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Let a self-adjoint operator $T:V\to V$ above $\mathbb{C}$, such that $\langle Tv,v \rangle \ge 0$ (so it's essentially a real number). We have learned before that for this kind of $T$, all it's eigenvalues are non-negative ($\ge 0$).

Show that $\det(\text{Id}+T)\ge 1+\det(T)$

I'd be glad for a guidance or an hint.

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Let $\lambda_i$ the eigenvalue of $T$ so we have $\lambda_i\ge0$ and $T$ is diagonalizable i.e. there's an invertible matrix $P$ such that $T=P\operatorname{diag}(\lambda_1,\ldots,\lambda_n)P^{-1}$ hence we have

$$\det(I+T)=\prod_{i=1}^n(1+\lambda_i)\ge1+\prod_{i=1}^n\lambda_i=1+\det(T)$$

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  • $\begingroup$ It is true because $T$ is also normal, and a normal matrix is diagonalizable. Right? $\endgroup$ – AlonAlon Jan 11 '15 at 13:16
  • $\begingroup$ Yes true and note that the proof remains correct even $T$ is only trigonalizable. $\endgroup$ – user63181 Jan 11 '15 at 13:19
  • $\begingroup$ Do you rely on the fact that $T$ has the same determinant as $\text{Diag}(\lambda_1,\ldots, \lambda_n)$? (Since they are similar matrices). $\endgroup$ – AlonAlon Jan 11 '15 at 13:41
  • $\begingroup$ Yes the determinant of $T$ is the product of its all eigenvalues. $\endgroup$ – user63181 Jan 11 '15 at 14:24

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