Question

show that every skew hermitian matrix a can be expressed uniquely as p -iq where p is a real skew symmetric and q a real symmetric matrix.
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Answer 1

Answer:

As we saw in Chap. 8, when V is a finite-dimensional vector space over F , then a linear mapping T:V→V is semisimple if and only if its eigenvalues lie in F and its minimal polynomial has only simple roots. It would be useful to have a result that would allow one to predict that T is semisimple on the basis of a criterion that is simpler than finding the minimal polynomial, which, after all, requires knowing the roots of the characteristic polynomial.

Answer 2

Given, P and q are a real symmetric matrix

If A is a square matrix, then we can write,[tex]A= \frac{1}{2} (A+A^{θ} +A−A^{θ} ) \\ A= \frac{1}{2} (A+A )+\frac{1}{2} (A−A θ) \\ A= \frac{1}{2} (A+A )+i \frac{1}{2i} (A−A θ) (1) \\ A=P-iQ (2) 1) From equation (1) and (2), \\ p=\frac{1}{2} (A+A θ)\\ Pθ =[ \frac{1}{2} (A^{θ}+(Aθ)θ) ] \\ pθ=\frac{1}{2} (A+A θ)\\\\ pθ=P\\ \\ Q=\frac{1}{2}(A-A θ)\\ Q =[ \frac{1}{2}(Aθ-(Aθ)θ) ]\\ Q= \frac {1}{2} Qθ=Q\\ \\\\ A=P-iQ [/tex]