is presented immediately below only once to reduce repetition. It also includes the easy part of the proof the Equality Characterization given above; that is, it proves that if and are linearly dependent then By definition, and are linearly dependent if and only if one is a scalar multiple of the other. If where is some scalar then which shows that equality holds in the . The case where for some scalar is very similar, with the main difference between the complex conjugation of If at least one of and is the zero vector then and are necessarily linearly dependent (just scalar multiply the non-zero vector by the number to get the zero vector; for example, if then let so that ), which proves the converse of this characterization in this special case; that is, this shows that if at least one of and is then the Equality Characterization holds. If which happens if and only if then and so that in particular, the Cauchy-Schwarz inequality holds because both sides of it are The proof in the case of is identical. Consequently, the Cauchy-Schwarz inequality only needs to be proven only for non-zero vectors and also only the non-trivial direction of the Equality Characterization must be shown. Proof 1 The special case of was proven above so it is henceforth assumed that The Cauchy–Schwarz equality (and the rest of the theorem) is an almost immediate corollary of the following : Equality is readily verified by elementarily expanding (via the definition of the norm) and then simplifying: This expansion does not require to be non-zero; however, must be non-zero in order to divide both sides by and to deduce the Cauchy-Schwarz inequality from it. Swapping and gives rise to: and thus Proof 2 The special case of was proven above so it is henceforth assumed that Let It follows from the linearity of the inner product in its first argument that: Therefore, is a vector orthogonal to the vector (Indeed, is the projection of onto the plane orthogonal to ) We can thus apply the Pythagorean theorem to which gives The Cauchy–Schwarz inequality follows by multiplying by and then taking the square root. Moreover, if the relation in the above expression is actually an equality, then and hence the definition of then establishes a relation of linear dependence between and The converse was proved at the beginning of this section, so