This applet visualizes a relationship between a basis in a plane and the dual basis.[br][br]Given a basis [math]\left\{\vec{v_1},\vec{v_2}\right\}[/math] in [math]\mathbb{R}^2[/math], the dual basis consists of two coordinate functionals [math]F_i:\mathbb{R}^2\to\mathbb{R}[/math], defined as [math]F_1\left(s\vec{v_1}+t\vec{v_2}\right)=s[/math] and [math]F_2\left(s\vec{v_1}+t\vec{v_2}\right)=t[/math]. [br][br]The red coordinate grid consists of level curves  [math]F_i\left(\vec{x}\right)=const[/math]. [br]The nullspace  Nul [math]F_i[/math] is just the line [math]F_i\left(\vec{x}\right)=0[/math].[br][br]Denote by "[math]\cdot[/math]" the dot product with respect to the standard basis [math]\left\{\vec{e_1},\vec{e_2}\right\}[/math] (not shown).[br][br]Suppose a linear functional  [math]F[/math] is written in the standard basis as [math]F\left(x\vec{e_1}+y\vec{e_2}\right)=ax+by[/math]. [br]Its  [i]gradient vector[/i]  [math]\nabla F=a\vec{e_1}+b\vec{e_2}[/math]  has the property [math]F\left(\vec{x}\right)=\nabla F\cdot\vec{x}[/math]. [br]Note that  [math]\nabla F[/math] is orthogonal to  Nul [math]F[/math].[br][br]Thus, we can visualize the dual basis [math]\left\{F_1,F_2\right\}[/math] as a pair of vectors  [math]\nabla F_1[/math] and [math]\nabla F_2[/math].

[b]Tasks[/b][br]1. Drag the endpoints of the black vectors and observe how the dual basis changes.[br]2. Position [math]\vec{v_1}[/math] in such way that [math]\nabla F_1[/math] would have length 1. [br]HINT: In that case, the projection of [math]\vec{v_1}[/math] onto [math]\nabla F_1[/math] will be [math]\nabla F_1[/math] itself.  [br]3. Suppose  [math]\vec{v_1}=1\vec{e_1}+0\vec{e_2}[/math] and [math]\vec{v_2}=2\vec{e_1}+1\vec{e_2}[/math]. What are the coordinates of the blue vectors then? [br]HINTS: Write the standard basis vectors as linear combinations of [math]\left\{\vec{v_1},\vec{v_2}\right\}[/math] first. Alternatively, the blue vectors [math]\vec{u_i}=\nabla F_i[/math] must satisfy to the  4 equations [math]\vec{v_i}\cdot\vec{u_j}=F_j\left(\vec{v_i}\right)[/math]. In matrix form, [math]\left[\vec{v_1}\:\vec{v_2}\right]^T\left[\vec{u_1}\:\vec{u_2}\right]=I_2[/math].