Connection between a symmetric linear transformation and the unit sphere

 

There is an interesting correspondence among the quadratic form of a symmetric linear transformation T:V \mapsto Von a real Euclidean space,the extreme values of the sphere and the eigenvectors of T

Let Q(x)=(T(x),x) be the quadratic form associated with a symmetric linar transformation which maps V into itself,then the set of elements u in V satisfying \langle u,u  \rangle=1 is called the unit sphere of V


 

It’s easy to ‘picture’ all of this so I’ll move on to the main theorem

Theorem

Let  Q(x)=\langle T(x),x \rangle be the quadratic form associated with a symmetric linear transformation T:V \mapsto V  on a real Euclidean space where V is finite dimensional.Assume that there exists an element u in V  such that Q(u) lies in the unit sphere and is an extremum with \langle u,u \rangle=1.Then,u is an eigenvector for T.

It quickly follows that Q(u) is the corresponding eigenvalue.

Proof

Assume first that Q attains its maximum at u

This means that Q(x) \leq Q(u)

\forall x | \langle x,x \rangle=1        (1)

Let Q(u)=\lambda .So, Q(u)=\lambda \langle x,x \rangle=\langle \lambda x,x \rangle since \langle x,x \rangle=1.

The inequality can be changed to \langle T(x),x \rangle \leq \langle \lambda x,x \rangle when \langle x,x \rangle=1.                                (2)

This result can be extended to all x in V by simply choosing x=ka where ||x||=k and ||a||=1(The proof is left to the reader)

The above inequality can be rewritten as:

\langle T(x)-\lambda x,x \rangle \leq 0

Define S(x)=\langle T(x)- \lambda x,x \rangle.

This means that \langle S(x),x \rangle \leq 0.     (3)

Also define Q'(x)=\langle S(x),x \rangle.From the inequalities (1),(2),it can be deduced that equality in (3) holds if x=u.

It is quite clear that Q'(x) \leq 0 for all x in V.Also, S is symmetric.Therefore if Q'(u)=0 then S(u)=0(this can be simply proved by taking x=a+tb,expanding Q'(x) through linearity and finding the extreme points by equating the derivative of the obtained quadratic polynomial to 0 )

Therefore, T(u)=\lambda u-which proves the result

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