Given that a is a matrix of order n and satisfies the equation A2 + 2A = 0, it is proved that the eigenvalue of a can only be 0 or - 2

Given that a is a matrix of order n and satisfies the equation A2 + 2A = 0, it is proved that the eigenvalue of a can only be 0 or - 2

Let a be the eigenvalue of A,
Then a ^ 2 + 2a is the eigenvalue of a ^ 2 + 2A
When a ^ 2 + 2A = 0, the eigenvalue of zero matrix can only be 0
So a ^ 2 + 2A = 0
So a (a + 2) = 0
So a = 0 or a = - 2
That is, the eigenvalue of a can only be 0 or - 2

I just saw an expert say that every n-order matrix has n eigenvalues. How to explain that det (λ I-A) = 0 has multiple roots?

It's a cliche that double roots are calculated by multiple numbers

In linear algebra, let a be a matrix of second order, and | 2e-a | = 0, | 3E + a | = 0, find the determinant of matrix A

|If 2e-a | = 0, then 2 is the eigenvalue of A
|3E + a | = 0, then | - 3) e-A | = 0, so - 3 is the eigenvalue of A
A is a square matrix of second order with only two eigenvalues
The product of eigenvalues is equal to | a |, so | a | = 2 × (- 3) = - 6