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Matrix A is diagonalizable if A = PDP⁻¹ where D is diagonal and P's columns are eigenvectors of A. Diagonalization simplifies computation: Aⁿ = PDⁿP⁻¹. An n × n matrix is diagonalizable iff it has n linearly independent eigenvectors, guaranteed if all eigenvalues are distinct.
From your work with eigenvalues and eigenvectors, you know that a vector v is an eigenvector of matrix A if Av = λv — multiplying by A just scales v by the eigenvalue λ. Diagonalization asks: what if *all* our basis vectors were eigenvectors? Then the matrix would be incredibly easy to work with, because it would just scale each coordinate independently.
Here is the construction. Suppose A is n × n and you can find n linearly independent eigenvectors v₁, v₂, …, vₙ with eigenvalues λ₁, λ₂, …, λₙ. Build matrix P by placing these eigenvectors as columns: P = [v₁ | v₂ | … | vₙ]. Build diagonal matrix D by placing the corresponding eigenvalues on the diagonal: D = diag(λ₁, λ₂, …, λₙ). Then A = PDP⁻¹. To verify: PDP⁻¹vᵢ = PD(eᵢ) = P(λᵢeᵢ) = λᵢvᵢ = Avᵢ, where eᵢ are standard basis vectors. The matrix P is the change-of-basis matrix from your basis of eigenvectors back to the standard basis.
The payoff is computation. Squaring A naively requires multiplying two full matrices. But A² = (PDP⁻¹)(PDP⁻¹) = PD²P⁻¹, and D² just squares the diagonal entries. More generally, Aⁿ = PDⁿP⁻¹, and Dⁿ is trivial: just raise each diagonal entry to the nth power. This turns computing the 100th power of a matrix from a nightmare into three multiplications. The same trick extends to matrix exponentials: eAt = PeDtP⁻¹, which is the key to solving systems of differential equations.
The condition for diagonalizability is that A has n linearly independent eigenvectors — enough to form a full basis. This is guaranteed when all n eigenvalues are *distinct* (no repeated eigenvalues), because eigenvectors from distinct eigenvalues are always linearly independent. When eigenvalues repeat, diagonalizability is not guaranteed — you need to check whether the geometric multiplicity (dimension of the eigenspace) equals the algebraic multiplicity (number of times the eigenvalue appears as a root of the characteristic polynomial) for each eigenvalue. If any eigenspace is "too small," A is not diagonalizable — though it may still have a Jordan normal form, the closest diagonal-like structure available.