Let a be a square matrix of second order, and the determinant of a = 1, a11 + A22 > 2. It is proved that a is similar to a diagonal matrix

Let a be a square matrix of second order, and the determinant of a = 1, a11 + A22 > 2. It is proved that a is similar to a diagonal matrix

A is similar to the minimum polynomial of diagonal matrix A, which has no multiple roots. λ e-A = λ - a11 - a12-a21, λ - a22d1 = 1D2 = | λ e-A | = (λ - a11) (λ - A22) - a21a12 = λ ^ 2 - (a11 + A22) λ + (a11a22-a21s12) = λ ^ 2 - (a11 + A22) λ + 1. Above is a quadratic function f (λ), and the opening upward discriminant △ = (a11 + A22)

Given that a and B are matrices of order 4, if AB + 2B = 0, R (b) = 2 and determinant a + e = a-2e = 0, the eigenvalue of a is obtained

It is known that a and B are matrices of order 4
AB+2B=0  ===》（A+2E）B=0
&R (b) = 2 = = = (R (a + 2e) is less than or equal to 2, = = (a) has eigenvalue - 2 and multiplicity is not less than 2
The determinant a + e = a-2e = 0, = = = "a has eigenvalues of 0,2
===The eigenvalues of a are 0,2, - 2, - 2

It is known that a is a matrix of order n, the square of a is a, and the rank (a) is R. it is proved that a can be similar diagonalized, and the similar diagonal form and determinant | a + e of a can be obtained|

I just answered the question
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The calculation method of 4-order determinant,
2 1/3 4 3
5 -1/2 1 -2
3 1/4 2 5
-4 1 0 5

C2 * 12
2 4 4 3
5 -6 1 -2
3 3 2 5
-4 12 0 5
Do r1-2r3, r3-2r1 in turn
-4 -2 0 -7
5 -6 1 -2
-7 15 0 9
-4 12 0 5
Expand by column 3 (- 1)*
-4 -2 -7
-7 15 9
-4 12 5
r4-r1, r2-2r1
-4 -2 -7
1 19 23
0 14 12
r1+4r2
0 74 85
1 19 23
0 14 12
Expand (- 1) by column 1*
Determinant = (74 * 12-85 * 14) / 12 = - 151 / 6
This determinant is troublesome, and there is no more convenient way to deal with it
Please accept if you are satisfied^_^