger H = H^2$ and so $H^\dagger u=0$ implies $H^2 u=0$. We have

\begin{align*}
\| H^\dagger u \|^2 = < H^\dagger u , H^\dagger u > = <u , H H^\dagger u >
\end{align*}

from which we have $H^2 u = 0$ implies $H^\dagger u = 0$. Thus

\begin{align*}
\ker (H^\dagger) = \ker (H^2)
\end{align*}

So from this and $\ker (H) = \ker (H^2)$:

\begin{align*}
\ker (H) = \ker (H^\dagger) .
\end{align*}Now, assume $u \not\in \ker (H)$, so that $Hu \not\in \ker H$ and $H^\dagger u \not\in \ker H$. Then from

\begin{align*}
<H u , H^2 e_\alpha > = <H^\dagger u , H^2 e_\alpha >
\end{align*}

we have

\begin{align*}
<H u - H^\dagger u , \lambda_\alpha e_\alpha > = 0 .
\end{align*}

Since the $\{ e_\alpha \}_{\alpha = 1,2, \dots , N}$ form a basis for $\ker (H)^\perp$,

\begin{align*}
H u = H^\dagger u \qquad \text{for arbitrary } u \in \ker (H)^\perp .
\end{align*}

As $H u = H^\dagger u$ for all $u \in V$,

\begin{align*}
H = H^\dagger .
\end{align*}

 

[julian]

Spoiler

Only if part:

Assume $H$ is Hermitian. Then

\begin{align*}
H^\dagger = H
\end{align*}

implies

\begin{align*}
H^\dagger H = H^2 .
\end{align*}

Meaning if $H^\dagger H \not= H^2$ then $H$ couldn't be Hermitian.

If part:

Define the Hermitian matrices

\begin{align*}
H_R = \frac{1}{2} (H + H^\dagger) \qquad \text{and} \qquad H_I = \frac{1}{2i} (H - H^\dagger)
\end{align*}

Then

\begin{align*}
H = H_R + i H_I .
\end{align*}

Note $H^\dagger = H$ if and only if $H_I = 0$.

The assumption that $H^2 = H^\dagger H$ implies

\begin{align*}
H_I H = 0
\end{align*}

which implies

\begin{align*}
i H_I^2 = - H_I H_R .
\end{align*}

From which we obtain

\begin{align*}
[H_R , H_I] = 2 i H_I^2 \qquad (*)
\end{align*}

As $H_I$ is Hermitian its eigenvectors form a basis for $V = \mathbb{C}^n$. Write $H_I e_i = \lambda_i^I e_i$. Then

\begin{align*}
0 = H^\dagger H_I e_i = H_R e_i \lambda_i^I - i (\lambda_i^I)^2 e_i .
\end{align*}

Assume $\lambda_i^I \not= 0$, then the previous equation implies

\begin{align*}
H_R e_i = i \lambda_i^I e_i .
\end{align*}

But then

\begin{align*}
[H_R , H_I] e_i = 0
\end{align*}

in contradiction to $(*)$. Therefore, we must have $\lambda_i^I = 0$.

Which implies $\lambda_i^I = 0$ for $i = 1,2, \dots n$. So that

\begin{align*}
H_I u = 0
\end{align*}

for all $u \in V$, implying:

\begin{align*}
H_I = 0 .
\end{align*}

 

[Infrared]

@julian I didn't fully read your posts, but it looks like you're making a bit of a mountain out of a molehill.

Spoiler

If $H$ is Hermitian, meaning $H=H^\dagger,$ then multiply both sides by $H$ on the right to get $H^2=H^\dagger H.$

On the other hand, if $H^2=H^\dagger H$, then to show that $H=H^\dagger$, you just have to check that $||Hx-H^\dagger x||^2=\langle Hx-H^\dagger x,Hx-H^\dagger x\rangle$ is always zero, which follows quickly from expanding the inner product and using the given relation.

Edit: Spoiler tags

 

[julian]

@julian I didn't fully read your posts, but it looks like you're making a bit of a mountain out of a molehill.

Spoiler

If $H$ is Hermitian, meaning $H=H^\dagger,$ then multiply both sides by $H$ on the right to get $H^2=H^\dagger H.$

On the other hand, if $H^2=H^\dagger H$, then to show that $H=H^\dagger$, you just have to check that $||Hx-H^\dagger x||^2=\langle Hx-H^\dagger x,Hx-H^\dagger x\rangle$ is always zero, which follows quickly from expanding the inner product and using the given relation.

Edit: Spoiler tags

I get:

\begin{align*}
\| Hx -H^\dagger x \|^2 = \langle (H H^\dagger - H^\dagger H) x , x \rangle
\end{align*}

Do we immediately know $H$ is a normal matrix?

 

[julian]

We can simplify my second proof:

Spoiler

Only if part:

Assume $H$ is Hermitian. Then $H^\dagger = H$ implies $H^\dagger H = H^2$. Meaning if $H^\dagger H \not= H^2$ then $H$ couldn't be Hermitian.

If part:

Define the Hermitian matrices

\begin{align*}
H_R = \frac{1}{2} (H + H^\dagger) \qquad \text{and} \qquad H_I = \frac{1}{2i} (H - H^\dagger)
\end{align*}

Then

\begin{align*}
H = H_R + i H_I .
\end{align*}

Note $H^\dagger = H$ if and only if $H_I = 0$.

As $H_I$ is Hermitian its eigenvalues are real and its eigenvectors form a basis for $V = \mathbb{C}^n$. Write $H_I e_i = \lambda_i^I e_i$. The assumption that $H^2 = H^\dagger H$ implies

\begin{align*}
H_I H = 0
\end{align*}

which implies

\begin{align*}
0 = H^\dagger H_I e_i = H_R e_i \lambda_i^I - i (\lambda_i^I)^2 e_i .
\end{align*}

Assume $\lambda_i^I \not= 0$, then the previous equation implies

\begin{align*}
H_R e_i = i \lambda_i^I e_i .
\end{align*}

But then this contradicts that the eigenvalues of an Hermitian matrix are real. Therefore, we must have $\lambda_i^I = 0$.

Which implies $\lambda_i^I = 0$ for $i = 1,2, \dots n$. So that

\begin{align*}
H_I u = 0
\end{align*}

for all $u \in V$, implying:

\begin{align*}
H_I = 0 .
\end{align*}

 

[Infrared]

@julian You're right, I just assumed it worked out and didn't check it :) Still, I think there is a bit of a simpler solution.

Spoiler

It's enough to separately check that $Hx=H^\dagger x$ when $x$ is in the column space of $H$ and that $Hx=H^\dagger x$ when $x$ is orthogonal to the column space of $H.$

The first case is true by the given condition. The second case holds because in that event, $||Hx||^2=\langle Hx,Hx\rangle=\langle x,H^\dagger H x\rangle=\langle x,H^2 x\rangle=0$ and also $H^\dagger x=0$ since the nullspace of $H^\dagger$ is the orthogonal complement of the column space of $H.$

 

[Structure seeker]

I'd like it if a simpler proof is possible. Here's my try on that:

Spoiler: Spoiler

$(H-H^\dagger)H=H^2-H^\dagger H$

So if $H=H^\dagger$ we have $H^2 = H^\dagger H$. If $H$ is invertible, multiply with the inverse for the converse statement. If not, remove from domain and image of the linear map that the matrix represents the eigenvector dimension with eigenvalue $0$ (it's clear the resulting linear map is hermitian iff $H$ is) and use induction on $n$. The case $n=1$ is left to the reader.

 

[Infrared]

@Structure seeker That doesn't look valid (or at least complete) to me. Can you say how to remove some subspace (I can't actually figure out exactly which subspace you mean exactly) from domain and range to reduce to the case of smaller matrices?

 

[Structure seeker]

You can restrict $H$ to the subspace orthogonal to the eigenvector $v$ by projecting on this subspace. Thus you remove linear combinations of $\{ v\}$

 

[Structure seeker]

And since the corresponding eigenvalue is $0$ and $H=H^\dagger$, $H$ simultaneously does nothing with $v$ and lands on something orthogonal to $v$

 

[Infrared]

I still don't fully understand. Here's what I understand your argument as, and correct me if I'm misreading.

It looks to me like you're taking a nonzero vector $v$ in the nullspace of $H$ and considering the subspace $S\subset\mathbb{C}^n$ that is orthogonal to $v$ and trying to restrict $H$ to a map $H:S\to S.$ I don't understand why if $x\in S$ then $H(x)$ is also in $S$ though. If you already know that $H$ was Hermitian, you could say $\langle Hx,v\rangle=\langle x,Hv\rangle=0$ but you can't use the inductive hypothesis yet because you haven't actually yet restricted your map to $n-1$ dimensional space.

Anyway, this seems very similar to the argument I gave in post 7, which I think is correct. Instead of breaking apart $\mathbb{C}^n$ into the nullspace of $H$ and its orthogonal complement, I break it apart into the columnspace of $H$ and its orthogonal complement, but this amounts to the same orthogonal decomposition in the end because $H$ is Hermitian.

 

[Structure seeker]

You can use ${H^\dagger}^2 V=H^\dagger H V=0$ to prove that the kernel $V$ of $H$ is also that of $H^\dagger$ since the rank of $H$ coincides with the rank of $H^\dagger$. That step should be inserted for arguing that $v$ is also a zero-eigenvalue eigenvector of $H^\dagger$.

 

[julian]

If you already know that $H$ was Hermitian, you could say $\langle H x,v \rangle =\langle x, Hv \rangle=0$

You use this in the proof of the spectral theorem for Hermitian matrices.

 

[Structure seeker]

Yeah but I did try to give an independent proof, infrared's point is crystal clear and very true.

 

[julian]

Yes.

 

[Structure seeker]

You can use ${H^\dagger}^2 V=H^\dagger H V=0$ to prove that the kernel $V$ of $H$ is also that of $H^\dagger$ since the rank of $H$ coincides with the rank of $H^\dagger$. That step should be inserted for arguing that $v$ is also a zero-eigenvalue eigenvector of $H^\dagger$.

But this still isn't right

 

[julian]

It would be good there was an easy way of proving $\ker (H) = \ker (H^\dagger)$ as this was the bulk of my solution in post #2.

If I understand you you are saying if $\dim \ker (H) = \dim \ker (H^\dagger)$ then does ${H^\dagger}^2 \ker (H) = H^\dagger H \ker (H)$ imply $\ker (H) = \ker (H^\dagger)$? I think one issue would be you could have $H u \not= 0$ but have $H^\dagger H u = 0$. But it was easy to establish that doesn't happen: We have $\| H u \|^2 = <Hu,Hu> = < u , H^\dagger H u >$ which means $H^\dagger H u = 0$ implies $Hu = 0$. So in fact, $\ker (H) = \ker (H^\dagger H)$.

 

[Structure seeker]

No I was skipping a crucial step, sorry for that. My argument only showed what you also showed in this post

 

[julian]

No, your idea worked I think.

There are no $u$ such that $H u \not=0$ and $H^\dagger H u =0$, so $\ker (H) = \ker(H^\dagger H)$. It follows then from ${H^\dagger}^2 = H^\dagger H$ that $\ker (H) = \ker ({H^\dagger}^2)$.

The second question is are there $u$ such that $H^\dagger u \not= 0$ but ${H^\dagger}^2 u = 0$?

(i) This would be the case if $\dim \ker ({H^\dagger}^2) > \dim \ker (H^\dagger)$,
(ii) this would not be the case if $\dim \ker ({H^\dagger}^2) = \dim \ker (H^\dagger)$, in that case we would have $\ker ({H^\dagger}^2) = \ker (H^\dagger)$.

As you noted $\text{rank} (H) = \text{rank} (H^\dagger)$. This implies $\dim \ker (H) = \dim \ker (H^\dagger)$. This together with $\ker (H) = \ker ({H^\dagger}^2)$ implies $\dim \ker ({H^\dagger}^2) = \dim \ker (H^\dagger)$. So, altogether we have

\begin{align*}
\ker (H) = \ker ({H^\dagger}^2) = \ker (H^\dagger) .
\end{align*}

 

[Structure seeker]

Thanks for working that out!

 

[julian]

If we use $\text{rank} (H) = \text{rank} (H^\dagger)$ we can simplify a bit my first proof, post #2:

Spoiler

Only if part:

Assume $H$ is Hermitian. Then $H^\dagger = H$ implies $H^\dagger H = H^2$. Meaning if $H^\dagger H \not= H^2$ then $H$ couldn't be Hermitian.

If part:

We have $\| H u \|^2 = <Hu,Hu> = < u , H^\dagger H u >$ which means $H^\dagger H u = 0$ implies $Hu = 0$. So $\ker (H) = \ker (H^\dagger H)$. It then follows from ${H^\dagger}^2 = H^\dagger H$ that $\ker (H) = \ker ({H^\dagger}^2)$. If $\dim \ker ({H^\dagger}^2) = \dim \ker (H^\dagger)$ we would have $\ker ({H^\dagger}^2) = \ker (H^\dagger)$. As $\text{rank} (H) = \text{rank} (H^\dagger)$, we have $\dim \ker (H) = \dim \ker (H^\dagger)$. This together with $\ker (H) = \ker ({H^\dagger}^2)$ implies $\dim \ker ({H^\dagger}^2) = \dim \ker (H^\dagger)$. So, altogether we have

\begin{align*}
\ker (H) = \ker ({H^\dagger}^2) = \ker (H^\dagger)
\end{align*}

Also, note $\ker (H^2) = \ker({H^\dagger}^2)$. As $H^2$ is Hermitian its eigenvectors form a basis for the vector space $V = \mathbb{C}^n$. Label the eigenvectors of $H^2$ with non-zero eigenvalues by $e_\alpha$ with $\alpha = i_1,i_2, \dots, i_N$ ($N \leq n$) and corresponding eigenvalues by $\lambda_\alpha$. These eigenvectors form a basis for $\ker (H)^\perp = \ker (H^\dagger)^\perp$.

Assume $u \not\in \ker (H)$, then $Hu \not\in \ker (H)$ and $H^\dagger u \not\in \ker (H^\dagger)$ (as $\ker (H) = \ker (H^2)$ and $\ker (H^\dagger) = \ker ({H^\dagger}^2)$). From $<u , H^\dagger H^2 e_\alpha > = <u , H^3 e_\alpha >$ we have:

\begin{align*}
<H u , H^2 e_\alpha > = <H^\dagger u , H^2 e_\alpha >
\end{align*}

or

\begin{align*}
<H u - H^\dagger u , \lambda_\alpha e_\alpha > = 0 .
\end{align*}

Since the $\{ e_\alpha \}_{\alpha = i_1,i_2, \dots , i_N}$ form a basis for $\ker (H)^\perp$,

\begin{align*}
H u = H^\dagger u \qquad \text{for arbitrary } u \in \ker (H)^\perp .
\end{align*}

As $H u = H^\dagger u$ for all $u \in V$,

\begin{align*}
H = H^\dagger .
\end{align*}

 

[Euge]

Spoiler

Given complex $n\times n$ matrices $A$ and $B$, let $(A,B) = \operatorname{trace}(A^\dagger B)$. This defines an inner product on the vector space of such matrices. Hence, if $H^2 = H^\dagger H$, then $$\|H - H^\dagger\|^2 = \|H\|^2 - 2\operatorname{Re}(H,H^\dagger) + \|H^\dagger\|^2 = 2\|H\|^2 - 2\operatorname{Re}\operatorname{trace}(H^2)$$ The condition $H^2 = H^\dagger H$ forces $\operatorname{trace}(H^2) = \|H\|^2$, so the right-most expression in the above equation equals $2\|H\|^2 - 2\|H\|^2$, i.e., zero.

Conversely, if $H$ is Hermitian, then $H^2 = HH = H^\dagger H$.