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.


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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.