I Unitary Operator: Can X Make L(A) Unitary?

[nojojo121]

TL;DR Summary

Unitary operator in complex matrix space equipped with standard inner product

Unitary operator

If an operator L(A) = [X, A], are there matrix X for which the operator is unitary? Keep in mind it is in a complex matrix space with standard inner product (A, B) = tr(A*B)

 

[fresh_42]

TL;DR Summary: Unitary operator in complex matrix space equipped with standard inner product

Unitary operator

If an operator L(A) = [X, A],...

How is $[\cdot,\cdot]$ defined?
And are we talking about a linear operator?

The use of unitary suggests that we speak about groups, in which case it would be $[X,A]=X^{-1}\cdot A^{-1}\cdot X\cdot A$. However, in that case, $L$ wouldn't be a linear operator.

It is also possible, that you meant $L(A)=X^{-1}AX,$ the adjoint operator $\operatorname{Ad}(X)$ of a matrix group. This would at least be a linear operator.

However, experience plus the term operator suggests that you use it as $L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X$ in which case the corresponding property would be skew-hermitian.

... are there matrix X for which the operator is unitary?

So you want to know whether there is an operator $L\, : \,A\longmapsto [X,A]$ such that
$$
\langle L(A),L(B) \rangle =\langle [X,A],[X,B] \rangle =\langle A,B \rangle
$$
I am not sure whether there is such a general answer to it, that it is completely irrelevant where $A,B$ are taken from and how $[\cdot,\cdot]$ is defined. My suspicion is a no. So please define the commutator and the domain the matrices are taken from.

In case $L(A)=[X,A]=X^{-1}A^{-1}XA$ the answer is no. $L$ isn't linear in that case and unitary requires linearity.
In case $L(A)=[X,A]=\operatorname{Ad}(X)(A)=X^{-1}AX$, the answer is yes, e.g. $L=X=1.$
In case $L(A)=\mathfrak{ad}X$ then $L=X=0$ is unitäry on all complex vector spaces.

Keep in mind it is in a complex matrix space with standard inner product (A, B) = tr(A*B)

This is neither standard nor generally an inner product at all. It suggests that you mean the Killing-form $(A,B)\longmapsto \operatorname{tr}(\mathfrak{ad}(A),\mathfrak{ad}(B)).$ This is an inner product in case $A,B$ are taken from a semisimple, in your case complex, Lie algebra.

Here is an overview that should help you navigate through the various set-ups:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/ usage of the terms.

 

[nojojo121]

How is $[\cdot,\cdot]$ defined?
And are we talking about a linear operator?

The use of unitary suggests that we speak about groups, in which case it would be $[X,A]=X^{-1}\cdot A^{-1}\cdot X\cdot A$. However, in that case, $L$ wouldn't be a linear operator.

It is also possible, that you meant $L(A)=X^{-1}AX,$ the adjoint operator $\operatorname{Ad}(X)$ of a matrix group. This would at least be a linear operator.

However, experience plus the term operator suggests that you use it as $L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X$ in which case the corresponding property would be skew-hermitian.

So you want to know whether there is an operator $L\, : \,A\longmapsto [X,A]$ such that
$$
\langle L(A),L(B) \rangle =\langle [X,A],[X,B] \rangle =\langle A,B \rangle
$$
I am not sure whether there is such a general answer to it, that it is completely irrelevant where $A,B$ are taken from and how $[\cdot,\cdot]$ is defined. My suspicion is a no. So please define the commutator and the domain the matrices are taken from.

In case $L(A)=[X,A]=X^{-1}A^{-1}XA$ the answer is no. $L$ isn't linear in that case and unitary requires linearity.
In case $L(A)=[X,A]=\operatorname{Ad}(X)(A)=X^{-1}AX$, the answer is yes, e.g. $L=X=1.$
In case $L(A)=\mathfrak{ad}X$ then $L=X=0$ is unitäry on all complex vector spaces.
This is neither standard nor generally an inner product at all. It suggests that you mean the Killing-form $(A,B)\longmapsto \operatorname{tr}(\mathfrak{ad}(A),\mathfrak{ad}(B)).$ This is an inner product in case $A,B$ are taken from a semisimple, in your case complex, Lie algebra.

Here is an overview that should help you navigate through the various set-ups:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/ usage of the terms.

Thanks for the reply. I should have been more specific. My definition for the operator is like you guessed, that is $L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X$ and $L\, : \,M{_{n\times n} } \longmapsto M{_{n\times n}}$. The inner product is $(A,B)= \operatorname{tr}(A{^\dagger} B).$ which apparently is standard according to Wikipedia. Nevertheless I am searching for X for which the defined operator is unitary.

 

[fresh_42]

Thanks for the reply. I should have been more specific. My definition for the operator is like you guessed, that is $L(A)=\mathfrak{ad}(X)(A)=[X,A]=X\cdot A-A\cdot X$ and $L\, : \,M{_{n\times n} } \longmapsto M{_{n\times n}}$. The inner product is $(A,B)= \operatorname{tr}(A{^\dagger} B).$ which apparently is standard according to Wikipedia.

Reference?

Nevertheless I am searching for X for which the defined operator is unitary.

$X=0$ does the job, and $(A,B)\longmapsto \operatorname{tr}(A^\dagger B)$ is definitely no inner product.
$$
\operatorname{tr}\left(\underbrace{\begin{bmatrix}1&0\\0&1\end{bmatrix}^\dagger}_{\neq 0} \cdot \underbrace{\begin{bmatrix}0&1\\1&0\end{bmatrix}}_{\neq 0}\right)=\operatorname{tr}\left(\begin{bmatrix}0&1\\1&0\end{bmatrix}\right)=0
$$

 

[Office_Shredder]

@fresh_42 I'm confused, aren't there always lots of pairs of vectors whose inner product is zero? Those are just orthogonal vectors.

I think this is just the standard entry-wise dot product written to look fancy

 

[fresh_42]

@fresh_42 I'm confused, aren't there always lots of pairs of vectors whose inner product is zero? Those are just orthogonal vectors.

I think this is just the standard entry-wise dot product written to look fancy

Yes, my bad. I was lost in translation. We use the term inner product to describe the dot product, and scalar product to describe the inner product.

Anyway, $L=X=0$ is still an answer.

Edit: ... and I automatically think about Lie algebras if someone uses $[\cdot,\cdot]$ and unitary.

 

[Infrared]

I'm confused- unitary operators are supposed to be invertible, but $L(X)=[X,X]=0$ represents a nontrivial kernel (unless $X=0$ but that case is just as easily dismissed). Am I missing something?

 

[fresh_42]

I'm confused- unitary operators are supposed to be invertible, but $L(X)=[X,X]=0$ represents a nontrivial kernel (unless $X=0$ but that case is just as easily dismissed). Am I missing something?

My very first statement was: define $[X,Y]$ since unitary belongs to groups and skew-hermitian is the corresponding non-regular property.

As I said, I associated Lie algebras since this is the most common use of $[\cdot,\cdot]$ on PF. Rarely anybody ever uses the group commutator. And the corresponding Lie group operation is the adjoint representation $\operatorname{Ad}$ on its tangent space. In that case, the group elements are commonly written by minuscules.

As long as unitary and presumably $[X,Y]=XY-YX$ as an operator occurs in one sentence, as long will I allow myself to consider the trivial operation.

However, you are all right. So I herewith answer: Yes. Set $L=1$ and $V=\mathbb{M}(1,\mathbb{C}).$

I only wanted to emphasize that the question was badly posed.

 

[nojojo121]

Reference?

$X=0$ does the job, and $(A,B)\longmapsto \operatorname{tr}(A^\dagger B)$ is definitely no inner product.
$$
\operatorname{tr}\left(\underbrace{\begin{bmatrix}1&0\\0&1\end{bmatrix}^\dagger}_{\neq 0} \cdot \underbrace{\begin{bmatrix}0&1\\1&0\end{bmatrix}}_{\neq 0}\right)=\operatorname{tr}\left(\begin{bmatrix}0&1\\1&0\end{bmatrix}\right)=0
$$

I poorly worded what is being searched. I need to find all the matrices $X$ that make the operator $L$ unitary. As someone else mentioned above I think there is a loss in translation but what I call inner product you call dot product. So essentially I have to show for which $X$ the operator $L$ is an isometry.

 

[fresh_42]

I poorly worded what is being searched. I need to find all the matrices $X$ that make the operator $L$ unitary. As someone else mentioned above I think there is a loss in translation but what I call inner product you call dot product. So essentially I have to show for which $X$ the operator $L$ is an isometry.

The important information is where $X$ and $A,B$ are taken from. You said that all complex matrices for $A$ and $B$ are allowed. And you said that the inner product in the definition of the unitary operator $L_X(A)=XA-AX$ is $\langle A,B\rangle = \operatorname{tr}(A^\dagger B).$ This runs down to (if I didn't make a mistake)
$$
\operatorname{tr}(A^\dagger X^\dagger XB-A^\dagger X^\dagger BX-X^\dagger A^\dagger XB+X^\dagger A^\dagger BX-A^\dagger B)=0\quad \forall \;A,B\in \mathbb{M}(n,\mathbb{C})
$$
This is a linear equation system in the coordinates $X=(x_{ij}).$ The solution is therefore a complex vector space. Since you are only interested in a bijective operator, we finally have to calculate all $X$ such that $A\longmapsto XA-AX$ is regular for all solutions $X.$

You can set $A=E_{ij},B=E_{kl}$ where $E_{rs}$ is the matrix with a $1$ at position $(r,s)$ and $0$ elsewhere in order to solve the system. Then see if $i \cdot E_{rs}$ gives even more equations.

I would start with $n=2,3$ and see how it goes. And I still think that there is something wrong here since you basically ask for regular elements in a Lie algebra. This is normally not of any interest.

 

[nojojo121]

The important information is where $X$ and $A,B$ are taken from. You said that all complex matrices for $A$ and $B$ are allowed. And you said that the inner product in the definition of the unitary operator $L_X(A)=XA-AX$ is $\langle A,B\rangle = \operatorname{tr}(A^\dagger B).$ This runs down to (if I didn't make a mistake)
$$
\operatorname{tr}(A^\dagger X^\dagger XB-A^\dagger X^\dagger BX-X^\dagger A^\dagger XB+X^\dagger A^\dagger BX-A^\dagger B)=0\quad \forall \;A,B\in \mathbb{M}(n,\mathbb{C})
$$
This is a linear equation system in the coordinates $X=(x_{ij}).$ The solution is therefore a complex vector space. Since you are only interested in a bijective operator, we finally have to calculate all $X$ such that $A\longmapsto XA-AX$ is regular for all solutions $X.$

You can set $A=E_{ij},B=E_{kl}$ where $E_{rs}$ is the matrix with a $1$ at position $(r,s)$ and $0$ elsewhere in order to solve the system. Then see if $i \cdot E_{rs}$ gives even more equations.

I would start with $n=2,3$ and see how it goes. And I still think that there is something wrong here since you basically ask for regular elements in a Lie algebra. This is normally not of any interest.

I think you are misunderstanding my usage of $B$, which is just an arbitrary matrix used to define the inner product. This is essentially the linear equation I am trying to solve:
$$
\operatorname{tr}(A^\dagger X^\dagger XA-A^\dagger X^\dagger AX-X^\dagger A^\dagger XA+X^\dagger A^\dagger AX)=\operatorname{tr}(A^\dagger A)
$$

 

[fresh_42]

I think you are misunderstanding my usage of $B$, which is just an arbitrary matrix used to define the inner product. This is essentially the linear equation I am trying to solve:
$$
\operatorname{tr}(A^\dagger X^\dagger XA-A^\dagger X^\dagger AX-X^\dagger A^\dagger XA+X^\dagger A^\dagger AX)=\operatorname{tr}(A^\dagger A)
$$

\begin{align*}
\langle L(A),L(B) \rangle &=\langle A,B \rangle\\
\operatorname{tr}(L(A)^\dagger \cdot L(B))&=\operatorname{tr}(A^\dagger\cdot B)\\
\operatorname{tr}([X,A]^\dagger \cdot [X,B])&=\operatorname{tr}(A^\dagger \cdot B)\\
\operatorname{tr}((XA-AX)^\dagger \cdot (XB-BX))&=\operatorname{tr}(A^\dagger \cdot B)\\
\ldots &\phantom{=} \ldots
\end{align*}
https://en.wikipedia.org/wiki/Unitary_operator

Where am I wrong?

 

[nojojo121]

\begin{align*}
\langle L(A),L(B) \rangle &=\langle A,B \rangle\\
\operatorname{tr}(L(A)^\dagger \cdot L(B))&=\operatorname{tr}(A^\dagger\cdot B)\\
\operatorname{tr}([X,A]^\dagger \cdot [X,B])&=\operatorname{tr}(A^\dagger \cdot B)\\
\operatorname{tr}((XA-AX)^\dagger \cdot (XB-BX))&=\operatorname{tr}(A^\dagger \cdot B)\\
\ldots &\phantom{=} \ldots
\end{align*}
https://en.wikipedia.org/wiki/Unitary_operator

Where am I wrong?

My mistake. You are completely right, unitary operators must preserve the inner product and not just the norm.

 

[fresh_42]

I'm not sure about the infinite-dimensional case. In finite dimensions, we have the following situation.
If $X,A,B \in \mathbb{M}(n,\mathbb{C}),$ i.e. complex linear transformations of $\mathbb{C}^n$ then we have $L(A)=(\mathfrak{ad}(X))(A)$ and $L\in \mathfrak{ad}(\mathfrak{gl}(\mathbb{C}^n))=\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n)\oplus \mathbb{C}\cdot \operatorname{Id}_{\mathbb{C}^n})=\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n))\cong \mathfrak{sl}(\mathbb{C}^n).$ Unitarity demands that $L\in \operatorname{SU}(\mathfrak{gl}(\mathbb{C}^n)).$

You are looking for matrices $X$ that have determinant $1$ and trace $0$ at the same time. This is unusual. May I ask about the context of your research?

 

[martinbn]

The important information is where $X$ and $A,B$ are taken from. You said that all complex matrices for $A$ and $B$ are allowed. And you said that the inner product in the definition of the unitary operator $L_X(A)=XA-AX$ is $\langle A,B\rangle = \operatorname{tr}(A^\dagger B).$ This runs down to (if I didn't make a mistake)
$$
\operatorname{tr}(A^\dagger X^\dagger XB-A^\dagger X^\dagger BX-X^\dagger A^\dagger XB+X^\dagger A^\dagger BX-A^\dagger B)=0\quad \forall \;A,B\in \mathbb{M}(n,\mathbb{C})
$$
This is a linear equation system in the coordinates $X=(x_{ij}).$ The solution is therefore a complex vector space. Since you are only interested in a bijective operator, we finally have to calculate all $X$ such that $A\longmapsto XA-AX$ is regular for all solutions $X.$

You can set $A=E_{ij},B=E_{kl}$ where $E_{rs}$ is the matrix with a $1$ at position $(r,s)$ and $0$ elsewhere in order to solve the system. Then see if $i \cdot E_{rs}$ gives even more equations.

I would start with $n=2,3$ and see how it goes. And I still think that there is something wrong here since you basically ask for regular elements in a Lie algebra. This is normally not of any interest.

@Infrared already said that $L_X$ always has a kernel, hence cannot be bijective.

 

[fresh_42]

@Infrared already said that $L_X$ always has a kernel, hence cannot be bijective.

The kernel of $L$ consists of the diagonal matrices with only one entry, complex multiples of $1.$
$$
\operatorname{ker}(L)=\{A\in \mathfrak{gl}(\mathbb{C}^n)\,|\,L(A)=0\}=\{A\,|\,XA=AX\;\forall \,X\in\mathfrak{gl}(\mathbb{C}^n)\} = \mathbb{C}\cdot \operatorname{Id}_{\mathbb{C}^n}
$$
Every choice of $X=S+\lambda\cdot \operatorname{Id}_{\mathbb{C}^n}$ with $S\in\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n))$ leads to $L(A)=\mathfrak{ad}(X)(A)=[X,A]=[S,A]=SA-AS.$ Thus $L(A)$ has a non-trivial kernel, the operator $L=\mathfrak{ad}S$ itself can be bijective.

 

[martinbn]

The kernel of $L$ consists of the diagonal matrices with only one entry, complex multiples of $1.$
$$
\operatorname{ker}(L)=\{A\in \mathfrak{gl}(\mathbb{C}^n)\,|\,L(A)=0\}=\{A\,|\,XA=AX\;\forall \,X\in\mathfrak{gl}(\mathbb{C}^n)\} = \mathbb{C}\cdot \operatorname{Id}_{\mathbb{C}^n}
$$
Every choice of $X=S+\lambda\cdot \operatorname{Id}_{\mathbb{C}^n}$ with $S\in\mathfrak{ad}(\mathfrak{sl}(\mathbb{C}^n))$ leads to $L(A)=\mathfrak{ad}(X)(A)=[X,A]=[S,A]=SA-AS.$ Thus $L(A)$ has a non-trivial kernel, the operator $L=\mathfrak{ad}S$ itself can be bijective.

No, because $L_X(X)=0$. If $S=X-\lambda$, then still $[S,X]=0$.

 

[fresh_42]

No, because $L_X(X)=0$. If $S=X-\lambda$, then still $[S,X]=0$.

Yes, I see it now. I'm getting too old for so little sleep. I was confused by the fact that $\mathfrak{sl}(V)$ contains regular matrices.

It had to be helpless. $\det = 1$ and $\operatorname{trace}=0$ doesn't fit very well.