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The characteristic polynomial of A is det(A − λI), whose roots are eigenvalues. A matrix is diagonalizable if its eigenvectors form a complete basis. Diagonalizable matrices A satisfy A = PDP-1, where D is diagonal and P is the eigenvector matrix. Similar matrices share eigenvalues and determinant.
From your study of eigenvalues and eigenvectors, you know that λ is an eigenvalue of A when Av = λv for some nonzero vector v — equivalently, when (A − λI)v = 0 has a nontrivial solution. This happens precisely when A − λI is singular, meaning det(A − λI) = 0. The characteristic polynomial is simply this determinant written as a function of λ: p(λ) = det(A − λI). Its roots are exactly the eigenvalues of A.
For an n×n matrix, p(λ) is a polynomial of degree n. For a 2×2 matrix [[a,b],[c,d]], p(λ) = (a−λ)(d−λ) − bc = λ² − (a+d)λ + (ad−bc). Notice what appears: the coefficient of λⁿ⁻¹ is always −tr(A) (the trace, sum of diagonal entries), and the constant term is always det(A). This means you can read off two important eigenvalue facts without solving anything: the sum of all eigenvalues equals tr(A), and the product of all eigenvalues equals det(A). The characteristic polynomial encodes these global properties as coefficients.
Diagonalization asks: can we choose a basis of eigenvectors? If A has n linearly independent eigenvectors v₁, …, vₙ with eigenvalues λ₁, …, λₙ, form the matrix P whose columns are these vectors and D = diag(λ₁, …, λₙ). Then AP = PD (multiply out: AP's i-th column is Avᵢ = λᵢvᵢ = PD's i-th column), so A = PDP⁻¹. This decomposition is powerful because Aᵏ = PDᵏP⁻¹, and raising a diagonal matrix to a power is trivial: just raise each diagonal entry to that power.
Not every matrix is diagonalizable. The obstruction arises when the geometric multiplicity (dimension of the eigenspace) is less than the algebraic multiplicity (multiplicity as a root of the characteristic polynomial) for some eigenvalue. For instance, the matrix [[2,1],[0,2]] has characteristic polynomial (λ−2)², so λ = 2 has algebraic multiplicity 2, but the eigenspace is only 1-dimensional — there is only one independent eigenvector. No basis of eigenvectors exists, so the matrix is not diagonalizable. Similar matrices (A and B = PAP⁻¹ for invertible P) always share the same characteristic polynomial and therefore the same eigenvalues, trace, and determinant — these are similarity invariants that capture intrinsic properties of the linear transformation regardless of which basis you choose.