Proving that T=T* for a Normal Linear Transformation

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To prove that T = T* for a normal linear transformation T satisfying T^2 = (1/2)(T + T*), it is essential to demonstrate that all eigenvalues of T are real. The discussion suggests starting with the matrix representation of T in an orthonormal basis and analyzing eigenvalues through the inner product <Ax, x>. Establishing that <Ax, x> is a real number would lead to the conclusion that T is Hermitian. The thread also references a theorem stating that if T is a normal transformation with a characteristic polynomial that can be factored into linear factors over the reals, then T is Hermitian, thus confirming T = T*. The user seeks assistance after struggling with the problem for two days.
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Homework Statement


let T:V \to V be a linear transformation which satisfies T^2 = \frac{1}{2} (T + T^*) and is normal. Prove that T=T^*.


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The Attempt at a Solution


I think we should start like this:
Let \mathbf{A}=[T]_B be the matrix representation of T in the orthonormal base B. We look at A as a matrix in the Complex plane. If we prove that for every Eigenvalue, \lambda, it happens that \lambda \in \mathbb{R}, then T = T^* by a theorem and we finished.
Now, be \mathbf{x} an Eigenvector that satisfies \mathbf{Ax} = \lambda \mathbf{x}, then &lt;\mathbf{Ax},\mathbf{x}&gt; = \lambda ||\mathbf{x}||^2 and \lambda = \frac{&lt;\mathbf{Ax},\mathbf{x}&gt;}{||\mathbf{x}||^2}. If only I could show that &lt;\mathbf{Ax},\mathbf{x}&gt; is a real number, I solved it.
 
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Prove that the transformation is hermitian, once you do it's pretty east to prove that eigenvalues of a hermitian transformation are real.
 
It doesn't help me, because I plan to show that the transformation is Hermitian by the following theorem:
"If T is a normal transformation whose Characteristic polynomial can be completely factored into linear factors over \mathbb{R}, then T is Hermitian".
And then it follows, of course, that T = T*
 
Can someone help me please? I work on it two days. Nothing works. Please. I spent hours.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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