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Let $T$ be a normal transformation on a finite-dimensional Hilbert space; that is, $TT^*=T^*T$, where $T^*$ is the adjoint of $T$.

Prove that if $T^3=\frac{1}{2}(T+T^*)$, then $T$ is self adjoint.

I have tried to do some math on $(Tv,u)$ but I was not successful in proving the following: $(Tu,v)=(u,Tv)$ which is what I need for self-adjoint transformation.

Edit: $(T^3,v)=\frac{1}{2}\left(\left(T+T^*\right),u \right)= \left( Tu,v\right)+ \left( T^*u,v\right)=\left( u,T^*v\right)+\left( u,Tv\right)=\left( u,T^3v\right)$ Therefore, $T^3$ is self adjoint. Does it mean that $T$ is self adjoint?

Thanks!

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    $\begingroup$ Are you dealing with finite-dimensional spaces, or with infinite-dimensional Hilbert spaces? $\endgroup$ – Daniel Fischer Feb 26 '16 at 14:55
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    $\begingroup$ That $T^3$ is self-adjoint doesn't automatically imply that $T$ is self-adjoint. But here, the equation $T^3 = \frac{1}{2}\bigl(T + T^{\ast}\bigr)$ gives more information. Suppose $\lambda$ is an eigenvalue of $T$. What does the equation tell you about $\lambda$? $\endgroup$ – Daniel Fischer Feb 26 '16 at 15:00
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    $\begingroup$ @DanielFischer Are you implying I should go with the direction of proving all the eigenvalues are real, therefore $T$ will be adjoint? $\endgroup$ – Alan Feb 26 '16 at 15:02
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    $\begingroup$ If $Tv = \lambda v$, then $T^{\ast} v = \overline{\lambda} v$, if $T$ is normal. Do you denote the complex conjugate with $\lambda^{\ast}$ rather than $\overline{\lambda}$? $\endgroup$ – Daniel Fischer Feb 26 '16 at 15:19
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    $\begingroup$ To get $\overline{\lambda}$, you can write "\overline{\lambda}". $\endgroup$ – Dustan Levenstein Feb 26 '16 at 15:22
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It is not true that if $T$ is normal and $T^3$ is selfadjoint, then $T$ is selfadjoint. Example: $$ T=\begin{bmatrix}e^{i\pi/3}&0\\0&e^{i\pi/3}\end{bmatrix}. $$ Then $T^3=I$ (or, even when $n=1$ and $T=e^{i\pi/3}$).

But here we know more. Since $T$ is normal, it is unitarily diagonalizable. So we may assume that $T$ is diagonal. We can then see that the equality $T^3=\text{Re}\,T$ will be satisfied by the eigenvalues of $T$. If $$\tag1 \lambda^3=\text{Re}\,\lambda $$ and we write $\lambda=re^{it}$, we immediately get $$\tag2 r^3e^{3it}=r\cos t. $$ There is the obvious solution $r=0$, i.e. $\lambda=0$. For $r\ne0$, the equality $(2)$ becomes $$\tag3 r^2\cos 3t=\cos t$$ and $$\tag4 r^2\sin 3t=0. $$ The equality $(4)$ gives us $t=k\pi/3$. And now we have $$\tag5 r^2=\frac{\cos k\pi/3}{\cos k\pi}=(-1)^k\cos k\pi/3. $$For $k=0,1,2,3,4,5$, the right-hand-side in $(5)$ is $$ 0,-\frac12,-\frac12,1,-\frac12,-\frac12. $$ As $r^2>0$, the only acceptable values of $k$ are $0$ and $3$, which give $\lambda=1$ and $\lambda=-1$ respectively.

In summary, the only possible eigenvalues of $T$ are $0,1,-1$, and thus $T$ (being normal) is selfadjoint.

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We have $2A^3 =A+A^*$ hence $A+A^* =(2A^3)^*=2(AAA)^* =2(A^* )^3$ thus the operator $A^3 $ is self-adjont. And therefore the operator $A=f(A^3 ) $ where $f(x) =\sqrt[3]{x} $ is self - adjont.

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  • $\begingroup$ I don't think you can take a cube root here; for example, if $A$ is a nonzero nilpotent matrix with $A^3=0$, then $A^3$ is indeed self adjoint, but $A$ is not. $\endgroup$ – Dustan Levenstein Feb 26 '16 at 15:26
  • $\begingroup$ I wasn't the downvoter, btw. $\endgroup$ – Dustan Levenstein Feb 26 '16 at 15:27
  • $\begingroup$ And neither was I. Thank you for you effort. $\endgroup$ – Alan Feb 26 '16 at 15:27
  • $\begingroup$ @DustanLevenstein you mean that the cubic root is not unique. You can always take the self-adjoint cubic root of a self-adjoint operator $A$, by applying the cubic root's Taylor series to $A$. $\endgroup$ – Oskar Limka Feb 26 '16 at 21:29
  • $\begingroup$ @OskarLimka But if $A = B^3$ is self-adjoint, that doesn't imply that $B$ is the self-adjoint cube root of $A$. $\endgroup$ – Daniel Fischer Feb 26 '16 at 22:33

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