smallest subset of $C$ which contains all the eigenvalues In each of the following cases, describe the smallest subset of $C$ which contains all the eigenvalues of every member of the set $S$.
(a) $S=\{A\in M_n(C) | A=BB^*$ for some $B\in M_n(C)\}$
(b) $ S=\{A\in M_n(C)| A=B+B^*$ for some $B\in M_n(C)\}$
(c) $ S=\{A\in M_n(C)| A+A^*=0 \}$
For $\lambda$ as an eigen value of $B$ , (a) will have $\lambda\lambda^*=|\lambda|^2\ge0.$ Hence The smallest subset will be $\{x\in R|x\ge0\}$.
(b) will have the set as $R$. 
(c)  will have the set as the set of purely imaginary numbers
 A: The given sets are all correct, but the argument for a) is not. Not all eigenvalues of $BB^\ast$ are $\lambda\cdot \overline{\lambda}$ for an eigenvalue $\lambda$ of $B$. For example,
$$B = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$$
has only the eigenvalue $1$, but
$$BB^\ast = \begin{pmatrix} 1 & 1\\0 & 1\end{pmatrix}\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}$$
has the eigenvalues $\dfrac{3\pm\sqrt{5}}{2}$. In general, all eigenvalues of $BB^\ast$ arise from eigenvalues of $B$ only if $B$ is normal, i.e. $BB^\ast = B^\ast B$.
To determine the sets of possible eigenvalues of matrices of the given forms, we use the standard inner product on $\mathbb{C}^n$,
$$\langle z,w\rangle = \sum_{k=1}^n z_k\overline{w_k}.$$
Then we have $\langle Az,w\rangle = \langle z, A^\ast w\rangle$ and $\langle A^\ast z,w\rangle = \langle z, Aw\rangle$ for all $A\in M_n(\mathbb{C})$ and $z,w\in \mathbb{C}^n$. So if $\lambda$ is an eigenvalue of $A$ and $z_\lambda$ an eigenvector to the eigenvalue $\lambda$, we have
$$\lambda \lVert z_\lambda\rVert^2 = \lambda \langle z_\lambda,z_\lambda\rangle = \langle \lambda z_\lambda,z_\lambda\rangle = \langle Az_\lambda,z_\lambda\rangle,$$
and this allows to determine the possible eigenvalues of matrices of the specified form.
a) For an eigenvalue $\lambda$ of $BB^\ast$ we have $$\lambda\lVert z_\lambda\rVert^2 = \langle BB^\ast z_\lambda,z_\lambda\langle = \langle B^\ast z_\lambda, B^\ast z_\lambda\rangle = \lVert B^\ast z_\lambda\rVert^2 \geqslant 0.$$ Since $z_\lambda \neq 0$ as an eigenvector, we can divide by the positive quantity $\lVert z_\lambda\rVert^2$ to obtain $\lambda \geqslant 0$. Conversely, for $t \geqslant 0$, we have $(\sqrt{t}I)(\sqrt{t}I)^\ast = tI$ with the eigenvalue $t$, so all non-negative real numbers occur as eigenvalues of some matrix of the form $BB^\ast$.
b) For an eigenvalue $\lambda$ of $B+B^\ast$, we have $$\lambda\lVert z_\lambda\rVert^2 = \langle (B+B^\ast)z_\lambda,z_\lambda\rangle = \langle B z_\lambda,z_\lambda\rangle + \langle B^\ast z_\lambda,z_\lambda\rangle = \langle B z_\lambda,z_\lambda\rangle + \langle z_\lambda, B z_\lambda\rangle = 2\operatorname{Re}\langle B z_\lambda,z_\lambda\rangle \in \mathbb{R},$$
so $\lambda\in \mathbb{R}$, since $\lVert z_\lambda\rVert^2 > 0$. Conversely, for $t\in \mathbb{R}$ we have $(\frac{t}{2}I) + (\frac{t}{2}I)^\ast = tI$ and $t$ is an eigenvalue of a matrix of the form $B+B^\ast$.
c) For an eigenvalue $\lambda$ of a skew-hermitian matrix $A$ (that is, $A+A^\ast = 0$, or $A^\ast = -A$), we have $$\lambda\lVert z_\lambda\rVert^2 = \langle Az_\lambda,z_\lambda\rangle = \langle z_\lambda, A^\ast z_\lambda\rangle = \langle z_\lambda, -A z_\lambda\rangle = \langle z_\lambda, -\lambda z_\lambda\rangle = -\overline{\lambda}\lVert z_\lambda\rVert^2,$$ hence $\lambda = -\overline{\lambda}$, and that means $\lambda \in i\mathbb{R}$. Conversely, every $it \in i\mathbb{R}$ is the eigenvalue of the skew-hermitian matrix $itI$.
