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1. Let $V = \R^2$, and let $v = (3,2)$ in the Cartesian basis of $\R^2$. Now, we choose another basis as follows $$ e_1 = (2,1), \quad e_2 = (0,1) $$. Expand $v$ in terms of $e_1, e_2$.
2. Let $V = \{a + bt + ct^2 \mid a, b, c \in \R\} $ be the space of polynomials of degree at most 2. Let $f_1(t), f_2(t), f_3(t)$ be three elements in $V$, given as follows $$ f_1(2) = 1, \quad f_1(3) = 0, \quad f_1(5) = 0 $$ $$ f_2(2) = 0, \quad f_2(3) = 1, \quad f_2(5) = 0 $$ $$ f_3(2) = 0, \quad f_3(3) = 0, \quad f_3(5) = 1 $$ Then
3. Consider the new coordiante $(u,v)$ on $\R^2$, related to $(x,y)$ by $$ x = \cosh(u) \cos(v), \quad y = \sinh(u) \sin(v) $$
You may use the formula $$ grad(f) = g^{ij}\, \d_i f \,\d_j $$ and $$ div(V^i \d_i) = \frac{1}{\sqrt{g}} \d_i(\sqrt{g} V^i) $$ where we used Einstein summation convention, and $\d_i = \frac{\d}{\d u_i}$, $u_1 = u, u_2 = v$.
1. Draw the picture to see. You should draw a parallogram with $v$ along the diagonal, and two sides parallel to $e_1,e_2$.
2. Since $V$ is 3 dim, suffice to show that $f_i$ are linearly independent. That is, if we have $$ 0 = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) $$ for all $t$, then we need to show $c_i=0$. Plug in $t=2,3,5$ to conclude this is indeed so.
Same thing for the next problem, suppose we try to find the expansion $$ f(t) = c_1 f_1(t) + c_2 f_2(t) + c_3 f_3(t) $$ then we can get $c_i$ by plug-in $t=2,3,5$.
You may worry about: what about $t$ equal to other values? Well, since we have an equation about degree two polynomials, if it holds at 3 points, then it holds for all $t$.
3. This is good place to introduce another trick, the relation about 2-dim orthogonal coordinate and holomorphic functions.
Suppose $z = f(w)$ is a holomorphic function, if we write $z = x+ i y$, $w = u + iv$, we get Cauchy-Riemmann equation $$ \begin{cases} \frac{\d x}{\d u} = \frac{\d y}{\d v} \cr \frac{\d x}{\d v} = - \frac{\d y}{\d u} \end{cases} $$ Hence, we have $$ g(\frac{\d }{\d u} ,\frac{\d }{\d u}) $$