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I have solved a question but I am not sure the last step of the question. If someone can verify it that would be great.

Let $V$ be a finite dimensional vector space with complex inner product. Let $T: V \rightarrow V$ be a linear transformation. If $\forall v\in V$, $\|Tv\|=\|T^*v\|$ then $T$ is a normal operator.

So what I did is,

Since $\forall v \in V$, $\|Tv\|=\|T^*v\|$, we will have $\forall v \in V$, $\langle Tv,Tv\rangle =\langle v,T^*Tv\rangle $ on the other hand $\forall v \in V$, $\langle T^*v,T^*v\rangle =\langle v,TT^*v\rangle $ and therefore $$\langle v,T^*Tv\rangle =\langle v,TT^*v\rangle \Rightarrow \langle v,(T^*T-TT^*)v\rangle =0\Rightarrow T^*T-TT^*=0 \Rightarrow T^*T=TT^*.$$ And therefore the operator is normal.

But then I started thinking, why can I actually say that the last two implications are correct? Can't be $T^*T-TT^*$ , the operator which sends v to it's perpendicular vector? Why wouldn't that be possible? I tried to answer this question but I couldn't.

I would be happy if someone can help me answer this question. Thanks!

(Reminder: Just in case we have different notations $T^*$ is the adjoint operator)

user642796
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Charles Carmichael
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2 Answers2

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You have to show that $$ \langle Sv,v\rangle =0 \quad \forall v\in V $$ implies $S=0$.

To prove this one has to use polarization identities. In the complex case this works for a general mapping $S$. In the real case, this holds only for self-adjoint $S$. However, the mapping $T^*T-TT^*$ is self-adjoint.

daw
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  • Maybe not the best way to see it, but the first thing that jumped out at me: by the spectral theorem, $S = T^T-TT^$ has an orthonormal basis of eigenvectors. But $\langle Sv,v\rangle = 0$ for all of those eigenvectors, which implies all the eigenvalues are $0$. – Steven Gubkin Jul 26 '14 at 17:12
  • I will try to prove that now. But isn't possible to create a linear functional such that $\forall v \in V <Sv,v>=0$ and $S \neq 0$ ? For example we can take the linear functional that makes 90 degrees rotation? I guess that would be on euclidean space, but can't we do it over a unitary space? I mean, now I know that the answer no, since you gave me a claim to prove, but I am trying to understand where does that fail? Thanks for the feedback by the way! – Charles Carmichael Jul 26 '14 at 19:27
  • @DavidC In your question, you are looking at complex vector spaces. So you have to make a distinction between "geometrically orthogonal" and "complex orthogonal". For example, $1$ and $i$ are "geometrically orthogonal", but their complex inner product is $i$ (or $-i$). Do you see how this complicates your "90 degree" intuition? – Steven Gubkin Jul 27 '14 at 19:14
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Suppose that $X$ is a complex inner-product space, and that $B(\cdot,\cdot)$ is a complex function on $X\times X$ such that $B$ is linear in the first coordinate and conjugate-linear in the second coordinate. Then $$ B(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}B(x+i^{n}y,x+i^{n}y). $$ A special case is the inner-product $B(x,y)=(x,y)$, which gives the usual polarization identity. If $B$ and $C$ are two such forms with the same trace $B(x,x)=C(x,x)$ for all $x\in X$, then $B=C$ because $B(x,y)=C(x,y)$ for all $x,y \in X$.

If $A$ and $B$ are linear operators on the complex inner-product space $X$ for which $(Ax,x)=(Bx,x)$ for all $x$, then $A=B$ because $(Ax-Bx,y)=0$ for all $x$, $y$.

So, in your case, $\|Tx\|^{2}=\|T^{\star}x\|^{2}$ for all $x\in X$ implies $$ (T^{\star}Tx,x)=(TT^{\star}x,x), \;\;\; x \in X, $$ So $T^{\star}T=TT^{\star}$ follows for a complex space.

Real spaces are different. There are easy-to-find operators linear operators on $\mathbb{R}^{2}$ such that $(Ax,x)=0$ for all $x$, even though $x \ne 0$. For example, let $R$ be the 90-degree rotation about the origin. Then $Rx \perp x$ for all $x$, which means $(Rx,x)=0$ for all $x$, even though $R\ne 0$. If you know that $(Ax,x)=(Bx,x)$ for all $x$ and if $A$, $B$ are symmetric (i.e., $(Ax,y)=(x,Ay)$ for all $x,y$,) then $A=B$. And that's a useful result on real spaces.

Disintegrating By Parts
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