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Consider $\mathbb{R}^{3}$ with the standard inner product. Let $a,b \in \mathbb{R}^{3}$ so that $\langle a,b \rangle = 2$. Define the linear map $L: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} : x \mapsto x- \langle x,a \rangle b$. Find the eigenvalues of $L$.

How can I solve this problem without using the matrix representation of $L$?

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    $\begingroup$ Hint If $x$ is an eigenvector of eigenvalue $\lambda$, then by definition we have $L(x) = \lambda x$, or after rearranging, $(1 - \lambda) x = \langle x, a \rangle b$. In particular, either (1) both sides are zero, or (2) both sides are nonzero, in which case $x$ and $b$ are parallel. $\endgroup$ – Travis Jan 25 '17 at 15:40
  • $\begingroup$ I understand the case where both sides are zero. But what does it say about the eigenvalues when both sides are non-zero? I guess I'm a bit confused by the expression of $\langle x,a \rangle b = x^{T}ab$. $\endgroup$ – simp Jan 25 '17 at 15:59
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    $\begingroup$ If both sides are nonzero, then, since $x$ and $b$ are parallel, $x = \mu b$ for some $\mu$---already this tells us that the candidate eigenspace for this case is just $\langle b \rangle$. What happens if we substitute this condition into the equation we derived? $\endgroup$ – Travis Jan 26 '17 at 12:37
  • $\begingroup$ You're welcome, glad to help. $\endgroup$ – Travis Jan 27 '17 at 13:11
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Hint: it is easy to see that $L': x \mapsto \langle x,a \rangle b$ has only one eigenvector with non-zero eigenvalue, $\langle a,b \rangle = 2$. Now try to show that if $\{\lambda_1, \dots, \lambda_n\}$ are eigenvalues of a linear map $L'$, then $\{1-\lambda_1, \dots, 1-\lambda_n\}$ are eigenvalues of $I-L': x \mapsto x - L'(x)$.

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