Let \(V\) be \(C^3\) with the standard inner product. Let \(T\) be the linear operator on \(V\) whose matrix in the standard ordered basis is defined by \[ A_{jk} = i^{j+k}, \quad (i^2 = -1). \] Find a basis for the null space of \(T^*\).
Solution
Goal. For \(T\) with matrix \(A_{jk} = i^{j+k}\) (\(j,k = 1,2,3\)), find a basis for \(\ker T^*\).
1. Compute the matrix \(A\).
1.1. \(A_{jk} = i^{j+k}\), so\[ A = \begin{pmatrix} i^2 & i^3 & i^4 \\ i^3 & i^4 & i^5 \\ i^4 & i^5 & i^6 \end{pmatrix} = \begin{pmatrix} -1 & -i & 1 \\ -i & 1 & i \\ 1 & i & -1 \end{pmatrix}. \]
\(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), \(i^5 = i\), \(i^6 = -1\).
2. \(T^*\) has matrix \(A^* = \bar A^T\).
the adjoint of an operator with matrix \(A\) (in an orthonormal basis) has matrix the conjugate transpose.
3. Compute \(A^*\).
3.1. \(\bar A = \begin{pmatrix} -1 & i & 1 \\ i & 1 & -i \\ 1 & -i & -1 \end{pmatrix}\).
conjugate each entry.
3.2. \(A^* = \bar A^T = \begin{pmatrix} -1 & i & 1 \\ i & 1 & -i \\ 1 & -i & -1 \end{pmatrix}\).
\(A\) is symmetric (\(A_{jk} = A_{kj}\)), so \(\bar A^T = \bar A\).
4. Find \(\ker A^*\).
4.1. Solve \(A^* x = 0\): the rows are \((-1, i, 1)\), \((i, 1, -i)\), \((1, -i, -1)\).
write out the system.
4.2. The rows are linearly dependent: row 3 = \(-1 \cdot\) row 1, and row 2 = \(-i \cdot\) row 1.
\((-1)(-1, i, 1) = (1, -i, -1)\) and \((-i)(-1, i, 1) = (i, 1, -i)\).
4.3. So the null space is the set of \(x\) with \(-x_1 + i x_2 + x_3 = 0\), i.e. \(x_3 = x_1 - i x_2\).
the single independent equation is \(-x_1 + i x_2 + x_3 = 0\).
4.4. A basis is \(\theset{(1, 0, 1), (0, 1, -i)}\).
setting \((x_1, x_2) = (1, 0)\) gives \((1, 0, 1)\); setting \((x_1, x_2) = (0, 1)\) gives \((0, 1, -i)\); these are independent and span the \(2\)-dimensional null space.
5. Q.E.D.
4.4 gives the basis \(\theset{(1,0,1), (0,1,-i)}\) for \(\ker T^*\).