If A is a orthogonal projection matrix on B and B is subspace of R^m, rank(A)=trace(A).
True:
Because A is orthogonal projection matrix on B, eigenvalues of A are 0s and1s.
For all x in B, Ax=x. So the gm(1)=dim(B). For all y in B^perb =Ax=0. So the gm(0)=dim(B^per).
dim(B)+dim(B^per)=m =gm(1)+gm(0)=m -> A is diagonalizable.
Because A is diagonalizable, trace(A)= gm(1)*1 + gm(0)*0=gm(1) = m - gm(0).
We know gm(0)=nullity(A),so trace(A)=gm(1)= m - nullity(A)=rank(A). In the conclusion, we get the rank(A) = trace(A).
請問上面證明,正確嗎?
還有如果一個矩陣P只是P^2=P (即idempotent matrix)
請問除了ker(P)與Im(P)呈直和,還有什麼性質嗎? 如eigenvalue , rank之類的。
而且應該沒法套用上面的論述吧?
即: If A is a idempotent matrix , rank(A)=trace(A). => False。
最後請問一個matrix P要被稱作orthogonal projection matrix,是不是要P^2=P, 且P^t=P呢?
謝謝~