I’m working through Axler’s Linear Algebra Done Right, 3rd edition. Problem 30 in chapter 5.A asks to prove there exists $x\in R^3$, satisfying $Tx-9x=(-4,5,\sqrt 7)$, if it’s known that -4,5, and $\sqrt 7$ are the eigenvalues of linear transformation T.

My initial reaction to this problem is that each of the three eigenvectors will be sent to scalar multiples of themselves, but my understanding breaks down a little after that. I don’t believe I’m allowed to assume x=c(-4,5,$\sqrt 7$) is an eigenvector, c some constant. Though if that were true, then clearly Tx would be sent to some scalar multiple of x, and the task simply becomes finding what scalar to plug in for c to get back x.

I’m uncertain how to make additional progress and rereading the chapter hasn’t proved fruitful. Am I overlooking some obvious theorem?

Also, I’m new here. Please let me know if I’m violating any rules, and point me toward that list, so that I can avoid breaking any in the future.

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    $\begingroup$ Hint: What are the eigenvalues of $T-9I$? Can it be singular? $\endgroup$ – hmakholm left over Monica Dec 26 '18 at 20:45
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    $\begingroup$ (Hint 2: The fact that the components of that vector are exactly the eigenvalues of $T$ is a red herring). $\endgroup$ – hmakholm left over Monica Dec 26 '18 at 20:45
  • $\begingroup$ Since 9 is not an eigenvector, T-9I has to be non-singular yeah? Would the eigenvalues of that transformation just be the eigenvalues of T shifted down by 9? $\endgroup$ – 7l-l04 Dec 26 '18 at 20:54
  • $\begingroup$ If that’s the case, then I could say something like, since T-9I is non-degenerate, there exists a vector x in R^3 such that T-9I sends x to (-4,5,root7) $\endgroup$ – 7l-l04 Dec 26 '18 at 20:56
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    $\begingroup$ x @Thor: Yes, exactly. $\endgroup$ – hmakholm left over Monica Dec 26 '18 at 21:03

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