Algebra and Geometry | Part IA, 2005

Given a non-zero vector viv_{i}, any 3×33 \times 3 symmetric matrix TijT_{i j} can be expressed as

Tij=Aδij+Bvivj+(Civj+Cjvi)+DijT_{i j}=A \delta_{i j}+B v_{i} v_{j}+\left(C_{i} v_{j}+C_{j} v_{i}\right)+D_{i j}

for some numbers AA and BB, some vector CiC_{i} and a symmetric matrix DijD_{i j}, where

Civi=0,Dii=0,Dijvj=0,C_{i} v_{i}=0, \quad D_{i i}=0, \quad D_{i j} v_{j}=0,

and the summation convention is implicit.

Show that the above statement is true by finding A,B,CiA, B, C_{i} and DijD_{i j} explicitly in terms of TijT_{i j} and vjv_{j}, or otherwise. Explain why A,B,CiA, B, C_{i} and DijD_{i j} together provide a space of the correct dimension to parameterise an arbitrary symmetric 3×33 \times 3 matrix TijT_{i j}.

Typos? Please submit corrections to this page on GitHub.