0
$\begingroup$

Let $T:V\to V$ be a linear transformation on a finite-dimensional inner product space $V$.Show that $\lambda$ is an eigenvalue of $T$ if and only if $\bar{\lambda}$ is an eigenvalue of $T^*$.

Proof given:

Let O be an orthonormal basis for V. Then $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is an eigenvalue of $[T]_O$ if and only if rank($[T]_O - \lambda I) \geq 1$ if and only if rank($([T]_O - \lambda I)^*) \geq 1$ if and only if rank($[T^*]_O - \bar{\lambda} I) \geq 1$ if and only if $\bar{\lambda}$ is an eigenvalue of $[T^*]_O$ if and only if $\bar{\lambda}$ is an eigenvalue of $T^*$.

My question is why $\lambda$ is an eigenvalue of a matrix, say A if and only if rank$(A- \lambda I) \geq 1$?

$\endgroup$
5
  • $\begingroup$ Explain your notation: what is $\;[T]_0\;$ ? And $\;\lambda\;$ is eigenvalue of an operator (or matrix) $\;T\;$ iff $\;\det (T-\lambda I)=0\iff \text{rank}\,(T-\lambda I)\;$ isn't full. What does the rank being greater than or equal one means? Oh, I see that's your question...perhaps they mean that rank is the dimension of the kernel... $\endgroup$
    – Timbuc
    Dec 1, 2014 at 3:52
  • $\begingroup$ I think that instead of rank, you should have "null" or "dim ker". Otherwise, the statements make no sense. $\endgroup$
    – Ben Grossmann
    Dec 1, 2014 at 3:54
  • $\begingroup$ $[T]_O$ is the matrix representing T in terms of basis O. $\endgroup$
    – user10024395
    Dec 1, 2014 at 3:54
  • $\begingroup$ Consider $A = 0$ on $\mathbb{R}$. You have $\lambda = 0$ is an eigenvalue and $rank(A) = 0$. Your statement is wrong. $\endgroup$
    – Empiricist
    Dec 1, 2014 at 3:55
  • $\begingroup$ So it should be kernel >= 1 instead? $\endgroup$
    – user10024395
    Dec 1, 2014 at 3:55

1 Answer 1

Reset to default
1
$\begingroup$

Perhaps for them $\;rank A=\dim\ker A\;$ , and then everything follows:

$$\lambda\;\;\text{eigenvalue of}\;\;A\iff \det (A-\lambda I)=0\iff \ker (A-\lambda I)\neq 0\iff $$

$$\dim\ker (A-\lambda I)>0\iff \text{rank} (A-\lambda I)\ge 1\,$$

Yet if rank is given the more usual meaning, then statement is dead wrong.

$\endgroup$
3
  • $\begingroup$ Thats what I thought too. I think the answer is wrong. Thanks $\endgroup$
    – user10024395
    Dec 1, 2014 at 3:56
  • $\begingroup$ I would sooner believe that there was a typo than believe that the textbook is using a different definition of rank. $\endgroup$
    – Ben Grossmann
    Dec 1, 2014 at 3:57
  • $\begingroup$ @Omnomnomnom, a rather huge and pretty misleading typo, indeed. $\endgroup$
    – Timbuc
    Dec 1, 2014 at 4:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .