Final answer:
The matrix b with the characteristic polynomial given is diagonalizable because the geometric and algebraic multiplicities of its eigenvalues match, including for the repeated eigenvalue, due to the rank of b - I being 3.
Step-by-step explanation:
The given characteristic polynomial of the matrix b is −(λ − 1)2(λ − 2)(λ − 3)(λ − 4). This indicates that the eigenvalues of matrix b are 1, 2, 3, and 4, with 1 being a repeated eigenvalue (multiplicity of 2). A key property for a matrix to be diagonalizable is that it should have a sufficient number of linearly independent eigenvectors to span the vector space; specifically, for each eigenvalue, the number of linearly independent eigenvectors (geometric multiplicity) should equal to the multiplicity of the eigenvalue in the characteristic polynomial (algebraic multiplicity).
It is given that the rank of b − I (where I is the identity matrix) is 3. Considering the multiplicity of the eigenvalue λ=1 is 2, and the rank of b − I is 3, we can deduce that the nullity of b - I (via the Rank-Nullity Theorem) is 2 (since 5 (the size of the matrix) - 3 (the rank of b - I) = 2). This means we have 2 linearly independent eigenvectors associated with the eigenvalue λ=1.
Since the algebraic and geometric multiplicities match for the eigenvalue λ=1, and the other eigenvalues are distinct (ensuring their geometric and algebraic multiplicities will also match), matrix b is indeed diagonalizable.