I need to show that if $T$ is an operator in an inner product space over the complex field and if $T^2=I$, then $T$ has to be normal.
$T^2=I$ implies that $T$ is a normal operator
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linear-algebra
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4This is false, e.g. $T(x,y)=(x+y,-y)$ on $\mathbb C^2$ with the standard inner product. – 2012-06-29
1 Answers
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This is false. Let $T = \left[ \begin{array}{cc} 1 & -2 \\\ 0 & -1 \end{array} \right].$
$T$ has eigenvalues $1, -1$ with eigenvectors $(1, 0), (1, 1)$ respectively, so satisfies $T^2 = I$. But the eigenspaces of $T$ are not orthogonal, so $T$ cannot be normal.