Show that if $\pm\lambda $ is an eigenvalue of $A$ then $\lambda^2$ is an eigenvalue of $A^TA$ and vice versa.

Question

Show that if $\pm\lambda $ is an eigenvalue of $A$ then $\lambda^2$ is an eigenvalue of $A^TA$ and vice versa.

If $\pm\lambda $ is an eigenvalue of $A$ then $Av=\pm\lambda v\implies v^TA^T=\pm \lambda v^T\implies v^TA^TAv=a^2v^Tv $

How to show that $a^2$ is an eigenvalue of $A^TA$ from above?

Conversely $A^TAv=a^2v\implies v^TA^TAv=a^2 v^Tv\implies \langle Av,Av \rangle =a^2\langle v,v\rangle\implies ||Av||=a^2||v||\implies ||Av||=\pm a||v||$

How to show that $Av=\pm av$ from here? Please help.

Answer 1

If $\lambda$ is a singular value of $A$, then there are orthonormal vectors $u$ and $v$ such that $Av = \lambda u$ and $A^\top u = \lambda v$. Thus, $$ (A^\top A) v = A^\top (A v) = A^\top (\lambda u) = \lambda (A^\top u ) = \lambda (\lambda v) = \lambda^2 v. $$ Since $||v||=1 \ne 0$, it follows that $\lambda^2$ is an eigenvalue of $A^\top A$ (similarly, $\lambda^2$ is an eigenvalue of $AA^\top$). Thus, your statement above is not quite correct as written.