# A problem on self adjoint matrix and its eigenvalues

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.

• When you say "set of eigenvalues" in case (a), are you counting them by geometric or algebraic multiplicity? – Robert Israel Jan 7 '13 at 7:49
• What did you try? What are your thoughts? – Did Jan 7 '13 at 7:58

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.
(b) Let $v=\frac{1}{\sqrt{n}}(1,\ldots,1)^T$. Extend $v$ to an orthonormal basis of $\mathbb{R}^n$. Put the basis vectors together as columns of a real orthogonal matrix $Q$. Then $A=Q\operatorname{diag}(\lambda_1,\ldots,\lambda_n)Q^T$ is self-adjoint and has eigenvalues $\lambda_1,\ldots,\lambda_n$. Furthermore, $Av=\lambda_1v$. Hence all entries of $Av$ are all equal to $\lambda_1/\sqrt{n}$. Therefore $A$ is not a diagonal matrix, or else it would be equal to some $D=\operatorname{diag}(d_1,\ldots,d_n)$, which means $(d_1,\ldots,d_n)$ must be a permutation of $(\lambda_1,\ldots,\lambda_n)$ and the entries of $Av=Dv=(d_1,\ldots,d_n)^T$ are not all equal.
(c) The same $A$ in (b) will do.