0
$\begingroup$

At the linear algebra course our professor was introducing the notion of vector spaces and we worked on some examples.
We proved that different kind of sets are vector spaces with regards to addition(that we defined) and multiplication by scalars.
He then gave us the following homework:
Let $T$:
$a_1,\quad a_2,\quad a_3,\quad \ldots, \quad a_n$
$ a_2-a_1,\quad a_3-a_2,\quad \ldots,$
$a_3-2a_2+a_1, \quad \ldots, $
$\quad \vdots $
Let $S$:
$b_1,\quad b_2,\quad b_3,\quad \ldots \quad b_n,$
$ b_2-b_1,\quad b_3-b_2,\quad \ldots,$
$b_3-2b_2+b_1, \quad \ldots, $
$\quad \vdots $
Then $L+S$ =
$a_1 + b_1,\quad a_2 + b_2,\quad a_3 + b_3,\quad \ldots, \quad a_n+b_n,$
$ a_2 +b_2-a_1-b_1,\quad a_3+b_3-a_2-b_2,\quad \ldots$
$a_3+b_3-2a_2-2b_2+a_1+b_1, \quad \ldots, $
$\quad \vdots $
And a simple multiplication of $T$ by a scalar. Proving that this is a vector space over the Real numbers is pretty easy.
Then we were asked to compute the dimension of $T_3(a), a = (a_1,a_2,a_3)$
I thought of writing $T_3$ like this:
$$ \begin{pmatrix} a_1 & a_2 & a_3 \\ a_2-a_1 & a_3-a_2 & 0 \\ a_3-2a_2+a_1& 0 & 0 \\ \end{pmatrix} = a_1 \begin{pmatrix} 1 & 0 & 0 \\ -1 & 0 & 0 \\ 1& 0 & 0 \\ \end{pmatrix} + a_2 \begin{pmatrix} 0 & 1 & 0 \\ 1 & -1 & 0 \\ -2& 0 & 0 \\ \end{pmatrix} + a_3 \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1& 0 & 0 \\ \end{pmatrix} $$
So then, the subspace of $T_3$ would be generated by the three difference tables:
$ \begin{Bmatrix} \begin{pmatrix} 1 & 0 & 0 \\ -1 & 0 & 0 \\ 1& 0 & 0 \\ \end{pmatrix} & \begin{pmatrix} 0 & 1 & 0 \\ 1 & -1 & 0 \\ -2& 0 & 0 \\ \end{pmatrix}& \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1& 0 & 0 \\ \end{pmatrix} \end{Bmatrix} $
These 3 matrices are linearly independendent, and I checked to see if any pair of them also is, but there aren't any. That means that this Set of three matrices/difference tables is a basis, so there for the dimension of the vector space $T_3$ is 3.
If this solution correct? Also, could this rationament be applied to higher variable difference table?
The thing that confused me the most was that he said we should use Newton interpolation by finite dimension. All that I could find on that subject is https://nptel.ac.in/courses/122104019/numerical-analysis/Rathish-kumar/rathish-oct31/fratnode5.htmlLink1 and https://nptel.ac.in/courses/122104019/numerical-analysis/Rathish-kumar/rathish-oct31/fratnode6.htmlLink2

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.