# Connection between a symmetric linear transformation and the unit sphere

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