Show R is a basis for Poly ; Find coordinate change matrix

Question

Given $Poly=\{ a_1 t^2 + a_2t + a_3\}$ ; and $B=\{t^2,t,1\}$ is a basis $Poly$.

a) Show that $R=\{t^2 +1 , t-2 , t+3\}$ is a basis for $Poly$

b) Find the coordinate change matrix $S$ from the basis $B$ to the basis $R$ . Now find $S^{-1}$, using S.

This is a seven part problem; I've worked on them kind of scattered; but these two I've been stuck on and are also holding up working on some of the other parts of the problem. These are Part's b and c ; Part a gave a vector $v$ (which was a quadratic) and asked to find the coordinates of it in respect to $R$ ; I don't see that information being related to these parts, but just FYI.

What confuses me if that everything I've read online and in my books gives the vectors in the basis to solve for when trying to do this.

So an example in a book (not my textbook) is:

Let $x= \begin{array}{lcr} \ 3 \\ \ {-2} \end{array} $

$B= \{u_1 , u_2 \} = \{ \begin{array}{lcr} \ 2 & 1 & \\ \ 7 & 4 & \\ \end{array} \}$ , set $ U=[u_1 u_2]$ (% What's U?) Find the change of basis matrix from $S$ to $B$.

I can type out this whole example from the book or take a picture of it to upload if you feel it would be beneficial to helping me, but as you can tell I'm not the greatest at typing matrices.

Like I said most of what I've looked up that has to do with change of basis matrices looks similar to this example; yet I'm failing to see how it relates to my problem.

For Part a I really don't know how to put $R$ in relation to $B$ ,other than by saying be know $a_1, a_2, a_3$ of $b$ and can therefor set the poly from $B=R$ ?

Like I said I'm pretty confused; and this is really putting on a damper on the rest of the assignment, any help would be appreciated.

Answer 1

For part a), as Poly has dimension $3$ and $R$ contains $3$ vectors, you have proved $R$ is a basis for Poly if you prove these $3$ vectors are linearly in,dependent.

So all you have to do is taking a linear relation among them: $$x(t^2+1)+y(t-2)+z(t+3)=0$$ and prove this implies $x=y=z=0$.

For part b) the change of coordinates matrix from basis $B$ to basis $R$ has for column vectors the coordinates of the vectors of the new basis $R$ in the old basis $B$.

To find $S^{-1}$, use row reductions until you obtain the unit matrix.In parallel, do the same reductions to the unit matrix. The final transformed matrix is $S^{-1}$.

Answer 2

In your first equation doesn't setting it equal to zero show that R is linearly independent ; but to show something is a basis we need to also show that it spans. So in order to show that there is only one solution to x,y, and z. We would set it equal to our poly to show there's one combination of x,y, and z for equaling the original a1, a2,a3. Please correct me if I'm wrong and let me know if I should show what I just described. (I will have to jump on a computer to do so)

Answer 3

Reversing the solution order presents a quick resolution

B

Column vectors of the matrix $\mathcal{S}$ are the coordinates of $\mathcal{R}$ in $\mathcal{B}$: $$ \mathcal{S} = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -2 & 3 \end{array} \right] $$ The inverse of transformation matrix is $$ \mathcal{S}^{-1} = \frac{\text{adj} \, \mathcal{S}}{\det \mathcal{S}}. $$ The determinant is $$\det \mathcal{S} = 5.$$ The adjugate matrix is the transpose of the cofactor matrix $\mathbf{C}$. The cofactor matrix is composed of the minors of $\mathcal{S}$ $$ \mathcal{C}_{row,col} = \left( -1 \right)^{row+col} M_{row,col}. $$ The inverse transformation is $$ \mathcal{S}^{-1} = \frac{1}{5} \left[ \begin{array}{rrr} 5 & 0 & 0 \\ 1 & 3 & -1 \\ -1 & 2 & 1 \end{array} \right] $$

Test

A quick check is valuable. The $\mathcal{R}$ polynomial $t^{2}+1$ is $\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]$ in the basis $\mathcal{B}$, and $\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]$ in the basis $\mathcal{R}.$ Therefore we expect $$ \mathcal{S}\, \left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]_{\mathcal{R}} = \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right]_{\mathcal{B}}. $$ This is just what we see: $$ \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -2 & 3 \end{array} \right] \left[ \begin{array}{rrr} 1 \\ 0 \\ 0 \end{array} \right]_{\mathcal{R}} = \left[ \begin{array}{rrr} 1 \\ 0 \\ 1 \end{array} \right]_{\mathcal{B}} $$ How would we express constants in the basis $\mathcal{B}$? $$ \frac{1}{5} \left[ \begin{array}{rrr} 5 & 0 & 0 \\ 1 & 3 & -1 \\ -1 & 2 & 1 \end{array} \right] \left[ \begin{array}{rrr} 0 \\ 0 \\ 1 \end{array} \right]_{\mathcal{B}} = \frac{1}{5} \left[ \begin{array}{rrr} 0 \\ -1 \\ 1 \end{array} \right]_{\mathcal{R}} $$ In functional form, $$ \frac{1}{5}\left( -(t-2) + (t+3)\right) = 1. $$

A

By inspection, the span of $\mathcal{S}$ is $\mathbb{R}^{3}$. This is verified by a row reduction: $$ \begin{align} \mathcal{S} &\to \mathbf{E}_{\mathcal{S}} \\ % \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -2 & 3 \end{array} \right] &\to \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] % \end{align} $$