2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • 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 $T = T^2$. $\endgroup$ – Sangchul Lee 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. :) $\endgroup$ – Sangchul Lee Nov 14 '15 at 2:30
  • $\begingroup$ @SangchulLee: See my answer for a counterexample to your comment. $\endgroup$ – Hurkyl 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. $\endgroup$ – Sangchul Lee Nov 14 '15 at 4:48
3
$\begingroup$

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 $W\neq V$, $p_W\neq\textrm{id}_V$.

$\endgroup$
0
$\begingroup$

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 $T^t T = \frac{T^t + T}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.