(a) With A = [{1, i, 0}, {i, 0, 1}], use elimination to solve Ax = 0

(b) Show that the nullspace you just computed is orthogonal to C(A^H) and not to the usual row space C(A^T). The four fundamental spaces in the complex case are N(A) and C(A) as before, and then N(AH) and C(A^H)

The nullspace I got from part a) does not fit with part b). ie, it is orthogonal to C(A^T), not C(A^H). the null space I got is {i,-1,1}.

Any help will be appreciated.

Question

(a) With A = [{1, i, 0}, {i, 0, 1}], use elimination to solve Ax = 0

(b) Show that the nullspace you just computed is orthogonal to C(A^H) and not to the usual row space C(A^T). The four fundamental spaces in the complex case are N(A) and C(A) as before, and then N(AH) and C(A^H)

The nullspace I got from part a) does not fit with part b). ie, it is orthogonal to C(A^T), not C(A^H). the null space I got is {i,-1,1}.

Any help will be appreciated.

anonymous · Answer

Since A has 2 columns, null space solution of x can be in terms of {x1, x2}. But you got {i,-1,1} which is in terms of {x1, x2, x3]. How is that possible??

helder_edwin · Answer

I assume the matrix is
$$ \large \begin{pmatrix}
                 1 & i\ i & 0\ 0 & 1
              \end{pmatrix} $$
which can be reduced to
$$ \large \begin{pmatrix}
                 1 & 0\ 0 & 1\ 0 & 0
              \end{pmatrix} $$
so the null space is
$$ \large \left\langle\begin{pmatrix}
                0\ 0\ 1
              \end{pmatrix}\right\rangle $$