Let $S = \{\lambda_1, \ldots , \lambda_n\}$ be an ordered set of $n$ real numbers, not all equal, but not all necessarily distinct. Pick out the true statements:
a. There exists an $n × n$ matrix with complex entries, which is not selfadjoint, whose set of eigenvalues is given by $S$.
b. There exists an $n × n$ self-adjoint, non-diagonal matrix with complex entries whose set of eigenvalues is given by $S$.
c. There exists an $n × n$ symmetric, non-diagonal matrix with real entries whose set of eigenvalues is given by $S$.
How can i solve this? Thanks for your help.
Best Answer
The general idea is to start with a diagonal matrix $[\Lambda]_{kj} = \begin{cases} 0, & j \neq k \\ \lambda_j, & j=k\end{cases}$ and then modify this to satisfy the conditions required.
1) Just set the upper triangular parts of $\Lambda$ to $i$. Choose $[A]_{kj} = \begin{cases} 0, & j>k \\ \lambda_j, & j=k \\ i, & j<k\end{cases}$.
2) & 3) Suppose $\lambda_{j_0} \neq \lambda_{j_1}$. Then rotate the '$j_0$-$j_1$' part of $\Lambda$ so it is no longer diagonal. Let $[U]_{kj} = \begin{cases} \frac{1}{\sqrt{2}}, & (k,j) \in \{(j_0,j_0), (j_0,j_1),(j_1,j_1)\} \\ -\frac{1}{\sqrt{2}}, & (k,j) \in \{(j_1,j_0)\} \\ \delta_{kj}, & \text{otherwise} \end{cases}$. $U$ is real and $U^TU=I$. Let $A=U \Lambda U^T$. It is straightforward to check that $A$ is real, symmetric (hence self-adjoint) and $[A]_{j_0 j_1}=\lambda_{j_1}-\lambda_{j_0}$, hence it is not diagonal.