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Basis of a vector space

  • Thread starter Coolster7
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  • #1
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Homework Statement



There is a standard basis, B = (1; z; z^2; z^3; z^4) where B is the basis of a R4[z] of real polynomials of at most degree 4.

I need to find another basis B' for R4[z] such that no scalar multiple of an
element in B appears as a basis vector in B' and also prove that this B' is a basis.

Can any help with this please?

Homework Equations





The Attempt at a Solution



I could only think to do a basis B'=(0,1,z,z^2,z^3) but no sure if this is correct.
 

Answers and Replies

  • #2
33,635
5,294

Homework Statement



There is a standard basis, B = (1; z; z^2; z^3; z^4) where B is the basis of a R4[z] of real polynomials of at most degree 4.

I need to find another basis B' for R4[z] such that no scalar multiple of an
element in B appears as a basis vector in B' and also prove that this B' is a basis.

Can any help with this please?

Homework Equations





The Attempt at a Solution



I could only think to do a basis B'=(0,1,z,z^2,z^3) but no sure if this is correct.
No, it isn't correct. The 0 vector can never be a vector in a basis for a vector space.

Here's a hint: the function 1 + z is not a linear multiple of any of the vectors in B, right?
 
  • #3
14
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No, it isn't correct. The 0 vector can never be a vector in a basis for a vector space.

Here's a hint: the function 1 + z is not a linear multiple of any of the vectors in B, right?
So you could have B' = (1+z, 1+z^2, 1+z^3, 1+z^4, 1+z^5) for example or would this not work because the highest degree for R4[z] is 4?

Thanks for your reply.
 
  • #4
33,635
5,294
Right, you can't have 1 + z5. Can't you think of any other combinations besides adding 1 to another vector in the basis?
 

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