Let \(f\) be the bilinear form defined in Exercise 6. Show that \(f\) is degenerate (not non-degenerate). Let \(V_1\) be the subspace of \(V\) consisting of the matrices of trace 0, and let \(f_1\) be the restriction of \(f\) to \(V_1\). Show that \(f_1\) is non-degenerate.
Solution
Exercise 6 defines \[ f(A,B) =n\operatorname{tr}(AB) -\operatorname{tr}(A)\operatorname{tr}(B) \] on \(V=M_n(\mathbb C)\).
1. The form \(f\) is degenerate on \(V\).
Let \(I\) be the identity matrix. It is nonzero, and for every \(B\in V\), \[ \begin{aligned} f(I,B) &=n\operatorname{tr}(IB) -\operatorname{tr}(I)\operatorname{tr}(B)\\ &=n\operatorname{tr}(B)-n\operatorname{tr}(B)\\ &=0. \end{aligned} \] Thus the nonzero vector \(I\) belongs to the radical of \(f\), so \(f\) is degenerate.
2. On the trace-zero subspace the form simplifies to a trace pairing.
If \(A,B\in V_1\), then \[ \operatorname{tr}(A)=\operatorname{tr}(B)=0. \] Hence \[ f_1(A,B)=n\operatorname{tr}(AB). \] Since \(n>0\), the scalar \(n\) is nonzero in \(\mathbb C\). Therefore it is enough to prove that \[ \operatorname{tr}(AB)=0\quad\text{for every }B\in V_1 \] forces \(A=0\).
3. Vanishing against all trace-zero matrices forces every off-diagonal entry of \(A\) to vanish.
Assume \(A\in V_1\) satisfies \[ \operatorname{tr}(AB)=0 \] for every \(B\in V_1\). For \(i\ne j\), the matrix unit \(E_{ji}\) has trace zero, so it belongs to \(V_1\). Moreover \[ \operatorname{tr}(AE_{ji})=A_{ij}. \] Thus \[ A_{ij}=0 \qquad(i\ne j). \] Hence \(A\) is diagonal.
4. Its diagonal entries are all equal, and trace zero then forces them all to vanish.
Assume first \(n\ge2\). For any \(i\ne j\), the matrix \[ B=E_{ii}-E_{jj} \] has trace zero. Therefore \[ 0=\operatorname{tr}(AB)=A_{ii}-A_{jj}. \] All diagonal entries of \(A\) are equal, say to \(c\). Since \(A\in V_1\), \[ 0=\operatorname{tr}(A)=nc. \] Thus \(c=0\) and \(A=0\).
If \(n=1\), then \(V_1=\{0\}\), so the restriction is non-degenerate vacuously. Consequently, for every positive integer \(n\), \[ \boxed{f_1\text{ is non-degenerate on }V_1.} \]