True or false? (linear algebra)?

True or false? If a diagonalizable operator has only the characteristic values 0 and 1, it is a projection (T^2 = T).





This seems pretty simple, but I just don't know. A diagonalizable operator has to have characteristic polynomial f = (x - c_1)^d_1 ... (x - c_k)^d_k. I basically take this to mean that the operator in question has to be 2 x 2, but I don't really know what other information I can take from this.

True or false? (linear algebra)?
Consider a diagonal matrix, D, with each diagonal element being either 0 or 1. For example:





[1 0 0 0]


[0 1 0 0]


[0 0 0 0]


[0 0 0 1]





would fit the bill.





For any of this kind of matrix, you can see that D^2 = D, since for diagonal matrices squaring the matrix just squares the diagonal elements and 0^2 = 0 and 1^2 = 1.





Now consider a diagonalizable operator, T, with characteristic values (eigenvalues) all 0 or 1. Then we can diagonalize the matrix into a diagonal matrix with 0s and 1s along the diagonal, like D. We can then write (Q' is the inverse of Q):





T = Q'DQ


T^2 = (Q'DQ)(Q'DQ)


= Q'DDQ (because Q'Q is the identity)


= Q'D^2Q


= Q'DQ (by the property of D).


= T





Which is the required result.