searching for a matrix, which brings $f(x)$ to $f(x-1)$ for polynomials up to degree of $2$ and considering the basis $(1,x,x^2)$

Question

I am trying to understand how to produce such a matrix. Do I have to write down a system of linear equations in order to see that? I do not see , how to solve it yet. Thanks

Answer 1

Basically you want to change basis from $\{1,x,x^2\}$ to $\{1, (x-1), (x-1)^2\}$. To do that you can express each of the polynomials in the old basis in terms of the new basis. For example: \begin{align*} 1 & = 1({\color{red}{1}})+0({\color{red}{x-1}})+0({\color{red}{(x-1)^2}})\\ x&=1({\color{red}{1}})+1({\color{red}{x-1}})+0({\color{red}{(x-1)^2}})\\ x^2 &=1({\color{red}{1}})+2({\color{red}{x-1}})+1({\color{red}{(x-1)^2}})\\ \end{align*} (As @John Hughes commented:) Assuming you are expressing the coordinates in column form. you get the matrix for conversion as $$S=\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}.$$ Now you want to express $f(x)=a+bx+cx^2$ in terms of new basis. So we identify $a+bx+cx^2$ with $\begin{bmatrix}a\\b\\c\end{bmatrix}$ and get $$[f]_{\mathcal{B}_{\text{new}}}=S\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}a+b+c\\b+2c\\c\end{bmatrix}.$$

Answer 2

Hint

You have a basis, $1, x, x^2$. Let's call those polynomials $b_0, b_1, b_2$.

What is $p_0 = f(b_0)$, as a polynomial (i.e., you should write something like $f(b_0) = x^2 + 3x + 1$, except that this isn't the right polynomial)?

Can you express that polynomial as a linear combination of $b_0, b_1, b_2$, so that the polynomial $p_0$ is written as $$ p_0 = c b_0 + d b_1 + e b_2 $$ where $c, d, e$ are real numbers? (Hint: yes. Because the $b$s form a basis).

Those three numbers are the entries in the first column of your matrix.

Now do the same for $b_1$ and $b_2$ to find the entries in the second and third columns.