# Thread: The Eigenvector of a Transpose Problem

1. ## The Eigenvector of a Transpose Problem

Let S $\displaystyle \in$ Mnxn(R) and D $\displaystyle \in$ Mnxn(R) be n $\displaystyle \times$ n matrices such that DTSD = S
where S is non degenerate. Let $\displaystyle \mu \neq 0$ be an eigenvalue of $\displaystyle D$ with eigenvector
x.
(i) Show that Sx is an eigenvector of DT and find its eigenvalue.
(ii) Show that $\displaystyle \frac{1}{\mu}$ is an eigenvalue of D.
[Hint: det(A) = det(AT ) for any square matrix A.]

This question has been giving me some real trouble and all help is welcome

2. ## Re: The Eigenvector of a Transpose Problem

Hi zook,

I'll try to get your going in the right direction.

i) Remember that for $\displaystyle \overline{x}$ to be an eigenvector of $\displaystyle M$ we must have two things: (1) $\displaystyle \overline{x}\neq \overline{0}$ and there is $\displaystyle \lambda$ (in whatever field you're working with, presumably the real numbers) such that $\displaystyle M\overline{x}=\lambda\overline{x}.$ This means that if we want to show $\displaystyle S\overline{x}$ is an eigenvector of $\displaystyle D^{T}$, we must establish (1) and (2).

To prove (1) we note that $\displaystyle \overline{x}\neq 0$ because it is an eigenvector of $\displaystyle D$ and so cannot be zero (by definition of eigenvector). Now also note that $\displaystyle S$ is assumed nonsingular/nondegenerate and so $\displaystyle S\overline{x}\neq 0,$ because $\displaystyle \overline{x}\neq 0.$

To prove (2) we us $\displaystyle D^{T}SD=S$. This implies $\displaystyle D^{T}SD\overline{x}=S\overline{x}.$ Since $\displaystyle \overline{x}$ is an eigenvector of $\displaystyle D$ with corresponding eigenvalue $\displaystyle \mu$ we have $\displaystyle D^{T}S(\mu\overline{x})=S\overline{x}.$ Pull the constant through to get $\displaystyle \mu D^{T}(S\overline{x})=S\overline{x}.$ Now divide by $\displaystyle \mu.$

Does this help? Let me know if anything is unclear. Good luck!

3. ## Re: The Eigenvector of a Transpose Problem

Yes that is brilliant Thank you