Let $V$ be a (complex) finite vector space equiped with an inner product and $T$ an operator on V. We say $T$ is positive if: $$\langle T(v), v \rangle \geq 0$$ for all $v$ in $V$.

We say $T$ is stricly positive if it is positive and the equality only holds when $v=0$.


I am trying to prove that, for any operator $T$, it is the case that $TT^*$ is positive and it is strictly positive if $T$ is invertible. I know $TT^*$ is self adjoint, so I can express it using its spectral decomposition as follows: $$TT^*=\sum_{i=1}^{n} \lambda_i P_i$$ So I have: $$\langle TT^*(v),v \rangle = \langle \sum_{i=1}^{n}\lambda_i P_i(v),\sum_{j=1}^{n}P_j(v) \rangle = \sum_{i,j=1}^{n}\lambda_i \langle P_i(v),P_j(v)\rangle = \sum_{i=1}^{n}\lambda_i \|P_i(v)\|$$ From here I don't know how to proceed, I know every $\lambda_i$ is real because $TT^*$ is self adjoint but I see no reason to assume it's non-negative. Perhaps this is not the best approach, what else can I use?

I'm also asking for some enlightment to prove the other statement (when $T$ is invertible), I can't figure out how to use the invertibility fact.

Thanks in advance

  • 1
    $\begingroup$ Hint: $\langle Su, w \rangle = \langle u, S^{*}w \rangle$. Try applying this to $u = T^{*}v$, $S = T$, $w = v$. $\endgroup$ Commented Jul 21, 2016 at 22:58

1 Answer 1


Given a linear map $T:\mathbb{C}^n\to \mathbb{C}^n$, for all $v\in\mathbb{C}^n$ we have $$\langle TT^*(v),v\rangle= \langle T^*(v),T^*(v)\rangle\geq 0,$$ because this is just the inner product of a vector with itself. Thus, $TT^*$ is positive.

Furthermore, if $T$ is invertible, $T^*$ is invertible, so equality holds in the above if and only if $v=0$. Therefore, if $T$ is invertible, $TT^*$ is strictly positive.

  • $\begingroup$ OMG I was overcomplicating it so stupidly, thanks! $\endgroup$
    – la flaca
    Commented Jul 21, 2016 at 23:08
  • $\begingroup$ Trust me, it's very easy to overlook a simple answer. Glad to help! $\endgroup$
    – Aweygan
    Commented Jul 21, 2016 at 23:10

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