A is a nonzero matrix of order n. given a ^ 2 + a = 0, can we deduce that - 1 is an eigenvalue of a?

A is a nonzero matrix of order n. given a ^ 2 + a = 0, can we deduce that - 1 is an eigenvalue of a?

Ha, this one has points!
Proof: 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

Matrix A is a square matrix of order 4, each element is 1, and non-zero eigenvalues are obtained

The practice on the third floor is a little too standard. In fact, this problem is obvious
Since the rank of a is 1, there are at least three zero eigenvalues
Using the sum of eigenvalues equal to tra, the nonzero eigenvalue is 4

A is an mxn matrix and the column vector x is a real number. It is proved that AX = 0 and ATA = 0 have the same solution
ATA is the transpose of a by A

Equation (1): ax = 0, equation (2): atax = 0. First of all, if X1 is the solution of (1), then it must also be the solution of (2), because substituting it into (2): atax1 = at (ax1) = at * 0 = 0. Secondly, it is proved that the solution of (2) is also the solution of (1). Let X1 be the solution of (2), then atax1 = 0 further has: x1tata X1 = 0, namely (ax1)