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Exercise 7.C.9

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Proof. It is true. We shall consider $𝔽 = ℝ$ for the sake of simplicity and it will be easy to see it generalizes to $ℂ$ as well.

Note that if $R$ is a self-adjoint square root of $I$ if and only if the matrix of $R$ is of the form

(\begin{matrix} a & b \\ b & c \end{matrix})

and

\begin{aligned} e_{1} & = R^{2} e_{1} \\ = R (a e_{1} + b e_{2}) \\ = a (a e_{1} + b e_{2}) + b (b e_{1} + c e_{2}) \\ = (a^{2} + b^{2}) e_{1} + b (a + c) e_{2} \end{aligned}

and

\begin{aligned} e_{2} & = R^{2} e_{2} \\ = R (b e_{1} + c e_{2}) \\ = b (a e_{1} + b e_{2}) + c (b e_{1} + c e_{2}) \\ = b (a + c) e_{1} + (b^{2} + c^{2}) e_{2} . \end{aligned}

Therefore, the problems falls down to finding $a, b, c \in 𝔽$ such that

\begin{aligned} a^{2} + b^{2} & = 1 \\ b^{2} + c^{2} & = 1 \\ b (a + c) & = 0 . \end{aligned}

As turns out, there are infinitely many solutions to this (just choose an $a \in [0, 1]$ , take $c = - a$ and solve for $b$ ). Each of this solutions define a different self-adjoint square root, hence the identity operator has infinitely-many self-adjoint square roots. □

celiopassos

2017-10-06 00:00

Exercise 7.C.9

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