Here, $\langle u,v\rangle$ denotes the inner product of $u$ and $v$.

Suppose $T$ is an operator in vectorspace $V$. Is it true that if $\langle Tv,v\rangle=\langle Tv,Tv\rangle$ for all $v\in V$ then $Tv=v$ for all $v$ in $V$?

It seems obviously true but it seems like something that is not immediately trivial to prove... or maybe I'm missing something really simple.


Your statement is not true. Counterexample:

\begin{align} T & = \begin{bmatrix}1&0\\0&0 \end{bmatrix},& v & = \begin{bmatrix} v_{1} \\ v_{2} \end{bmatrix} & \implies & & T\,v = \begin{bmatrix} v_{1} \\ 0 \end{bmatrix} \end{align} \begin{align} \left\langle T\,v ,\,v \right\rangle & = \begin{bmatrix} v_{1} \\ 0 \end{bmatrix} \begin{bmatrix} v_{1} & v_{2} \end{bmatrix} = v_{1}^2 & \left\langle T\,v ,\,T\,v \right\rangle & = \begin{bmatrix} v_{1} \\ 0 \end{bmatrix} \begin{bmatrix} v_{1} & 0 \end{bmatrix} = v_{1}^2 \end{align}

All you can say is that $\,T\,$ is orthogonal projection operator.

  • 3
    $\begingroup$ To complement this answer, recall the fact that an operator $T$ is orthogonal projection if and only if $T = T^*$ and $T^2 = T$. Then it is clear that this implies the given condition. Conversely, from polarization argument we can check that the given condition implies $T = T^* T$, which then shows that $T = T^*$ and $T = T^2$. $\endgroup$ Nov 14 '15 at 2:19
  • $\begingroup$ @SangchulLee Thank you for elaborating on this! $\endgroup$
    – Vlad
    Nov 14 '15 at 2:21
  • $\begingroup$ You're welcome! I am pretty sure you already know this, but I just wanted to assure myself of the fact by leaving a comment. :) $\endgroup$ Nov 14 '15 at 2:30
  • $\begingroup$ @SangchulLee: See my answer for a counterexample to your comment. $\endgroup$
    – user14972
    Nov 14 '15 at 4:38
  • $\begingroup$ @Hurkly, Initially I only considered complex inner product spaces, and then thought that the proof also works for real inner product spaces. Now I found that it doesn't. So you're right, in real inner product spaces we have a counterexample. $\endgroup$ Nov 14 '15 at 4:48

Counterexample 1. Let $T$ defined by $Tv=0$ for all $v\in V$, $T$ is an operator of $V$ such that: $$\forall v\in V,\langle Tv,v\rangle=0=\langle Tv,Tv\rangle.$$ Nonetheless, if $V\neq\{0\}$, there exists $v\in V$ such that: $$Tv\neq v.$$

Counterexample 2. Let $W$ be a proper close subvector space of $V$, then one has : $$V=W\overset{\perp}{\oplus} W^{\perp}.$$ Let $p_{W}$ be the orthographic projection of $V$ on $W$ and of kernel $W^{\perp}$, notice that: $$\forall v\in V,p_W(v)\in W\textrm{ and }v-p_W(v)\in W^{\perp}.$$ Hence : $$\forall v\in V,\langle p_W(v),v-p_W(v)\rangle=0.$$ Therefore : $$\forall v\in V,\langle p_W(v),v\rangle=\langle p_W(v),p_W(v)\rangle.$$ Since $W\neq\{0\}$, $p_W\neq 0$ and since $W\neq V$, $p_W\neq\textrm{id}_V$.


Here's another failure mode; let the inner product be the dot product, and define

$$T = \left( \begin{matrix}\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{matrix} \right) $$

Writing $v = \left( \begin{matrix} x \\ y \end{matrix} \right)$, we have $ Tv = \left( \begin{matrix} \frac{x+y}{2} \\ \frac{y-x}{2} \end{matrix} \right) $, and consequently,

$$ \langle Tv, v \rangle = \frac{1}{2} (x^2 + y^2) = \langle Tv, Tv \rangle $$

The thing that makes this counterexample work is that

$$ T^t T = \left( \begin{matrix}\frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{matrix} \right) $$

and the fact that the expression $v^t A v$ depends only on the symmetric part of $A$; in the case of $2 \times 2$ matrices, it depends only on the top-left entry, the bottom-right entry, and the sum of the antidiagonal entries.

In general, the counterexamples are precisely the non-identity matrices $T$ satisfying $T^t T = \frac{T^t + T}{2}$.


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