I've just been reading on the differential geometry concept of the derivative operator $D$, after years of thinking mostly in terms of differentiation on $\mathbb R^n$ rather than on general manifolds.
Now I am wondering: in the space $\mathbb R^n$, we can say that a function $f$ has a minimum at point $p$, if $Df_p=0$, and if $D^2f_p$ as a matrix is positive definite.
The latter is defined by the criterion that for any $\mathbb R^n$ column vector $x$, we have $x^T(D^2f_p)x\geq 0$.
However, this definition relies on a choice of basis so far as I understand, since $x^T$ is only defined for a choice of basis. How do we define this property without choosing a basis?
It is true that the expression (how it looks) for a quadratic form depends on the relative basis that you choose to work with. Say $A:=D^2f_p$ is the matrix representing the quadratic form $x^TAx$ relative to some basis $\{e_1,...,e_n\}$ (wlog say the space is $n$-dimensional Euclidean). Then the entries $(a_{ij})_{i,j}=A$ are given by $a_{ij}=e^T_iAe_j$ for $i,j\in\{1,2,...,n\}$. Suppose we decide to work with some other basis $\{f_1,f_2,...,f_n\}$ and let $f_i=b_{1i}e_1+...+b_{ni}e_n$ for every $i\in\{1,2,...,n\}$. Let $B:=(b_{ij})_{i,j}$ be the matrix representing the transformation of basis from the first to the latter one. It can be shown (you can try this as an exercise) that now the quatratic form can be represented by the matrix $C=B^TAB$ relative to the new basis $\{f_1,...,f_n\}$. Moreover for all $x$ we have $$x^TCx=x^TB^TABx=(Bx)^TA(Bx)=y^TAy\geqslant 0$$ i.e. the positive definite property of a quadratic form is independent of the basis you work with.