ok i got quick questions i have a question that asks me to prove that a complex vector times its complex conjugate transpose is real. I know that this is true just need a little help on notation, what is the proper way to define the complex vector?
and the second question what property of a matrix would give you atleast one zero eigenvalue the matrix in question looks almost diagonal...so would dominatly diagonal be the property? and then i need to prove this property.
I guess that since a real vector is an element of vector space R^n over R.Originally Posted by action259
We can say that a complex vector is an element of vector space C^n over C.
Thus, if,
v=(a1+ib1,a2+ib2,...,an+ibn)
It conjugate vector is,
u=(a1-ib1,a2-ib2,...,an-ibn)
Thus,
vu
Is the product of all ye components.
Since, for all k from 1 to n
(ak+ibk)(ak-ibk)=(ak)^2+(bk)^2 element of R
We see that each component becomes real.
Since the identity matrix is diagonaly dominant, and it has no non-zero
e-values, the property you seek cannot be that it is diagonaly dominant.
I suspect the property you need is that its determinant is zero.
(A possible way of approaching this maybe to look at the Jordan Canonical
form of your matrix, it must have a Jordan block with 0's on the diagonal
so its determinant must be zero)
RonL
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