I have got the below question as part of an homework assignment:

Let $T$ be a linear operator on the finite-dimensional inner product space $V$, and suppose $T$ is both positive and unitary. Prove $T = I$.

I tried to prove it the following way:

$T$ is normal $\Rightarrow V$ has an orthonormal basis consisting of characteristic vectors of $T$

$T$ is unitary $\Rightarrow$ every characteristic value $|\lambda| = 1$.

  • To get a contradiction, assume that there exists some characteristic value $\lambda = -1$.

$\Rightarrow$ there is a vector $v \in V $ that $T(v) = -v$ and $v$ is not zero

looking on the inner product: $\langle T(v),v\rangle \ =\ \langle -v,v\rangle \ =\ -\langle v,v\rangle $

and $ \ \langle v,v\rangle \ = \ ||v||^2 > 0$ so $\ \langle T(v), v\rangle \ < \ 0$ and this contradicts the fact that $T$ is positive.

$\Rightarrow$ all the characteristic values $\lambda_i = 1$.

Can anyone suggest another way of proving it? I'd love to get some feedback too.

  • 1
    $\begingroup$ I think your proof is correct. $\endgroup$
    – xpaul
    Apr 20, 2013 at 0:02
  • 1
    $\begingroup$ This is true in the complex case too. Your operator is normal, so it is diagonalizable in an orthonormal basis. Then positive and unitary means that the spectrum is both positive and contained in the unit circle. So the spectrum is $\{1\}$. And $I$ is the unique diagonalizable operator with such a spectrum. $\endgroup$
    – Julien
    Apr 20, 2013 at 20:18
  • $\begingroup$ A single suggestion is to change the beginning saying instead of "$T$ is normal", $T$ is self-adjoint, that comes from positivity. $\endgroup$
    – Ilovemath
    Feb 22, 2019 at 11:15


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