MHB Linear independence of polynomial set.

Click For Summary
The discussion centers on determining the values of c for which the set {1+cx, 1+cx^2, x-x^2} serves as a basis for the polynomial space P2. The original poster concludes that no values of c allow this set to be a basis, as they find it to be linearly dependent. They demonstrate this by constructing and reducing an augmented matrix, leading to a conclusion that one polynomial can be expressed as a linear combination of the others. Additionally, the conversation suggests verifying if the standard basis {1, x, x^2} can be generated from the given set to confirm their basis status. Ultimately, the set fails to meet the criteria for being a basis due to linear dependence.
bamuelsanks
Messages
3
Reaction score
0
Hi guys,

I've been working on a question which is as follows:

For which real values of c will the set $\{1+cx, 1+cx^2, x-x^2\}$ be a basis for $P_2$?

I'm coming up with the answer as no values of c, but am I really wrong?
I've only checked linear independence, because it would imply that it spans $P_2$ (right?)

I figure one would just create the augmented matrix:
$\left( \begin{array}{ccc} 0 & c & 1 \\ 1 & 0 & 1 \\-1 & 1 & 0\end{array} \right)$

And reduce:
$\left( \begin{array}{ccc} 1 & 0 & \frac{1}{c} \\ 0 & 1 & \frac{1}{c} \\ 0 & 0 & 0\end{array} \right)$

Thanks in advance,
SB
 
Physics news on Phys.org
for any c real number
1+cx = 1+cx^2 + c(x-x^2 )

that means the first one can be written from the other two elements (linear compination from other elements)thats means they can't be a basis.
other way to prove that they are basis or not try to generate the standard basis {1 , x , x^2 } from the given basis if that can be done then they are basis
 
Last edited:
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
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K