This is a visually large applet, best viewed on a computer.[br][br]On the left panel, you can control two vectors, [math]u[/math] (purple) and [math]v[/math] (orange). You can also toggle "arrows" vs. "functions", but leave that alone for now.[br]On the right panel, you can drag a point to control a pair of coefficients.[br][br]The coefficients you choose on the right panel are also printed near the top of the left panel. They are used to make a LINEAR COMBINATION of [math]u[/math] and [math]v[/math], with the result being called [math]w[/math] (blue). Some algebraic steps are shown so that you can see how your coefficients affect the calculation and the result.[br][br]Think of [math]u[/math] and [math]v[/math] as ingredients that make an unnamed function whose[br][list][*]input is the coordinate vector on the right panel, and whose [/*][*]output is [math]w[/math] on the left panel.[/*][/list][br]Notice that if you choose the coordinate vector [math]\left(0,0\right)[/math], we get [math]w=\left(0,0\right)[/math], no matter where you set [math]u[/math] and [math]v[/math].[br][br]You can click a checkbox to show/hide a lattice. All this means is, a selection of 49 integer pairs (in a 7x7 grid) are automatically tried in place of the coefficients, and all 49 results are plotted. This might give you a sense of connection between the coordinate grid in the right panel and a parallelogram grid on the left panel. You might get the feeling that linear combinations of [math]u[/math] and [math]v[/math] can be used to reach any point in the plane. You'd be right...usually.[br][br]If [math]u[/math] and [math]v[/math] line up with the origin, we call them "linearly dependent". Two important things happen:[br][list=1][*]Their lattice collapses to a line or even just a point--can you get that to happen? Either way, the output [math]w[/math] is trapped on this line (or point) and can't reach other parts of the plane.[/*][*]It becomes possible to reach the origin through a non-trivial linear combination. That is, there's a place [i]other than[/i] [math]\left(0,0\right)[/math] where you can put the coefficient vector (right panel) so that [math]w=\left(0,0\right)[/math]. Now when you add that funny coefficient vector to any other, [math]w[/math] doesn't move. In other words: anywhere [math]w[/math] can be, it can be in infinitely many ways.[/*][/list][br]If [math]u[/math] and [math]v[/math] DON'T line up with the origin, we call them "linearly independent". Several important things happen:[br][list=1][*]Their lattice does not collapse. (Using linear combinations of them, you can get anywhere in the plane. We say [math]u[/math] and [math]v[/math] "span the plane".) The unnamed function we've been talking about is "onto the plane" or "surjective".[/*][*]There's no input coefficient vector other than [math]\left(0,0\right)[/math] that produces the output [math]w=\left(0,0\right)[/math]. This property is contagious: [i]no point in the plane[/i] is reachable in multiple ways. Different input coefficient vectors always produce different output vectors. The unnamed function we've been talking about is "one to one" or "injective".[/*][*]Since the unnamed function is one to one and onto, it is invertible.[/*][*]In this case, the set [math]\left\{u,v\right\}[/math] is linearly independent and spans the plane. We call it a [i]basis[/i] for the plane.[/*][/list][br]________________________________________________________________________________________________[br][br]You may want to experiment with the checkboxes on the right panel. How does locking one of the input coordinates restrict the movement of the output?[br][br]Note:[br]If each coefficient is either unrestricted or locked at 0, the set of points [math]w[/math] can reach is called a [b]vector space[/b].[br][br]________________________________________________________________________________________________[br][br]Although most people are introduced to vector spaces with geometric examples (often with vectors represented as arrows), the concept of a vector space is much broader than this.[br][br]You can flip the switch from Arrows to Functions to explore the span of two functions, [math]f[/math] and [math]g[/math]. Some recommendations:[br][list=1][*]Use the input coefficient vector [math]\left(0,0\right)[/math]. The output function [math]h[/math] should be the Zero Function, the most boring function of all. Can you produce the Zero Function by using any [i]other[/i] input coefficient vector? This should be impossible (can you prove it?) with the default functions, which are linearly independent. Try again with the functions [math]f\left(x\right)=x+1[/math] and [math]g\left(x\right)=2x+2[/math]. What do these "linearly dependent" functions have in common, either algebraically or in their graphs?[/*][*]Lock one coordinate or the other at 0 to see multiples of just one function.[/*][*]The space [math]P_1[/math] of polynomials of degree at most 1 is a 2-dimensional vector space. That means, if [math]f[/math] and [math]g[/math] are linearly independent polynomials of degree at most 1, they'll span [math]P_1[/math]. (Initially, they are, but maybe you've changed that.) Are you able to produce [b]every[/b] polynomial function of the form [math]y=ax+b[/math] (their graphs are non-vertical lines) by choosing coefficients?[/*][*]You might be interested in [math]P_2[/math], the space of polynomials of degree at most 2. That space has dimension 3, so we can't span it with just two functions; however, you might set [math]f\left(x\right)=x[/math] (or some other simple degree 1 polynomial function) and [math]g\left(x\right)=x^2[/math] (or some other simple degree 2 polynomial function). Now play with the input coefficient vector and see what polynomials you can produce.[/*][*]Play with [math]f\left(x\right)=\cos\left(x\right)[/math] and [math]g\left(x\right)=\sin\left(x\right)[/math]. Notice how linear combinations of these two functions all have basically the same shape.[/*][*]Play with [math]f\left(x\right)=\cos\left(x\right)[/math] and [math]g\left(x\right)=\sin\left(2x\right)[/math]. Why is this situation so different from #5?[br][/*][/list]