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I got stuck on this exercise and its really frustrating :/

It says something like this:

Given W={p(x) ∈ P₂[R] : P(1)=P(-1)} a linear transformation f:P₂[R] --> P₂[R] is required such that Ker(f) will be W

1)What should i know about W to know if i can get any linear transformation with the previous information?

2)Is it possible to get the linear transformation such that the polynomial h[x]= X will belong to Im(f)?

I dont understand the exercise.. take a look:

For 1) i think i should know the basis of W because its a linear transformation

I did this:

P₂[R]=ax^2 + bx + c

P(1)-P(-1)=0

a + b + c - (a - b + c)=0

a + b + c - a + b - c = 0

2b = 0

b=0

Then,

P₂[R]=ax^2 + c P₂[R]=a(x^2) + c(1)

Basis=(x^2,1)

What should i do now? I have no idea what to do :(...

What about point 2)?

Thanks!!

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  • $\begingroup$ 1) is not quite right. I mean, it's obvious in this simple case, but first you should check that $W$ is a subspace. Check it is closed for addition, multiplacation etc. THEN you find a basis. $\endgroup$ – tommy1996q Feb 20 '18 at 20:41
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You just have to define an appropriate application on the basis vectors. For instance, you want $W$ to be the kernel, right? Then take the basis $\{x^2,x,1 \}$ and send $x^2$ and $1$ to $0$ and $x$ in $x$. This way the kernel is $W$ and $Span(x)$ is the image.

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  • $\begingroup$ Thank you for replying! I still dont get it, dude.. :( $\endgroup$ – JustToKnow Feb 20 '18 at 20:40
  • $\begingroup$ You're welcome. Let's do this step-by-step. 1) You know that every linear function is completely defined when defined over a basis 2) You find a basis (for instance the one I wrote above) 3)You send things in a way that gives you what is required. You want $Span(x^2,1)$ to be the kernel? No problem, I just define the application so that they go to zero. Want $Span(x)$ to be the image? No problem, I have another vector in the basis that I need to send somewhere. I just choose to send him in the right place $\endgroup$ – tommy1996q Feb 20 '18 at 20:43
  • $\begingroup$ Okay, please.. its really frustrating. I feel stupid :( $\endgroup$ – JustToKnow Feb 20 '18 at 20:44
  • $\begingroup$ @JustToKnow don't worry, linear algebra is hard at first, but once understood it gets easier. What of the steps above you don't get? $\endgroup$ – tommy1996q Feb 20 '18 at 20:47
  • $\begingroup$ How can i know if with the given information i will be able to determine a linear transformation?How can i get it?Why did you choose span(x^2,1) ONLY with 2 elements? $\endgroup$ – JustToKnow Feb 20 '18 at 20:51

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