There are two possible answers. First, the determinant of a matrix is the same as the product of all of its eigenvalues. Second, if A satisfies that equation, its eigenvalues must also. You can manipulate the equation to a quadratic that has two solutions. Since A is 3 by 3, it has three eigenvalues. Since only two numbers sastisfy that equation, one of them must be a duplicate eigenvalue. The determinant depends upon which of them is the duplicate.
But that is not possible. We have . That is , is an annihilator polynomial of being and . As , and . This means that the minimal polynomial of is just and are eigenvalues of as you said. The third one should be necessarily real and is diagonalizable in . So, is similar (in ) to and as a consequence:
Again, as you said or which implies (contradiction).
how [COLOR=rgb(0, 0, 0)]lambda3=lambda1 or lambda3= lambda2[/COLOR] is possible? lambda_3 must be real as the matrix is 3 by 3.Is that problem is wrong or something? Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me
Plainly: does not exist.
The minimal polynomial of divides to any annihilating polynomial of so or or . As and we conclude that is the minimal polynomial of .Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me