Eigenvector

Diagonalization and eigendecomposition

Suppose the eigenvectors of A form a basis of ⁠${\displaystyle \mathbb {C} ^{n}}$⁠, or equivalently A has n linearly independent eigenvectors v₁, v₂, ..., v_n (with associated eigenvalues λ₁, λ₂, ..., λ_n). The eigenvectors need not be orthogonal to one another, and the eigenvalues need not be distinct. Define the square matrixQ whose columns are the n linearly independent eigenvectors of A, $${\displaystyle Q={\begin{bmatrix}\mathbf {v} _{1}&\mathbf {v} _{2}&\cdots &\mathbf {v} _{n}\end{bmatrix}}.}$$ Since each column of Q is an eigenvector of A, right multiplying A by Q scales each column of Q by its associated eigenvalue: $${\displaystyle AQ={\begin{bmatrix}\lambda _{1}\mathbf {v} _{1}&\lambda _{2}\mathbf {v} _{2}&\cdots &\lambda _{n}\mathbf {v} _{n}\end{bmatrix}}.}$$