1
$\begingroup$

Consider $\mathbb{R}^{3}$ with standard inner product. Let $a,b \in \mathbb{R}^{3}$ be defined such that $\langle a,b \rangle=2$. Now define:

$L:\mathbb{R}^{3} \rightarrow \mathbb{R}^{3}: x \mapsto x-\langle x, a \rangle b$

Find all eigenvalues of $L$. (Hint: Don't use the matrix notation of $L$.)

This is a question from my textbook. The only solution I have found is: $-1$, because when $x=b$, $b$ maps to $-b$. Am I missing something here?

Also, if I am missing something, I would prefer if you would give me a hint, rather than the answer.

  • 1
    yes, you're missing something. Suppose $v$ is a generic eigenvector, meaning $Lv= \lambda v$. What can you say on $v$? and on $\lambda$? – 2017-01-25
  • 0
    Or, a different Hint: $L$ is the sum of Identity and a rank 1 linear operator – 2017-01-25
  • 2
    Look at vectors $x$ with $x \perp a$ – 2017-01-25
  • 0
    Your proposals helped me find one more eigenvalue: $1$, I however don't feel like I'm finding these according to a proper method of sorts. Is a good way to go about this to try and find all $x$ for which $\langle x , a \rangle$ has known vaules? – 2017-01-25

1 Answers 1

0

the eigen-value equation is this: $$ L[x]=x-\langle {x,a} \rangle b=\lambda x $$ which can be written as: $$ (1-\lambda)x=\langle {x,a} \rangle b $$ which has two solutions: either $x\perp a \land \lambda=1$ or $x\propto b$. In such case: $$ x=\alpha b \implies\alpha(1-\lambda)b=\alpha\langle {b,a} \rangle b\implies \lambda=1-\langle {b,a} \rangle=-1 $$