How do I express ordered bases for polynomials as a matrices? Linear Algebra.

Question

I am given 2 ordered bases for polynomials of degree 1 or less. I am asked to find the change of basis matrix that converts vectors with respect to basis B1 into vectors with respect to basis B. I need help with determining how to express basis B1 as a matrix.

B = {1, x} .

B1 = {x - 4, 1}.

If I'm not mistaken, basis B expressed as a matrix would be:

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

---

I am having trouble with how to express basis B1 into a matrix form. Would it be:

$$ \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} $$ or $$ \begin{bmatrix} -4 & 1 \\ 1 & 0 \end{bmatrix} $$

---

I tried solving this problem using $$ \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} $$ The change of basis matrix I got was $$ \begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} $$ Thanks.

Answer 1

Let be $\vec v \in \mathbb{R^2}$ a vector and consider two different basis:

$$B_v:{\vec v_1,\vec v_2}$$ and $$B_w:{\vec w_1,\vec w_2}$$

Vector $\vec v $ can be represented in two ways:

$$\vec v =a_1 \cdot \vec v_1+a_2 \cdot \vec v_2 =x_1 \cdot \vec w_1+x_2 \cdot \vec w_2$$

or in matrix form:

$$\vec v =V \cdot a=W\cdot x$$

NOTE matrices V and W have the corresponding vectors of the basis as columns

Now suppose we know the components of $\vec v$ with respect to basis $B_v$ and we want to find the $x$ components of $\vec v$ with respect to basis $B_w$, thus:

$$V \cdot a=W\cdot x \implies x=W^{-1}Va=Ma$$

matrix $M=W^{-1}V$ is the matrix of change of basis from $B_v$ to $B_w$.

NOTE

This concept can be extended to vectors $\vec v \in \mathbb{R^n}$.

In your case, assuming $B$ as the canonical basis:

W= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

and

V=\begin{bmatrix} -4 & 1 \\ 1 & 0 \end{bmatrix}

thus

$$M=W^{-1}V=IV=V=\begin{bmatrix} -4 & 1 \\ 1 & 0 \end{bmatrix}$$

Answer 2

If $V$ is a a vector space with ordered basis $B$ and $B_1$, then the change of basis matrix (from $B_1$ to $B$) is the matrix of linear transformation of the identity map denoted by $[I]_{B_1}^{B}$.

In particular when, $V=P_1(\Bbb R)$ and $B,B_1$ are as above then,

$I:V \to V $ then the change of basis from $B_1$ to $B$ is obtained as,

$I(x-4)=x-4=-4(1)+1(x)$ and $I(1)=1=1(1)+0(x)$

$[I]_{B_1}^{B}= \begin{bmatrix} -4 & 1 \\ 1 & 0 \end{bmatrix} $