Let \(f\) be a form on a finite-dimensional vector space \(V\). Look at the definition of left non-degeneracy given in Exercise 6. Define right non-degeneracy and prove that the form \(f\) is left non-degenerate if and only if \(f\) is right non-degenerate.
Solution
The form is called right non-degenerate if the only vector \(\beta\) such that \[ f(\alpha,\beta)=0 \] for every \(\alpha\in V\) is \(\beta=0\).
Choose an ordered basis of \(V\) and let \(A\) be the matrix of \(f\) in that basis. With coordinate columns \(X,Y\), the form has the shape \[ f(X,Y)=Y^*AX. \]
1. Left non-degeneracy is equivalent to \(\ker A=\{0\}\).
Fix \(X\). The equality \[ f(X,Y)=Y^*AX=0 \] for every \(Y\) holds exactly when \(AX=0\). Thus the left radical corresponds to \(\ker A\). Hence left non-degeneracy is equivalent to \[ \ker A=\{0\}. \]
2. Right non-degeneracy is equivalent to \(\ker A^*=\{0\}\).
Fix \(Y\). We have \[ f(X,Y)=Y^*AX=(A^*Y)^*X. \] This vanishes for every \(X\) exactly when \[ A^*Y=0. \] Thus the right radical corresponds to \(\ker A^*\), and right non-degeneracy is equivalent to \[ \ker A^*=\{0\}. \]
3. The two conditions are equivalent.
For every finite matrix, \[ \operatorname{rank}A^*=\operatorname{rank}A. \] Since \(A\) and \(A^*\) are both \(n\times n\), rank-nullity gives \[ \dim\ker A^*=\dim\ker A. \] Therefore \[ \ker A=\{0\} \iff \ker A^*=\{0\}. \] Combining steps 1 and 2, \[ \boxed{f\text{ is left non-degenerate}\iff f\text{ is right non-degenerate}.} \]