Linear Algebra - Linear Operators

[Victor Feitosa]

Homework Statement

True or false?
If T: ℙ₈(ℝ) → ℙ₈(ℝ) is defined by T(p) = p', so exists a basis of ℙ₈(ℝ) such that the matrix of T in relation to this basis is inversible.

Homework Equations

The Attempt at a Solution

So i think that my equations is of the form:
A.x = x'
hence A is non-singular and therefore is inversible. But my exercise anwser say the opposite.
Could somebody elucide me?

P.s: (sorry for bad english, I'm from Brazil and still learning it.)

 

[Mark44]

Homework Statement

True or false?
If T: ℙ_s → ℙ_s is defined by T(p) = p', so exists a basis of ℙ_s such that the matrix of T in relation to this basis is inversible.

Homework Equations

The Attempt at a Solution

So i think that my equations is of the form:
A.x = x'
hence A is non-singular and therefore is inversible.

You need to give a reason that A is nonsingular, if in fact this is a true statement.

BTW, we say "invertible" if an inverse exists.

But my exercise anwser say the opposite.
Could somebody elucide me?

What does this transformation do to the standard basis for P_s? Also, what is P_s? Was this supposed to be P₅, the space of polynomials of degree < 5?

P.s: (sorry for bad english, I'm from Brazil and still learning it.)

 

[Victor Feitosa]

Hello, my friend! Thanks for your feedback.

I thought that A is non-singular because if it was, A.x = x' could not be true, because it would have infinite solutions. But thinking right now i don't see how this is true at all.
No, it's not P₅, it's P₈. I thought it was a generic polynomial space.
P₈ is the space of real polynomials of degree < 8.
I'll edit first post.

 

[Victor Feitosa]

I see why this is false!

My matrix A is a echeloned matrix with trace 0. So it's det is 0 and it's not invertible.
Sorry for not posting how i found this. I will try to edit and post my anwser!

Thanks PF.

 

[HallsofIvy]

Hello, my friend! Thanks for your feedback.

I thought that A is non-singular because if it was, A.x = x' could not be true, because it would have infinite solutions.

You seem to have "singular" and "non-singular" reversed! This linear operator is singular precisely because it is not one- to- one.
If the difference between polynomials p and q is a constant, then Ap= p'= q'= Aq so A inverse of p' is not unique.

But thinking right now i don't see how this is true at all.
No, it's not P₅, it's P₈. I thought it was a generic polynomial space.
P₈ is the space of real polynomials of degree < 8.
I'll edit first post.

 

[Victor Feitosa]

You seem to have "singular" and "non-singular" reversed! This linear operator is singular precisely because it is not one- to- one.
If the difference between polynomials p and q is a constant, then Ap= p'= q'= Aq so A inverse of p' is not unique.

Oooooh, so logic and true! Thank you.
So, a matrix is singular if it's determinant is zero hence it have no inverse, is it right?

 

[Mark44]

Oooooh, so logic and true! Thank you.
So, a matrix is singular if it's determinant is zero hence it have no inverse, is it right?

Yes.
Regarding the problem in your first post, you need to show that if B is any basis for P₈, the matrix for the transformation is not invertible. What you've shown is the matrix in terms of the standard basis is noninvertible.

 

[Victor Feitosa]

Thanks everyone who helped! I think i figured how to do it and will try soon and post my result!