Singular Value Decomposition: A = UΣVᵀ

Diagonalisation asked for an eigenvector basis of $A$ itself — impossible when $A$ isn't square, and unreliable even when it is (a defective matrix has no such basis). This note previews the fix: instead of one basis serving both the domain and codomain of $A$, use two — an orthonormal basis $V$ for the domain and a different orthonormal basis $U$ for the codomain — chosen so that $A$ maps each domain basis vector to a scalar multiple of the matching codomain basis vector. This is the singular value decomposition, and it exists for every matrix, with no exceptions.

Learning objectives. By the end of this note you should be able to: