matrices can be factorized in this way. For example, the defective matrix (which is a shear matrix) cannot be diagonalized. The eigenvectors are usually normalized, but they need not be. A non-normalized set of eigenvectors, can also be used as the columns of . That can be understood by noting that the magnitude of the eigenvectors in gets canceled in the decomposition by the presence of . If one of the eigenvalues has more than one linearly independent eigenvectors (that is, the geometric multiplicity of is greater than 1), then these eigenvectors for this eigenvalue can be chosen to be mutually orthogonal; however, if two eigenvectors belong to two different eigenvalues, it may be impossible for them to be orthogonal to each other (see Example below). One special case is that if is a normal matrix, then by the spectral theorem, it's always possible to diagonalize in an orthonormal basis . The decomposition can be derived from the fundamental property of eigenvectors: The linearly independent eigenvectors with nonzero eigenvalues form a basis (not necessarily orthonormal) for all possible products , for , which is the same as the image (or range) of the corresponding matrix transformation, and also the column space of the matrix . The number of linearly independent eigenvectors with nonzero eigenvalues is equal to the rank of the matrix , and also the dimension of the image (or range) of the corresponding matrix transformation, as well as its column space. The linearly independent eigenvectors with an eigenvalue of zero form a basis (which can be chosen to be orthonormal) for the null space (also known as the kernel) of the matrix transformation . Example The 2 × 2 real matrix may be decomposed into a diagonal matrix through multiplication of a non-singular matrix Then for some real diagonal matrix . Multiplying both sides of the equation on the left by : The above equation can be decomposed into two simultaneous equations: Factoring out the eigenvalues and : Letting this gives us two vector equations: And can be represented by a single vector equation involving two solutions as eigenvalues: where represents the two eigenvalues and , and represents the vectors and . Shifting to the left hand side and factoring out Since is non-singular, it is essential that is nonzero. Therefore, Thus giving us the solutions of the eigenvalues for the matrix as or , and the resulting