Linear algebra linearly independent

  • Thread starter Mdhiggenz
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  • #1
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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




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Answers and Replies

  • #2
33,722
5,418

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.
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.
 
  • #3
jbunniii
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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?
 
  • #4
327
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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?
 
  • #5
327
1
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.
 
  • #6
33,722
5,418
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.
 
  • #7
327
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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?
 
  • #8
33,722
5,418
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.
 

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