- What is a bilinear form on a vector space \(V\) over a field \(K\)?
- When are two bilinear forms equivalent (congruent)?
- What is an orthogonal matrix \(Q \in \operatorname{O}(n)\), and what are its special geometric and algebraic properties?
Solution
A bilinear form on a \(K\)-vector space \(V\) is a map \[ B:V\times V\to K \] that is linear in each argument. In a basis \(e_1,\dots,e_n\), its Gram matrix is \[ M=(B(e_i,e_j)), \] and if \(x,y\) are coordinate columns then \[ B(x,y)=x^TM y. \]
Two bilinear forms are equivalent if there is \(P\in\operatorname{GL}_n(K)\) carrying one to the other. Their matrices are then related by congruence: \[ M_2=P^TM_1P. \] Thus congruence, not similarity, is the natural change-of-basis relation for bilinear forms.
Over \(\mathbb R\), an orthogonal matrix is a matrix \(Q\) satisfying \[ Q^TQ=I, \] equivalently \(Q^{-1}=Q^T\). It preserves the Euclidean inner product: \[ \langle Qx,Qy\rangle=x^TQ^TQy=\langle x,y\rangle. \] Hence it preserves lengths, distances, and angles; its columns and rows are orthonormal; and \[ (\det Q)^2=1, \] so \(\det Q=\pm1\). Its complex eigenvalues have absolute value \(1\).
For an orthogonal change of basis, congruence and similarity have the same matrix expression, \[ Q^TAQ=Q^{-1}AQ. \] This identity is one reason orthogonal diagonalization is especially natural for real symmetric matrices.