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i am not sure about this question.

given two real matrices P,Q (nxn) with real componponents(real valued), and U= P+iQ. $D=\begin{pmatrix}P&-Q\\ Q&P\end{pmatrix}$

a)prove that if U is hermitian matrix, then D is symmetrical matrix

b)prove that if U is unitary matrix, then D is orthogonal matrix.

what i did:

i tried to use the properties of a)hermitian matrix(if a given matrix is hermetian and its components are real valued then both terms should be equivalent, since an hermitian matrix is $a^H=a$, so $x^Hax \in R$ for every $x \in C^{nx1}$, so we get that if eigen values of given matrix is in R then a is hermitian ($eig(a) \in R$).

b)if a given matrix is orthogonal, then $a^ta=aa^t=I$. now, since a unitary matrix is unitary iff it applies $a*a=aa*=I$. both terms are close and related to each other, but i don't know how to prove it.

i've tried to put it all together, but i don't know how to show that they apply, i never get to it. i'm stuck on it for quite a while.

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Part a) is easy: $$ D^T = \pmatrix{P^T & Q^T\\-Q^T & P^T}, \qquad U^H = P^T - iQ^T $$ verify that if $U = U^H$, then $D = D^T$.

Part b) is a little trickier. Note that $$ U^HU = (P^TP + Q^TQ) + i(P^TQ - Q^TP) $$ So, if $U$ is unitary, then $P^TP + Q^TQ = I$ and $P^TQ - Q^TP = 0$. Now, using block-matrix multiplication, verify that $$ D^TD = \pmatrix{P^TP + Q^TQ & P^TQ - Q^TP\\Q^TP - P^TQ & P^TP + Q^TQ} $$ so that if $U$ is unitary, we have $D^TD = I$ as desired.

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    $\begingroup$ Alternatively, we could make quick work of this problem using the properties of the Kronecker product, noting that $$ D = I \otimes P + \pmatrix{0&1\\-1&0}\otimes Q $$ $\endgroup$ – Omnomnomnom Nov 13 '17 at 22:08

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