Let \(E\) be a finite-dimensional vector space over a field \(F\). Suppose \(B:E\times E\to F\) is a bilinear map, not necessarily symmetric. Define \[ E_1=\{x\in E:B(x,y)=0\text{ for all }y\in E\}, \] \[ E_2=\{y\in E:B(x,y)=0\text{ for all }x\in E\}. \] Prove that \(\dim E_1=\dim E_2\).
Solution
Let \(n=\dim E\) and choose a basis \(e_1,\ldots,e_n\) of \(E\). Let \[ M=(B(e_i,e_j))_{i,j} \] be the matrix of the bilinear form in this basis, and for \(x\in E\) let \([x]\) be its coordinate column vector.
1. \(\dim E_1=\nullity(M^T)\).
Write \(x=\sum_i x_i e_i\). For each basis vector \(e_j\), \[ B(x,e_j)=\sum_i x_i B(e_i,e_j). \] Thus the column whose \(j\)th entry is \(B(x,e_j)\) is \(M^T[x]\). By linearity of \(B\) in the second variable, \[ x\in E_1 \iff B(x,e_j)=0\text{ for every }j \iff M^T[x]=0. \]
2. \(\dim E_2=\nullity(M)\).
For \(y=\sum_j y_j e_j\), the column whose \(i\)th entry is \(B(e_i,y)\) is \(M[y]\). By linearity of \(B\) in the first variable, \[ y\in E_2 \iff B(e_i,y)=0\text{ for every }i \iff M[y]=0. \]
3. One has\[ \boxed{\dim E_1=\dim E_2}. \]
A matrix and its transpose have the same rank, so rank-nullity gives \[ \nullity(M)=n-\operatorname{rank}(M) =n-\operatorname{rank}(M^T) =\nullity(M^T). \] Steps 1 and 2 give the claim.
4. Q.E.D.
Step 3 is the required equality.