Linear algebra linearly independent

Click For Summary

Homework Help Overview

The discussion revolves around determining the linear independence of a set of vectors in the context of polynomials, specifically within the space P3, which some participants suggest refers to polynomials of degree less than or equal to three. The vectors in question are 1, x², and x² - 2.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants explore the definition of P3 and question whether it refers to polynomials of degree less than three or those of degree three or less. They discuss the setup of the problem and the method of forming a matrix to analyze linear dependence.

Discussion Status

Some participants have provided guidance on recognizing linear dependence through observation, while others are engaged in verifying the correctness of their approaches and reasoning. There is an ongoing exploration of the implications of the definitions and the relationships between the polynomials.

Contextual Notes

There is uncertainty regarding the definition of P3, with participants noting inconsistencies in notation across different textbooks. This lack of clarity may influence the interpretation of the problem and the subsequent analysis.

Mdhiggenz
Messages
324
Reaction score
1

Homework Statement


Determine whether the following vectors are linearly independent in P3 "not sure what P3 stands for maybe polynomial of third degree?"

1,x2,x2-2

let p1(x)=1
p2(X)=x2
p3=x2-2

c1p1(x)+c2p2(x)+c3p3(x)=z

where z=0x2+0x+0

I then create a matrix using the above relation where I get

(c2+c3)x2+c1-2c3

The matrix I'm thinking if solving looks like this [0 1 1;0 0 0; 1 0 -2] I know the zero must be at the bottom so I switch things up and get [1 0 -2; 0 1 1;0 0 0]

When I actually solve it I get c2=-c3
which is a nontrivial solution, so I state that it is linearly dependent.

Is my work correct, and or is there a faster way to come up with the conclusion that it is linearly dependent?

Thanks




Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Mdhiggenz said:

Homework Statement


Determine whether the following vectors are linearly independent in P3 "not sure what P3 stands for maybe polynomial of third degree?"
It's probably "polynomials of degree less than 3."

The notation isn't consistent from textbook to textbook, in my experience.
Mdhiggenz said:
1,x2,x2-2

let p1(x)=1
p2(X)=x2
p3=x2-2

c1p1(x)+c2p2(x)+c3p3(x)=z

where z=0x2+0x+0

I then create a matrix using the above relation where I get

(c2+c3)x2+c1-2c3

The matrix I'm thinking if solving looks like this [0 1 1;0 0 0; 1 0 -2] I know the zero must be at the bottom so I switch things up and get [1 0 -2; 0 1 1;0 0 0]

When I actually solve it I get c2=-c3
which is a nontrivial solution, so I state that it is linearly dependent.

Is my work correct, and or is there a faster way to come up with the conclusion that it is linearly dependent?

Thanks
This one is simple enough that you can tell that the set of functions is linearly dependent merely by observation. p3(x) is pretty obviously a linear combination of the other two functions.
 
Mdhiggenz said:

Homework Statement


Determine whether the following vectors are linearly independent in P3 "not sure what P3 stands for maybe polynomial of third degree?"
Most likely the set of all polynomials of degree 3 or less. The polynomials of degree 3 don't form a vector space. Nonetheless, shouldn't you find out for sure before attempting the problem?
 
Mark44 said:
It's probably "polynomials of degree less than 3."

The notation isn't consistent from textbook to textbook, in my experience.

This one is simple enough that you can tell that the set of functions is linearly dependent merely by observation. p3(x) is pretty obviously a linear combination of the other two functions.

How so mark?

Would it simply be because the first two could be represented as x2-2(1) which would give you px3?
 
jbunniii said:
Most likely the set of all polynomials of degree 3 or less. The polynomials of degree 3 don't form a vector space. Nonetheless, shouldn't you find out for sure before attempting the problem?

True.
 
Mdhiggenz said:
How so mark?

Would it simply be because the first two could be represented as x2-2(1) which would give you px3?

I understand what you're trying to say, but you aren't saying it very well. It is because p3(x) = (1) * p2(x) + (-2) * p1(x). This shows that p3 is a linear combination of p1 and p2.
 
Mark44 said:
I understand what you're trying to say, but you aren't saying it very well. It is because p3(x) = (1) * p2(x) + (-2) * p1(x). This shows that p3 is a linear combination of p1 and p2.

Oh! So just like in differential equations then? the same logic is being used in this problem?
 
Mdhiggenz said:
Oh! So just like in differential equations then? the same logic is being used in this problem?
Yes. And the term "linear combination" means the same thing in both areas.
 

Similar threads

Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K