Prove if $\lambda$ is eigenvalue of X, then $\lambda$ is an eigenvalue of $X^T$

Question

Question I'm trying to solve is in the title.

My work:

If $Xv = \lambda v$ for vector v, then $\lambda$ is an eigenvalue of X.

We have $Xv = \lambda v$, multiplying both sides of the equation by $X^T$ yields:

$X^TXv =X^T \lambda v$

and

$X^TX \implies I$

Therefore, $Iv =X^T \lambda v$, which is

$v =X^T \lambda v$

which gives me

$\lambda^{-1}v =X^Tv$

but this implies that $\frac{1}{\lambda}$ is an eigenvalue of $X^T$, not $\lambda$.

Did I mess up somewhere?

Answer 1

Let $\chi_A$ the characteristic polynomial of $A$. The eigen values of A are the roots of $\chi_A$ (https://en.wikipedia.org/wiki/Characteristic_polynomial)

As $$\chi_X = \chi_{X^T}$$ so we have the result.