Old qual question:
Let $V$ be a finite dimensional positive definite inner product space over $\mathbb{R}$ under $\langle,\rangle$ and let $A:V\to V$ be a linear map. Define $B=A^tA$. Now define a new bilinear form $\{,\}$ by $$\{v,w\}=\langle Bv,w\rangle.$$ Assume that $\ker A=m$.
- Prove that $\{,\}$ is symmetric.
- Calculate the index of positivity, negativity, nullity of the bilinear form $\{,\}$. (Recall that the index of positivity (resp. negativity, nullity of $\{,\}$ means the number of the positive (resp. negative, zero) eigenvalues of the corresponding symmetric matrix.)
OK so part 1 is easy, and I did this by putting $V$ in an ONB and then the adjoint of $A$ is just $A^t$. However, I'm not sure how to get going on part 2. My friend suggested showing that $\ker A=\ker A^tA$, and I can do that, but I'm still not sure where to go.
Thank you.
For any vector $v\in V$, $$v^tA^tAv=(Av)^t Av\geq 0.$$ This implies that the index of negativity is $0$. Further, the index of nullity is $m$ since $(Av)^tAv=0$ iff $Av=0$. Finally, together these imply that the index of positivity is $n-m$.