Let \(V\) be finite-dimensional over a field \(F\), and let \[ B:V\times V\to F \] be bilinear, not necessarily symmetric. Define \[ V_1=\{x\in V:B(x,y)=0\text{ for all }y\in V\}, \] \[ V_2=\{y\in V:B(x,y)=0\text{ for all }x\in V\}. \] Prove that \[ \dim V_1=\dim V_2. \]
Solution
Let \(n=\dim_F V\), and fix a basis of \(V\).
1. There is a matrix \(A\in M_n(F)\) such that, in the chosen coordinates,\[ B(x,y)=x^TAy \]for all \(x,y\in V\).
If \(e_1,\ldots,e_n\) is the chosen basis, define \[ A_{ij}\coloneqq B(e_i,e_j). \] Bilinearity then gives the displayed formula for arbitrary coordinate vectors \(x\) and \(y\).
2. In these coordinates,\[ V_1=\ker A^T. \]
An element \(x\) lies in \(V_1\) exactly when \[ x^TAy=0 \] for every \(y\). This is equivalent to the row vector \(x^TA\) being zero, hence to \[ A^Tx=0. \]
3. In these coordinates,\[ V_2=\ker A. \]
An element \(y\) lies in \(V_2\) exactly when \[ x^TAy=0 \] for every \(x\). This is equivalent to \[ Ay=0. \]
4. The kernels of \(A\) and \(A^T\) have the same dimension.
The rank of a matrix equals the rank of its transpose. Hence, by rank-nullity, \[ \begin{aligned} \dim\ker A &=n-\operatorname{rank}A\\ &=n-\operatorname{rank}A^T\\ &=\dim\ker A^T. \end{aligned} \]
5. Therefore\[ \boxed{\dim V_1=\dim V_2}. \]
Combine steps <1>2–<1>4.
6. Q.E.D.
Step 5 is the required equality.