行列式
線形代数学がたまに金融工学に役に立ちます。たとえば主成分分析が固有ベクトルを使用します。固有ベクトルを計算するために以下の数式を使います。
| A-lambda*I | =0 |
This characteristic equation basically indicates that non-zero eigenvectors exist only if the determinant of A-lambda*I is 0, where lambda is an eigenvalue. The premise of this is the equation:
Ax = lambda*x => x(A-lambda*I) = 0
If A-lambda*I is invertible, the only solution to the above is x=0. By a property of the determinant, the matrix is not invertible only if the determinant is 0. I was wondering why this might be so.
One property of the determinant is that |AB|=|A||B| (not proven here, but based on use of a function |TS|/|S| to compute |T|).
If A is invertible, there exists B such that AB = I. If det(A) = |A| = 0, then det(AB)=0. However, det(I) = 1 (can be proven by induction from 2x2 case). Thus, A cannot be invertible if det(A)=0. I have yet to prove that, if A is not invertible, then det(A)=0.
Actually, it seems that this proof is simple, if the definition of an inverse matrix as A^-1 =
[c11/|A| c21/|A| c31/|A| ... cn1/|A|]
[... ]
[c1n/|A| c2n/|A| c3n/|A| ... cnn/|A|]
If the determinant of A is non-zero, the matrix above exists. How to prove that the inverse actually is defined as above is non-trivial. In the 2x2 case, A^-1 * A = I does hold. Perhaps some induction is possible.
