MHB Finding a 3rd polynomial to create a basis.

  • Thread starter Thread starter Harambe1
  • Start date Start date
  • Tags Tags
    Basis Polynomial
Click For Summary
To find a third polynomial \( t_3 \) in the space \( T \) such that \( \{t_1, t_2, t_3\} \) forms a basis, the polynomials must satisfy the condition \( t(1)=0 \). The existing polynomials \( t_1 \) and \( t_2 \) are already defined, and since \( T \) is a subspace of \( P_3(\mathbb{R}) \) with dimension 3, it requires a third polynomial that is linearly independent from \( t_1 \) and \( t_2 \). The discussion emphasizes the need to identify a polynomial that meets these criteria while adhering to the specified condition. Finding \( t_3 \) involves ensuring it complements the existing polynomials to span the subspace \( T \).
Harambe1
Messages
5
Reaction score
0
Hi,

I am struggling with the following problem:

"Let $V=P_3(\Bbb{R})$ and let $t_1=3x^3-x-2$ and $t_2=x^3-3x+2$ with $T=\left\{ t\in V \:|\: t(1)=0 \right\}$. Find ${t_3}\in\left\{T\right\}$ such that $\left\{t_1, t_2, t_2\right\}$ is a basis of T.

Not sure where to go as each column matrix will have 4 elements but then there is only 3 polynomials in the basis.Thanks.
 
Physics news on Phys.org
Harambe said:
Hi,

I am struggling with the following problem:

"Let $V=P_3(\Bbb{R})$ and let $t_1=3x^3-x-2$ and $t_2=x^3-3x+2$ with $T=\left\{ t\in V \:|\: t(1)=0 \right\}$. Find ${t_3}\in\left\{T\right\}$ such that $\left\{t_1, t_2, t_2\right\}$ is a basis of T.

Not sure where to go as each column matrix will have 4 elements but then there is only 3 polynomials in the basis.Thanks.

Hi Harambe! Welcome to MHB! ;)

Indeed, there are 4 elements, so a basis of $V$ will have 4 polynomials.
However, $T$ has the restriction $t(1)=0$, meaning we will be left with a basis of 3 polynomials.

Do you have any ideas how we might find that 3rd polynomial?
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 39 ·
2
Replies
39
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
27
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
8
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K