A is a nonzero matrix of order n. given a ^ 2 + a = 0, can we deduce that - 1 is an eigenvalue of a? Yes A^2+A=0 So f (x) = x ^ 2 + X is a zeroing polynomial of matrix A, The eigenvalue of a can only be the root of the zeroing polynomial f (x), that is, 0 or - 1, Because a is a nonzero matrix, it is impossible to have all zero eigenvalues, So there must be an eigenvalue - 1 A is a non-zero matrix, so when the eigenvalues are impossible, they are all zero? For example, a = [0 01; 0 00; 0 00]

A is a nonzero matrix of order n. given a ^ 2 + a = 0, can we deduce that - 1 is an eigenvalue of a? Yes A^2+A=0 So f (x) = x ^ 2 + X is a zeroing polynomial of matrix A, The eigenvalue of a can only be the root of the zeroing polynomial f (x), that is, 0 or - 1, Because a is a nonzero matrix, it is impossible to have all zero eigenvalues, So there must be an eigenvalue - 1 A is a non-zero matrix, so when the eigenvalues are impossible, they are all zero? For example, a = [0 01; 0 00; 0 00]

Ha, no point this time!
You're right. There's something wrong with it
This proves that:
Because a ^ 2 + a = 0
So (a + e) a = 0
So the column vectors of a are solution vectors of (a + e) x = 0
And because a is not zero
So (a + e) x = 0 has nonzero solution
So | a + e | = 0
So - 1 is an eigenvalue of A