Let $T$ be a linear transformation in a complex finite dimensional vector space equipped with a positive definite inner product. Suppose that $TT^* = 4T - 3I,$ where $I$ is the identity and $T^*$ is the adjoint of $T$.

Prove that $T$ is positive definite and find all possible eigenvalues of $T.$

I have a feeling that using Tr(T) might come in handy here... but I haven't been able to show either result. Any help is greatly appreciated.


Hint: The identity $TT^*=4T-3I$ implies that $T={1\over 4 }(TT^*+3I)$

You have $\langle T(x),x\rangle=\langle{1\over 4}(TT^*+3I)(x),x\rangle={1\over 4}(\langle TT^*(x)\rangle+\langle 3x,x\rangle)={1\over 4}(\langle T(x),T(x)\rangle+3\langle x,x\rangle)\geq 0 $.

$(TT^*)^*=(4T^*-3I)=TT^*=4T-3I$ this implies that $T=T^*$. You have $T^2=4T-3I$ thus the eigenvalues of $T$ are the roots of $X^2-4X+3$ which are $1$ or $3$.

  • $\begingroup$ you forget a $x$ in the last line $\endgroup$ – Hamza Jun 25 '16 at 19:04

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