If is a matrix, and satisfy .
show that:
vice verse, If matrix satisfy the condition : ,
then not always Idempotent matrix, such as: (corrected)
can we add some conditions to such that is a Idempotent matrix
Well, if the matrix is the it is true since if , then
And since from the condition it follows that the minimal pol. of divides , we get what we want (if the matrix has rank 2 then zero can't be one of its eigenvectors thus its characteristic pol. is
For matrices of order greater than 2 I'm not sure but I doubt it.
Tonio
Think of A as acting as a linear transformation on an n-dimensional vector space V. Then and . The reason for this is that if then . Also, if then .
Now let be a basis for ran(A) (so that k = rank(A)), and let be a basis for ker(A). Then is a basis for V, and the matrix of the linear transformation A with respect to this basis is a diagonal matrix with k 1s in the first k places on the diagonal, and 0s in the remaining places on the diagonal. This matrix obviously has trace k. But it is similar to the original matrix A, and so has the same trace as A. Therefore trace(A) = k = rank(A).
If B is an arbitrary n×n matrix with nonzero trace, then there is a scalar multiple of B, namely , that has trace k. So knowing that a matrix has trace k tells you practically nothing about the structure of the matrix, and it will be no help at all in determining whether or not B is an idempotent.