Let \(T\) be the linear operator on \(C^2\) defined by \(T\epsilon_1 = (1 + i, 2)\), \(T\epsilon_2 = (i, i)\). Using the standard inner product, find the matrix of \(T^*\) in the standard ordered basis. Does \(T\) commute with \(T^*\)?
Solution
The matrix of \(T\) in the standard ordered basis is obtained from the two given images as its columns: \[ [T]=A= \begin{pmatrix} 1+i&i\\ 2&i \end{pmatrix}. \] For the standard inner product on \(\mathbb C^2\), the matrix of the adjoint is the conjugate transpose [@HK71]. Hence \[ \boxed{ [T^*]=A^*= \begin{pmatrix} 1-i&2\\ -i&-i \end{pmatrix}.} \]
1. Compute the two products.
Direct multiplication gives \[ AA^*= \begin{pmatrix} 3&3+2i\\ 3-2i&5 \end{pmatrix}, \] whereas \[ A^*A= \begin{pmatrix} 6&1+3i\\ 1-3i&2 \end{pmatrix}. \] These matrices are different; for example, their \((1,1)\) entries are \(3\) and \(6\).
2. Therefore \(T\) does not commute with its adjoint.
Matrix multiplication represents composition in the standard basis, so \(TT^*=T^*T\) would force \(AA^*=A^*A\). Step 1 shows this fails. Thus \[ \boxed{TT^*\ne T^*T.} \] In particular, \(T\) is not normal.