An important application is the representation of the scalar product of an interior product space by the associated induced norm . Conversely, one can ask whether a given norm is induced by a scalar product. This is exactly the case when the norm fulfills the parallelogram equation, the scalar product can then be determined from the square of the norm by means of polarization.
${\ displaystyle \ | v \ | = {\ sqrt {\ langle v, \, v \ rangle}}}$

The real case (symmetrical bilinear form)

Let there be a vector space over the body and a symmetric bilinear form , i.e. H.
${\ displaystyle V}$${\ displaystyle \ mathbb {R}}$${\ displaystyle \ alpha \ colon V \ times V \ to \ mathbb {R}}$

for all , .
${\ displaystyle v, v_ {1}, v_ {2}, w \ in V}$${\ displaystyle c \ in \ mathbb {R}}$

Its associated square shape is then defined by
${\ displaystyle q \ colon V \ to \ mathbb {R}}$

${\ displaystyle q (v): = \ alpha (v, v), \; v \ in V.}$

Conversely, the symmetrical bilinear shape is clearly determined by its square shape. This expresses the polarization formula: It applies
${\ displaystyle \ alpha}$

${\ displaystyle {\ begin {aligned} \ alpha (v, w) & = {\ frac {1} {2}} \ left (q (v + w) -q (v) -q (w) \ right) , \ quad v, w \ in V \\ & = {\ frac {1} {2}} \ left (q (v) + q (w) -q (vw) \ right), \ quad v, w \ in V \\ & = {\ frac {1} {4}} \ left (q (v + w) -q (vw) \ right), \ quad v, w \ in V. \ end {aligned}}}$

The following example shows that this does not apply to any (including not symmetrical) bilinear forms. With the help of the matrices

${\ displaystyle A: = {\ begin {pmatrix} 1 & 1 \\ - 1 & 1 \ end {pmatrix}} \ quad {\ text {and}} \ quad B: = {\ begin {pmatrix} 1 & 0 \\ 0 & 1 \ end { pmatrix}}}$

let the bilinear forms be given by
${\ displaystyle \ alpha, \ beta \ colon \ mathbb {R} ^ {2} \ times \ mathbb {R} ^ {2} \ to \ mathbb {R}}$

${\ displaystyle \ alpha (v, w): = v ^ {T} Aw \ quad {\ text {and}} \ quad \ beta (v, w): = v ^ {T} Bw, \ quad v, w \ in \ mathbb {R} ^ {2}.}$

Then and are different, but define the same square shape.
${\ displaystyle \ alpha}$${\ displaystyle \ beta}$

The complex case (sesquilinear form)

Let there be a vector space over the body and a (not necessarily Hermitian ) sesquilinear form . Its associated square shape is defined by, as in the real case
${\ displaystyle V}$${\ displaystyle \ mathbb {C}}$${\ displaystyle \ alpha \ colon V \ times V \ to \ mathbb {C}}$${\ displaystyle q \ colon V \ to \ mathbb {C}}$

${\ displaystyle q (v): = \ alpha (v, v), \; v \ in V.}$

A sesquilinear form is also clearly defined by its square shape. For sesquilinear forms the polarization formula is:

${\ displaystyle \ alpha (v, w) = {\ frac {1} {4}} \ left (q (v + w) -q (vw) \ right) - {\ frac {i} {4}} \ left (q (v + iw) -q (v-iw) \ right), \ quad v, w \ in V,}$

if is semilinear in the first argument and
${\ displaystyle \ alpha}$

${\ displaystyle \ alpha (v, w) = {\ frac {1} {4}} \ left (q (v + w) -q (vw) \ right) + {\ frac {i} {4}} \ left (q (v + iw) -q (v-iw) \ right), \ quad v, w \ in V,}$

if is semilinear in the second argument.
${\ displaystyle \ alpha}$