# 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

## 2 Answers

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.

(a) Consider an upper triangular matrix with the eigenvalues on the diagonal and at least one nonzero entry in the strictly upper triangular part.

(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.