Let \[ A = \begin{bmatrix} 1 & i \\ -i & 2 \end{bmatrix} \] and let \(g\) be the form (on the space of \(2 \times 1\) complex matrices) defined by \(g(X, Y) = Y^*AX\). Is \(g\) an inner product?
Solution
Yes. The form \(g\) is an inner product.
1. The matrix \(A\) is Hermitian.
We have \[ A^* =\begin{pmatrix}1&i\\-i&2\end{pmatrix} =A. \] Therefore \[ g(Y,X)=X^*AY =\overline{Y^*A^*X} =\overline{g(X,Y)}, \] so \(g\) is Hermitian. Sesquilinearity follows directly from the matrix formula \(g(X,Y)=Y^*AX\).
2. The form is positive definite.
Let \[ X=\binom{z_1}{z_2}. \] Then \[ \begin{aligned} g(X,X) &=X^*AX\\ &=|z_1|^2+2|z_2|^2 +i\overline{z_1}z_2-i\overline{z_2}z_1\\ &=|z_1+iz_2|^2+|z_2|^2. \end{aligned} \] This is nonnegative. If it is zero, then \(z_2=0\) and \(z_1+iz_2=z_1=0\), so \(X=0\). Hence \[ g(X,X)>0 \] for every nonzero \(X\).
Thus \(g\) is a positive Hermitian sesquilinear form, i.e. an inner product.