Can I please receive feedback on my first proof and help proving the other? Thank you for your time and consideration.
- a. Let $A$ be a bounded linear operator on a complex Hilbert space $\mathscr{H}$. I wish to prove $\sigma(A A^*) \subset \sigma(A^* A) \cup \{0\}$.
$\textit{Proof.}$ Let $A$ be a bounded linear operator on a complex Hilbert space $\mathscr{H}$. Let that $\sigma(A A^*) \subset \sigma(A^* A) \cup \{0\}$ using the polar decomposition of $A$ in checking possible invertibility of $(\lambda I - A A^\ast)$ $$(Tx)(t) = \alpha(t) \cdot x(t).$$ $(T-\lambda)$ is injective always $$\sigma_p(T) = \phi.$$ So, $$x(t) = \frac{y(t)}{\alpha(t) - \lambda}$$ $$\lambda I - A A^\ast = 1$$ $$\lambda I = 1 + AA^\ast.$$ Thus, $\displaystyle{I = \frac{1 + AA^\ast}{\lambda}}.$
- b. Construct an operator $A$ for which $\sigma(A A^*) \not \subset \sigma(A^* A)$.
For $1b)$ consider the Hilbert space $\ell^2(\mathbb{N})$ and the shift operator $s:\ell^2(\mathbb{N})\to \ell^2(\mathbb{N})$ given by $\sum_{i\in\mathbb{N}}\lambda_i e_i\mapsto \sum_{i\in\mathbb{N}}\lambda_i e_{i+1}$. One computes that the adjoint is given by $\sum_{i\in\mathbb{N}}\lambda_i e_i\mapsto \sum_{i>1}\lambda_i e_{i-1}$. Now $s^{\ast}s=id$ and $ss^{\ast}$ is the projection onto the subspace generated by $\{e_i\}_{i>1}$. As $e_1$ is orthonormal to this subspace $e_1$ lies in the kernel of this map so $0\in \sigma(s s^{\ast})$ and $0\notin \sigma(s^{\ast}s)=\sigma(id)$. Which shows that $\sigma (ss^{\ast})\not\subset \sigma(s^{\ast}s)$.