Symmetrical endomorphisms and quadratic forms (This last part of my linear algebra course is causing me quite a bit of headaches, so please be patient)
Let $V$ be a vector space over the real field, and we'll indicate with $(\cdot,\cdot)$ its scalar product (that we'll assume defined positive). If $T:V\rightarrow\mathbb{R}$ is a symmetrical endomorphism, the function $\phi:V\rightarrow\mathbb{R}$ defined by $\phi(v)=(T(v),v)$ is a quadratic form.
Ok, so this is a theorem on my linear algebra textbook. ...What? Wasn't a quadratic form one of the form $\phi(v)=(v,v)$? What does it mean to arbitrarily put $T(v)$ there?
Thank you in advance.
 A: Choose an orthonormal basis $\;\{v_1,...,v_n\}\;$ of $\;V\;$ consisting of eigenvectors of $\;T\;$ , say $\;Tv_i=\lambda iv_i\;,\;\;\lambda_i\in\Bbb R\;$ (why does there exist such a basis?), and express $\;v=\sum_{k=1}^na_kv_k\in V\;$ , then
$$Tv=\sum_{k=1}^na_kTv_k=\sum_{k=1}^na_k\lambda_kv_k\implies $$
$$\langle Tv,v\rangle=\sum_{k,j=1}^na_ka_j\langle Tv_k,v_j\rangle=\sum_{k,j=1}^na_ka_j\lambda_k\delta_{k,j}=\sum_{k=1}^na_k^2\lambda _k$$
and there you go....
A: A quadratic form on a real vector space $V$ is a map $Q:V\to\Bbb R$ such that there exists some symmetrical bilinear form $\phi:V\times V\to \Bbb R$ with the property that for all vectors $X\in V$, 
$$Q(X)=\phi(X,X)$$
This symmetrical bilinear form doesn't have to be a scalar product, nor the scalar product with which you are currently working with.
To convince yourself that the theorem is true, you only need to convince yourself that the map
$$\phi:V\times V\to \Bbb R,\quad (X,Y)\mapsto\phi(X,Y)=\langle TX,Y\rangle$$
is indeed a symmetrical bilinear form. Bilinearity is obvious, and symmetry follows at once from the definition of a symmetric endomorphism, and symmetry of the underlying scalar product (which I renamed $\langle\cdot,\cdot\rangle$ to distinguish it from the parantheses of pairs of elements.)
