There is an interesting correspondence among the quadratic form of a symmetric linear transformation on a real Euclidean space,the extreme values of the sphere and the eigenvectors of

Let be the quadratic form associated with a symmetric linar transformation which maps into itself,then the set of elements in V satisfying is called the unit sphere of

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

**Theorem**

Let be the quadratic form associated with a symmetric linear transformation on a real Euclidean space where is finite dimensional.Assume that there exists an element in such that lies in the unit sphere and is an extremum with .Then, is an eigenvector for .

It quickly follows that is the corresponding eigenvalue.

**Proof**

Assume first that attains its maximum at

This means that

(1)

Let .So, since .

The inequality can be changed to when (2)

This result can be extended to all in 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:

Define .

This means that . (3)

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

It is quite clear that for all in .Also, is symmetric.Therefore if then (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, -which proves the result