Question

(9) Let E R" and let A E L(R"). Define a map f : R" -> R" by f (x) A,)v. Here (is the Euclidean inner product (a) Prove that f is a C1 map and find f'(x) (b) Prove that there exist two that f U V is a bijection on R" neighborhoods of the origin in R", U and V, such

Answer 1

(a) The map f is a polynomial in $x_i$ where $x = (x_i)_{n \times 1}$ . Hence it is a $C^1$ function.

To find the derivative of f we look at partially differentiating it with respect to the variables. That is

$f^{\prime}(x) = \Bigg( \dfrac{\partial f}{ \partial x_1} , \dfrac{\partial f}{ \partial x_2} , \dots , \dfrac{\partial f}{ \partial x_n} \Bigg)$

Now  $\dfrac{\partial f}{ \partial x_i} = \dfrac{\partial}{ \partial x_i} (x+\langle Ax,x \rangle v)$

$= \dfrac{\partial}{ \partial x_i} (x) + \dfrac{\partial}{ \partial x_i} (\langle Ax,x \rangle v) = \dfrac{\partial}{ \partial x_i} (x) + \langle\dfrac{\partial}{ \partial x_i} Ax, x \rangle + \langle Ax, \dfrac{\partial}{ \partial x_i} x \rangle$

$= e_i + \langle \begin{pmatrix} a_{1i} \\ .\\ .\\ .\\ a_{ni} \end{pmatrix} , x \rangle v + \langle Ax, e_i \rangle v = e_i + \sum_{k=1}^n x_k (a_{ki} + a_{ik})) v$

Where $e_i$ is a column matrix with 1 in the ith place and 0 else where and

This is the derivative of f at x.

(b) At x=0, and the derivative $f^{\prime}(0) = \Big( e_1, e_2, \dots , e_n \Big) = I$ . is invertible hence by inverse function theorem, there are neighborhoods and of the origin such that $f : U \rightarrow V$ is invertible. Therefore a bijection.