Let \(f\) be the bilinear form defined in Exercise 6, and let \(V_2\) be the subspace of \(V\) consisting of all matrices \(A\) such that \(\operatorname{trace}(A) = 0\) and \(A^* = -A\) (\(A^*\) is the conjugate transpose of \(A\)). Denote by \(f_2\) the restriction of \(f\) to \(V_2\). Show that \(f_2\) is negative definite, i.e., that \(f_2(A, A) < 0\) for each non-zero \(A\) in \(V_2\).
Solution
1. From Exercise 6, \(f(A, B) = n\operatorname{tr}(AB) - \operatorname{tr}(A)\operatorname{tr}(B)\).
the definition of \(f\).
2. For \(A \in V_2\), \(\operatorname{tr}(A) = 0\), so \(f(A, A) = n\operatorname{tr}(A^2)\).
substitute \(\operatorname{tr}(A) = 0\) into 1.
3. Since \(A^* = -A\), \(A\) is skew-Hermitian, so \(A\) is diagonalizable with purely imaginary eigenvalues \(i\lambda_1, \ldots, i\lambda_n\) (with \(\lambda_j \in \RR\)).
a skew-Hermitian matrix is normal, hence unitarily diagonalizable, and its eigenvalues are purely imaginary.
4. \(\operatorname{tr}(A^2) = \sum_j (i\lambda_j)^2 = -\sum_j \lambda_j^2 \le 0\).
the trace is the sum of the eigenvalues of \(A^2\), which are \((i\lambda_j)^2 = -\lambda_j^2\).
5. \(\operatorname{tr}(A^2) < 0\) for \(A \neq 0\).
if \(A \neq 0\), then at least one eigenvalue \(i\lambda_j \neq 0\), so \(\lambda_j \neq 0\) and \(-\sum_j \lambda_j^2 < 0\).
6. Hence \(f(A, A) = n\operatorname{tr}(A^2) < 0\) for every nonzero \(A \in V_2\).
2, 4, and 5.
7. Q.E.D.
6.