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Let $S = \{\lambda_1, \cdots, \lambda_n\}$ 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 $n\times n$ matrix with complex entries, which is not self-adjoint, whose set of eigenvalues is given by $S$.

b. There exists an $n\times n$ self-adjoint, non-diagonal matrix with complex entries whose set of eigenvalues is given by $S$.

c. There exists an $n\times n$ symmetric, non-diagonal matrix with real entries whose set of eigenvalues is given by $S$.


a) No Idea.

b) as hermitian matrices(self adjoint) matrices has real eigen values only so $b$ may be true..

c) same logic as $b$. please help.

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For any given set of real numbers, one can find a hermitian (self-adjoint) matrix with those entries as eigenvalues. One easy way to see this is, stack all your real numbers in the diagonal of a diagonal matrix $D$. Let $U$ be any unitary matrix (means with complex entries). Then $A=UDU^H$ is a hermitian (self-adjoint) matrix with the diagonal entries of D as its eigenvalues.

Now consider any non self-adjoint invertible matrix $P$. Then $B=PAP^{-1}$ is also a complex matrix which is not self-adjoint but has same eigenvalues as $A$ (why?).

Now consider any orthonormal matrix $Q$ (distinguish orthonormal and unitary), then $C=QDQ^T$ is a symmetric matrix (with real entries) with the given real numbers as eigenvalues.

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    This answer fails to ensure the negative requirements of the question. For instance $A$ could still be diagonal. And even if $P$ is not unitary (which is the right condition, self-adjoint for a conjugating matrix is not relevant), then it might happen that $PAP^{-1}$ is still self-adjoint.2014-12-14