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Suppose 𝑇 ∈ ℒ (𝑉). Prove that 9 is an eigenvalue of $𝑇^2$ if and only if 3 or
−3 is an eigenvalue of 𝑇.
SOLUTION
First suppose 3 is an eigenvalue of 𝑇.
Then there exists 𝑣 ∈ 𝑉 with 𝑣 ≠ 0 such that 𝑇𝑣 = 3𝑣 .
Applying 𝑇 to both sides, we get $𝑇^2$𝑣 = 3𝑇𝑣 = 9𝑣.
Thus 9 is an eigenvalue of $𝑇^2$
Similarly, if −3 is an eigenvalue of 𝑇 , then 9 is an eigenvalue of $𝑇^2$
.
Now suppose 9 is an eigenvalue of $𝑇^2$.
Thus $𝑇^2$− 9𝐼 is not injective.
Because $𝑇^2$ − 9𝐼 = (𝑇 + 3𝐼)(𝑇 − 3𝐼) , this implies that 𝑇 + 3𝐼 is not injective or 𝑇 − 3𝑖 is
not injective.
Thus −3 is an eigenvalue of 𝑇 or 3 is an eigenvalue of 𝑇.
Edward Frenkel
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