- Put the eigenvectors side by side as the columns of V. Applying A to each column gives back that column times its eigenvalue: A V = V D. (D is diagonal, so multiplying it on the right scales column i by λᵢ.)
- Multiply both sides on the right by V⁻¹ — allowed because the eigenvectors are independent, so V is invertible: A = V D V⁻¹.
- The three factors are all easy to apply: two coordinate changes and a diagonal stretch. That is the whole idea of a change of basis.
A = [[2, 1], [1, 2]], V = (1/√2)·[[1, 1], [1, −1]], D = [[3, 0], [0, 1]]
V⁻¹ = V here (this V is its own inverse: V @ V = I):
V D V = ½·[[1, 1], [1, −1]] @ [[3, 0], [0, 1]] @ [[1, 1], [1, −1]]
= ½·[[3, 1], [3, −1]] @ [[1, 1], [1, −1]]
= ½·[[4, 2], [2, 4]] = [[2, 1], [1, 2]] ✓
in eigen-coordinates the same A is just diag(3, 1):
anything along [1, 1] gets ×3, anything along [1, −1] gets ×1.
A symmetric matrix has perpendicular eigenvectors, so its V is a rotation or reflection and V⁻¹ = Vᵀ — the friendliest case, and the one PCA lives in.