$T: V \rightarrow V$ be a normal linear operator on a finite dimensional inner product space.

linear algebra

I did an undergrad in mathematics but I've been working in the financial sector for a few years and it seems my math skills have all but deteriorated… I'm going problems in my old linear algebra textbook for fun and I've been trying to figure out these three factoids for awhile now… Would appreciate some help, thanks

Let $T: V \rightarrow V$ be a normal linear operator on a finite dimensional inner product space. I'm trying to understand why the following things are true:

$(i):$ $||T(x)|| = ||T^*(x)||$ for all $x \in V$

$(ii):$ If $\lambda$ is an eigenvalue for $T$ with corresponding eigenvector $x$, then $T^*(x) = \bar{\lambda}x$, where $\bar{\lambda}$ is the complex conjugate of $\lambda$.

$(iii):$ If $\lambda,\mu$ are distinct eigenvalues, with corresponding eigenvectors $x,y$, then $\langle x,y \rangle = 0$

My thoughts:

$(i)$: I was thinking that this should be true, since the $^*$ just takes the transpose and the conjugate, so the norm should remain the same… Would obviously like to understand this deeper though.

$(ii):$ Seems obvious but I don't know how to show it…

$(iii)$ I have no clue

If somebody could help me out with this and provide some details that'd be great, my math skills arn't what they used to be…

Best Answer

  • $\|Tv\|^2 = \langle Tv, Tv \rangle = \langle v, T^* T v \rangle$. Now use the fact that $T$ is normal to show that the right-hand side equals $\|T^* v\|^2$.
  • Show that $T-\lambda I$ is also normal (note that its conjugate transpose is $T^* - \bar{\lambda} I$). Then note $0=\|(T-\lambda I) x\|^2$ and use the result in (i).
  • $\bar{\lambda} \langle x, y \rangle = \langle \lambda x, y \rangle = \langle Tx, y\rangle = \langle x, T^* y \rangle = \langle x, \bar{\mu} y \rangle = \bar{\mu} \langle x, y \rangle$, where we have used the result in (ii).
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