Want to confirm a problem, the number of linearly independent eigenvectors = the number of different eigenvalues plus the multiplicity of multiple roots = the rank of the matrix, right?

Want to confirm a problem, the number of linearly independent eigenvectors = the number of different eigenvalues plus the multiplicity of multiple roots = the rank of the matrix, right?

There's nothing right
such as
2 0 0
0 1 1
0 0 1
The number of linearly independent eigenvectors = 2
The number of different eigenvalues plus the multiplicity of multiple roots = 2 + 2 = 4
Rank of matrix = 3

If the eigenvalues of matrix A and B are the same (including multiplicity), then why is it wrong that the two matrices are similar?
A. If the eigenvalues of matrix A and B are the same (including multiplicity), then the two matrices are similar
B. If the eigenvalues of a and B are the same (including multiplicity), then the two matrices are congruent
C. If real matrices A = a transpose, B = B transpose and have the same eigenvalues (including multiplicity), then a and B are congruent
D. If a and B are equivalent, then a and B must be similar

To negate a proposition, we only need a counterexample
such as
A =
1 0
0 1
B =
1 1
0 1

In linear algebra, are all eigenvalues of two matrices the same, including multiplicity, and their characteristic polynomials are the same

Oh, yes
The characteristic polynomial is the product (λ - λ I)