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I'm stuck on this problem for a long time. I can't find a proper solution for this without using complex calculations. I hope some one can help me with this problem:

Let $ (\;|\;)$ be the standard inner product on $\mathbb C^{2}$. Prove that there is no non-zero linear operator $T$ on $\mathbb C^{2}$ such that $(\alpha|T\alpha) = 0\;$ for every $\alpha$ in $\mathbb C^{2}$. Generalize.

Kishalay Sarkar
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le duc quang
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    Well, what is your method with using complex calculations? – Chris Eagle Feb 27 '13 at 15:32
  • Oh, I have to consider the matrix of $T$ in standard basis and matrix of standard inner product , suppose 2 matrix are $A$ and $B$. Then we must have $\alpha*.BA\alpha = 0$. But it's too complex, so I stop and try to find another solution, but not succeed... – le duc quang Feb 27 '13 at 15:35
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    See http://math.stackexchange.com/questions/57350/inner-product-and-matrices for a method to solve this, and also http://math.stackexchange.com/questions/298353/norm-of-bounded-operator-on-a-complex-hilbert-space?lq=1. If $\langle Tx,x\rangle = 0$ for all $x$, then $\langle Tx,y\rangle=0$ for all $x$ and $y$, hence $T=0$. – Jonas Meyer Feb 27 '13 at 17:15
  • @JonasMeyer: It's not clear enough for me. Can you clarify more??? – le duc quang Feb 28 '13 at 00:14

2 Answers2

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This is not possible in $\mathbb{C}^2$ and this is slightly tricky.

Note that the links provided by Jonas Meyer show you the most straighforward route: if $(Tx,x)=0$ for all $x$, then $(Tx,y)=0$ for al $x,y$ so $T=0$.

Now here is another approach which is an opportunity to see two important facts about bounded linear operator on a complex Hilbert space.

Claim 1: a bounded operator on a complex Hilbert space is self-adjoint ($T^*=T$) if and only if $(Tx,x)\in\mathbb{R}$ for all $x$.

Remark: this fails in the real case as shows the example down below. Note the use of $iy$ and semilinearity of the inner product in the last step of the proof.

Proof: If $T^*=T$ first then $$ (Tx,x)=(x,T^*x)=(x,Tx)=\overline{(Tx,x)} $$ is real.

Conversely, we will prove that $(x,Ty)=(Tx,y)$ for all $x,y$, which is clearly equivalent to $T$ being self-adjoint since $(Tx,y)=(x,T^*y)$.

Since $(Tx+Ty,x+y)$ is real, we have $$ (Tx+Ty,x+y)=\overline{(Tx+Ty,x+y)}=(x+y,Tx+Ty). $$ Expanding the lhs and the rhs yields $$ (Tx,y)+(Ty,x)=(x,Ty)+(y,Tx). $$ Applying this to $x$ and $iy$ yields $$ (Tx,y)-(Ty,x)=(x,Ty)-(y,Tx). $$ Summing these two equalities, we get $(Tx,y)=(x,Ty)$ as desired.

End of the proof.

Claim 2: If $T$ is a self-adjoint bounded operator on a Hilbert space, then $$ \|T\|=\sup_{\|x\|=1}|(Tx,x)|. $$

Proof: the inequality $\geq$ is trivial. The other direction use the fact that $$ \|T\|=\sup_{\|x\|=1}\|Tx\|=\sup_{\|x\|=\|y\|=1}|(Tx,y)| $$ which follows directly from the fact that $\|z\|=\sup_{\|y\|=1}|(x,y)|$, a simple consequence of Cauchy-Schwarz.

From here, you can use the polarization identity and the parallelogram law to conclude. But I will only give details if you ask since this is already pretty long.

End of proof.

As an application of the above claims, if $(Tx,x)=0$ for all $x$, then $T$ is self-adjoint so $$ \|T\|=\sup_{\|x\|=1}|(Tx,x)|=0, $$ hence $T=0$.

So such a nonzero $T$ does not exist.

But it is possible in $\mathbb{R}^2$: $$\left(\matrix{0& 1\\-1&0} \right).$$

Julien
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  • Thank you so much. Can you generalize this problem for $C^{n}$. I think it's possible, because your proof seems to not depend so much on dimension of space, right??? – le duc quang Feb 28 '13 at 00:17
  • @leducquang Yes, exactly the same argument works for complex Hilbert space. In patrticular for $\mathbb{C}^n$. – Julien Feb 28 '13 at 02:13
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Claim 1: If for two complex linear operators $T,S$ we have $\langle{Tx,x \rangle}=\langle{Sx,x \rangle}\forall x \in \mathbb C^{2}$ then $\langle{Tx,y \rangle}=\langle{Sx,y \rangle} \forall x,y \in \mathbb C^{2}$ (analogous result holds with result to linear operator being in the second argument also)

Proof: This has been answered here

Claim 2: If for a complex linear operator $T$ we have $\langle{x,Ty \rangle}=0 \forall x,y \in \mathbb C^{2}$ then $T \equiv O$ (zero operator).

Proof: Put $x=Ty$, then $\langle{x,Ty \rangle}=\langle{Ty,Ty \rangle}=0$

$\implies Ty=0 \in \mathbb C^{2}$

$\implies T \equiv O$

Main Proof: $\langle{x,Tx \rangle}=0=\langle{x,Ox \rangle},\forall x \in \mathbb C^{2}$ ($O$ is the zero operator)

$\implies \langle{x,Ty \rangle}=\langle{x,Oy \rangle}=\langle{x,O \rangle}=0 \forall x,y \in \mathbb C^{2}$(from claim 1)

$\implies T \equiv O$ (from claim 2)

So no non-zero complex linear operator satisfies the given condition.

user600016
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