Maths WorkshopMaths problems with full solutions
Exercise 557Linear algebra › Quadratic formsHigh schoolAdded 3 Nov 2026

Polar form of a quadratic form

#polar form#symmetric bilinear form#quadratic form

Statement

Given the quadratic form Q(x,y)=x2+4xy+3y2Q(x,y)=x^2+4xy+3y^2, compute its polar form φ(u,v)\varphi(u,v) using the polarization identity
φ(u,v)=12(Q(u+v)−Q(u)−Q(v)).\varphi(u,v)=\frac12\left(Q(u+v)-Q(u)-Q(v)\right).

Hint

Compute Q(u+v)Q(u+v), Q(u)Q(u), Q(v)Q(v), then combine.

Show solutionHide solution

Solution

Notation. u=(x1,y1)u=(x_1,y_1), v=(x2,y2)v=(x_2,y_2).

Compute.
Q(u+v)=(x1+x2)2+4(x1+x2)(y1+y2)+3(y1+y2)2.Q(u+v)=(x_1+x_2)^2+4(x_1+x_2)(y_1+y_2)+3(y_1+y_2)^2.
Expand:
=x12+2x1x2+x22+4x1y1+4x1y2+4x2y1+4x2y2+3y12+6y1y2+3y22.=x_1^2+2x_1x_2+x_2^2+4x_1y_1+4x_1y_2+4x_2y_1+4x_2y_2+3y_1^2+6y_1y_2+3y_2^2.

Subtract Q(u)+Q(v)Q(u)+Q(v). The pure terms cancel:
Q(u+v)−Q(u)−Q(v)=2x1x2+4x1y2+4x2y1+6y1y2.Q(u+v)-Q(u)-Q(v)=2x_1x_2+4x_1y_2+4x_2y_1+6y_1y_2.

Divide by 2.
φ(u,v)=x1x2+2x1y2+2x2y1+3y1y2.\varphi(u,v)=x_1x_2+2x_1y_2+2x_2y_1+3y_1y_2.

Check. φ(u,u)=x12+4x1y1+3y12=Q(u)\varphi(u,u)=x_1^2+4x_1y_1+3y_1^2=Q(u). Correct.

Conclusion. The polar form is φ(u,v)=x1x2+2x1y2+2x2y1+3y1y2\varphi(u,v)=x_1x_2+2x_1y_2+2x_2y_1+3y_1y_2.