Having the dual base $\beta^*=$ {$\phi_1,\phi_2,\phi_3$} where $\phi_1(x,y,z)=x-y$, which is $\beta=$ { $v_1,v_2,v_3$ }?

Question

Let's suppose that I have $\beta=$ { $v_1,v_2,v_3$ } a base of $\mathbb{R^3}$ and its dual base $\beta^*=$ {$\phi_1,\phi_2,\phi_3$} where $\phi_1(x,y,z)=x-y$.

Which is a base $\beta=$ { $v_1,v_2,v_3$ }$\space $? (As $\beta$ is not uniquely determined, there are more that one base, so I have to give one of them)I mean, I have to find which are the values of $v_1,v_2$ and $v_3$.

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What I have done is:

Defining $v_1=(a,b,c), v_2=(d,e,f)$ and $v_3=(g,h,i)$

$(x,y,z)=\lambda_1 (a,b,c)+\lambda_2(d,e,f)+\lambda_3(g,h,i)$

And from here I tried to solve the system giving aleatory values to $a,b,c,d,e,f,g,h$ and $i$ (knowing that $\lambda_1=x-y$). But I didn't reach to any solid conclusion... How can I solve this problem?

Answer 1

The only thing we can deduce from the given information is that there are numbers $a,b,c,d,e,f$ for which $$ v_1 = (a+1,a,b), \quad v_2 = (c,c,d), \quad v_3 = (e,e,f), $$ and the vectors $v_1,v_2,v_3$ are linearly independent. Note that these vectors will be independent for "most" (or "randomly selected") numbers $a,b,c,d,e,f$.

Answer 2

Okay, so we know that we have to solve the set of equations (which define the dual basis) \begin{align} \phi_1(v_1)=1 \\ \phi_1(v_2)=0\\ \phi(v_3)=0, \end{align} subject to $v_2$ and $v_3$ being linearly independent (note that, by linearity, $v_1$ will automatically be linearly independent of $\{v_2,v_3\}$).

So let's start with the latter two. That is, solve

$$ \begin{pmatrix} 1 &-1 & 0\end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix}=0 $$

We can see that this is solved by any $v\in \{(t_1,t_1,t_2)|\; t_2\in \mathbb{R}\},$ which is clearly a two-dimensional subspace of $\mathbb{R}^3$, so this must also be the complete set of solutions (since $\phi_1$ is not the $0$ functional).

Now, a "natural" choice of basis for this subspace is $v_2=(1,1,0), \; v_3=(0,0,1)$ and furthermore, this can clearly be extended to a basis of $\mathbb{R}^3$ by choosing $v'_1=(1,-1,0)$. Now, $\phi_1(v_1')=2$, so setting $v_1=\frac{1}{2} v_1'$ gives one basis solving the problem.

However, in general, $\beta$ is any set of vectors of the form $\{v_1+\alpha_1 v_2+\beta_1 v_3,\alpha_2 v_2+\beta_2 v_3, \alpha_3 v_2+\beta_3 v_3\}$ where the $\alpha_j$ and $\beta_j$ are scalars subject to the condition that $$ \begin{pmatrix} \alpha_2 & \beta_2 \\ \alpha_3 & \beta_3 \end{pmatrix} $$ is invertible.

Answer 3

Write $\phi_i(x,y,z)=a_{i1}x+a_{i2}y+a_{i3}z$, and form $M=(a_{ij})$ (so row i gives the coefficients of $\phi_i$).

Note that the first row is $(1 \ -1 \ \ 0)$. Now let $P$ be the matrix with columns $v_1,v_2,v_3$. Then, you can easily see that $MP=(\phi_i(v_j))$. By definition of a dual basis, you want $MP=I_3$. So, what you want is to choose the second and third row of $M$ such that $M$ is invertible. Then the basis you seek will be given by the columns of $M^{-1}$.

I advise you to take $M$ as simple as possible (eg triangular). For example $M=\pmatrix{1 & -1 & 0 \cr 0 & 1 & 0\cr 0 & 0 & 1}$ works ( This means that you define $\phi_2,\phi_3$ as $\phi_2(x,y,z)=y, \phi_3(x,y,z)=z$). Its inverse is $M=\pmatrix{1 & 1 & 0 \cr 0 & 1 & 0\cr 0 & 0 & 1}$ .

With this example, you get $v_1=(1,0,0),v_2=(1,1,0),v_3=(0,0,1)$.