1. ## Hermitian matrices

(a) Show that if C is in Mn(C) is Hermitian and x*Cx = 0 for all x is in Cn, then C = 0.
(b) Show that for any A is in Mn(C) there are ( unique ) Hermitian matrices B and C for which A = B + iC.
(c) Show that if x*Ax is real for all x is in Cn, then A is Hermitian.
Thank you!

2. Originally Posted by mivanova
(a) Show that if C is in Mn(C) is Hermitian and x*Cx = 0 for all x is in Cn, then C = 0.
(b) Show that for any A is in Mn(C) there are ( unique ) Hermitian matrices B and C for which A = B + iC.
(c) Show that if x*Ax is real for all x is in Cn, then A is Hermitian.
For (a) I would use a polarisation identity, $\displaystyle y^*Cx = \tfrac14\bigl((x+y)^*C(x+y) + i(x+iy)^*C(x+iy) - (x-y)^*C(x-y) i(x-iy)^*C(x-iy)\bigr)$. If $\displaystyle x^*Cx=0$ for all x, then each term the right side of the identity will be 0, and so $\displaystyle y^*Cx=0$ for all x and y. It easily follows that C=0. Note that this result is true for all n×n complex matrices, they don't need to be hermitian.

(b) is kind of obvious. Just take $\displaystyle B=\tfrac12(A+A^*),\ C=\tfrac1{2i}(A-A^*)$. (Of course, you have to check that B and C are hermitian.)

For (c), choose some particular values for x. If x has a 1 for the j'th coordinate and zeros everywhere else then x*Ax is just the (j,j)-element of A. So if that is real then A has real elements on the diagonal. Now take x to have two nonzero elements, in the j'th and k'th coordinates. If both these coordinates are 1, then the condition x*Ax is real tells you that the imaginary part of the sum of the (j,k)- and (k,j)-elements is zero. If the j'th coordinate of x is 1 and the k'th coordinate is i, it tells you that the real parts of the (j,k)- and (k,j)-elements of A are the same. Put those results together and they tell you that A is hermitian.