The existence matrix has a double root eigenvalue, which only corresponds to the eigenvalue of a linearly independent eigenvector

The existence matrix has a double root eigenvalue, which only corresponds to the eigenvalue of a linearly independent eigenvector

yes , we have.
Such as a=
1 1
0 1
1 is the double eigenvalue of A
Because R (A-E) = 1
So there are only 2 - R (a - E) = 1 linearly independent eigenvectors belonging to eigenvalue 1

If a matrix can be diagonalized in linear algebra, then if one of its eigenvalues is a k-fold eigenvalue, it corresponds to K linearly independent eigenvectors
Linear algebra problem
If a matrix is diagonalizable, then if one of its eigenvalues is a k-fold eigenvalue, then it corresponds to K linearly independent eigenvectors

Yes, and all linearly independent eigenvectors of all different eigenvalues can be used as a basis of linear space, under which the matrix can be reduced to diagonal matrix

Let a = (α, - γ 2, γ 3, - γ 4), B = (β, γ 2, - γ 3, γ 4), where α, β, γ 2, γ 3, γ 4 are all 4-dimensional sequence vectors
If we know that determinant a = 4 and determinant B = 1, what is determinant A-B equal to?

If the matrix C is a square matrix of order n, then | KC | = k ^ n * | C | 1) | - B | = | B | = 1; 2) - B = (β, - γ 2, γ 3, - γ 4) is the same as the last three columns of A. 3) A-B = (α - β, - 2 * γ 2, 2 * γ 3, - 2 γ 4), except that the last three columns are all multiplied by