There is an interesting correspondence among the quadratic form of a symmetric linear transformation \(T:V \mapsto V\)on 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