[Linear Algebra] Kernel and range

  • #1
member 545369

Homework Statement


Let P2 be the vector space of all polynomials of a degree at most 2 with real coefficients. Let T: P2→ℝ be the functioned defined by:

##T(p(t)) = p(2) - p(1)##

a) Find a non-zero element of the Kernel of T. (I think I figured this one out, but I'm not too sure).
b) Find a specific element of p(t) for which T(p(t)) = π. Describe the image of T as an explicit subset of ℝ. ("Image" and "Range" mean the same.)

Homework Equations


##p(t) = a_0 + a_1t + a_2t^2##

The Attempt at a Solution


[/B]
a) What I understood by the "non-zero element of ker(T)" part of the question was that they didn't want me to just write f(x) = 0. So, this is what I wrote down:

##ker(T) = \{f(x) ∈ P^2 : f(2) = f(1)\}##

b) I'll write down my attempt at a solution:

If we're look for a ##p(t)## such that ##T(p(t)) = π##, then we're going to need an equation that follows the requirements of ##p(2) - p(1) = π##. I was thinking solving the following linear system:

##a_0 + 2a_1 + 4a_2t = π + 2##
##a_0 + a_1 + a_2 = 2##

But I feel like I'm overcomplicating things, because the margin given to me in this question doesn't permit the space required to solve this system of linear equations. There must be a clever way of doing this, no? I was thinking about substituting ##a_0 = 0##, since I was asked for a specific solution but I don't know… It seems as if I'm over-complicating things.
 

Answers and Replies

  • #2
15,536
13,634

Homework Statement


Let P2 be the vector space of all polynomials of a degree at most 2 with real coefficients. Let T: P2→ℝ be the functioned defined by:

##T(p(t)) = p(2) - p(1)##

a) Find a non-zero element of the Kernel of T. (I think I figured this one out, but I'm not too sure).
b) Find a specific element of p(t) for which T(p(t)) = π. Describe the image of T as an explicit subset of ℝ. ("Image" and "Range" mean the same.)

Homework Equations


##p(t) = a_0 + a_1t + a_2t^2##

The Attempt at a Solution


[/B]
a) What I understood by the "non-zero element of ker(T)" part of the question was that they didn't want me to just write f(x) = 0. So, this is what I wrote down:

##ker(T) = \{f(x) ∈ P^2 : f(2) = f(1)\}##

b) I'll write down my attempt at a solution:

If we're look for a ##p(t)## such that ##T(p(t)) = π##, then we're going to need an equation that follows the requirements of ##p(2) - p(1) = π##. I was thinking solving the following linear system:

##a_0 + 2a_1 + 4a_2t = π + 2##
##a_0 + a_1 + a_2 = 2##

But I feel like I'm overcomplicating things, because the margin given to me in this question doesn't permit the space required to solve this system of linear equations. There must be a clever way of doing this, no? I was thinking about substituting ##a_0 = 0##, since I was asked for a specific solution but I don't know… It seems as if I'm over-complicating things.

You shouldn't switch between notations, like ##P_2## and ##P^2##.
With respect to a): I understood it as to give an example of such a vector ##f##. So which polynomial ##f(t) \neq 0## would satisfy ##f(2) = f(1)##?
It might help to write the conditions on kernel and range in terms of your unknowns: ##a_0 \, , \, a_1\, , \, a_2 \,.##

As to b). The same advice here. What is ##T(p(t))## written out? If you do so, it should be easy to find values, such that ##T(p(t)) = \pi.##
(And you have a "t" too many in your equations.)
 
  • #3
pasmith
Homework Helper
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Hint for part (2): What is T(p) for p(t) = at?
 
  • #4
vela
Staff Emeritus
Science Advisor
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Homework Statement



a) Find a non-zero element of the Kernel of T. (I think I figured this one out, but I'm not too sure).

The Attempt at a Solution


[/B]
a) What I understood by the "non-zero element of ker(T)" part of the question was that they didn't want me to just write f(x) = 0. So, this is what I wrote down:

##ker(T) = \{f(x) ∈ P^2 : f(2) = f(1)\}##
ker(T) is a set, which is what you wrote down. The question asks you to find a non-zero element of this set.

By the way, note that the term is spelled kernel; you misspelled it (with an a) in the thread title. I corrected it.
 
  • #5
member 545369
You shouldn't switch between notations, like ##P_2## and ##P^2##.
With respect to a): I understood it as to give an example of such a vector ##f##. So which polynomial ##f(t) \neq 0## would satisfy ##f(2) = f(1)##?
It might help to write the conditions on kernel and range in terms of your unknowns: ##a_0 \, , \, a_1\, , \, a_2 \,.##

As to b). The same advice here. What is ##T(p(t))## written out? If you do so, it should be easy to find values, such that ##T(p(t)) = \pi.##
(And you have a "t" too many in your equations.)

Excuse the typos… So if f(2) = f(1), then: ##a_0 + a_1 + a_2 = a_0 + 2a_1 + 4a_2##… Cancel out the ##a_0##s, and you'd get: ##a_1 + a_2 = 2a_1 + 4a_2##

With a little bit of isolation, you'd get:
##a_1 = 2a_1##
##a_2 = 4a_2##

So, ##ker T= \{f(x) ∈ P_2 : a_1 = 2a_1; a_2 = 4a_2; a_0 = a_0\}## ?

EDIT: Forgot to add the fact that ##a_0 = a_0##
 
  • #6
15,536
13,634
Excuse the typos… So if f(2) = f(1), then: ##a_0 + a_1 + a_2 = a_0 + 2a_1 + 4a_2##… Cancel out the ##a_0##s, and you'd get: ##a_1 + a_2 = 2a_1 + 4a_2##

With a little bit of isolation, you'd get:
##a_1 = 2a_1##
##a_2 = 4a_2##

So, ##ker T= \{f(x) ∈ P_2 : a_1 = 2a_1; a_2 = 4a_2\}## ?

EDIT: Forgot to add the fact that ##a_0 = a_0##
You must not separate the ##a_i## because your codomain ##\mathbb{R}## is one-dimensional. Thus you can only say ##3a_2 + a_1 = 0## for polynomials ##p(t) = a_2 t^2 +a_1t +a_0 \in \ker(T)##. With that you can easily find a ##p(t) \in \ker(T) - \{0\}##.
The range can be treated similar.
 
  • #7
member 545369
Hint for part (2): What is T(p) for p(t) = at?

Ok so ##T(p(t)) = p(2) - p(1)## and if we allow ##p(t) = at## then: ##T(p(t)) = 2a - a## If we want it to equal π, then we would set ##2a - a = π##. Obviously, ##a = π## in this example.

I'm just gonna type up my attempt at extending what you said up to the actual question:

##T(p(t)) = 4a_2 + 2a_1 + a_0 - a_2 - a_1 - a_0 = 3a_2 + a_1 = π##

So, ##img T= \{f(x) ∈ P_2 : 3a_2 + a_1 = π\}##?
 
  • #8
member 545369
You must not separate the ##a_i## because your codomain ##\mathbb{R}## is one-dimensional. Thus you can only say ##3a_2 + a_1 = 0## for polynomials ##p(t) = a_2 t^2 +a_1t +a_0 \in \ker(T)##. With that you can easily find a ##p(t) \in \ker(T) - \{0\}##.
The range can be treated similar.

I'm sorry, I don't understand what you mean (we haven't taken codomain.) After solving the image part (hopefully I did so correctly?) I can understand where I took a weird route in isolating the variables, but why isn't that a valid move? Anyway, here is how I'd solve it using the method that (I think) you're describing:

##4a_2 + 2a_1 + a_0 = a_0 + a_1 + a_2##
##3a_2 - a_1 = 0##

Thus: ##ker T = \{f(x) ∈ P_2 : 3a_2 - a_1 = 0\}##
 
  • #9
15,536
13,634
So, ## img \; T=\{f(x)∈P_2:3a_2+a_1=π\}##
No. ##img \; T## denotes the entire image, range of ##T##, not only the ones which end up in ##\pi##. It is not a set of polynomials anymore since you evaluated them at points ##2## and ##1## and got a real number out of it. So the image is a set of real numbers.
Which are the real numbers, that can be written as ## f(2) - f(1) = 3a_2 + a_1##?
Are there real numbers which can't be written this way?

For the example with ##\pi## you have to find numbers ##a_2## and ##a_1## such that ##3a_2 + a_1 = \pi##.
pasmith's hint meant: you can choose ##a_2=0## and only bother with ##a=a_1##.


##4a2+2a1+a0=a0+a1+a2##
##3a2−a1=0##

Thus: ##\ker T=\{f(x)∈P_2:3a2−a1=0\}##
It has to be ##+a_1##, but yes. And it's better to say what the ##a_i## mean in this context, so
##ker \, T = \{ f(x) \in P_2 : f(x) = a_2x^2 + a_1 x + a_0 \wedge 3a_2 +a_1 = 0\}##

codomain is the english term for the set which you map to. In your case ##\mathbb{R}##.
 
  • #10
member 545369
No. ##img \; T## denotes the entire image, range of ##T##, not only the ones which end up in ##\pi##. It is not a set of polynomials anymore since you evaluated them at points ##2## and ##1## and got a real number out of it. So the image is a set of real numbers.
Which are the real numbers, that can be written as ## f(2) - f(1) = 3a_2 + a_1##?
Are there real numbers which can't be written this way?

For the example with ##\pi## you have to find numbers ##a_2## and ##a_1## such that ##3a_2 + a_1 = \pi##.
pasmith's hint meant: you can choose ##a_2=0## and only bother with ##a=a_1##.



It has to be ##+a_1##, but yes. And it's better to say what the ##a_i## mean in this context, so
##ker \, T = \{ f(x) \in P_2 : f(x) = a_2x^2 + a_1 x + a_0 \wedge 3a_2 +a_1 = 0\}##

codomain is the english term for the set which you map to. In your case ##\mathbb{R}##.


Aaah! Silly me, of course the img isn't what I stated! So since the question is asking for a specific element, I just need to formulate a polynomial… right? So I'll just let ##a_2 = π## and ##a_1 = -2π##. So, the specific element would be: ##πt^2 - 2πt = p(t)##

And yeah the minus was an accident in the kernel. thanks for the help, I think I got it!
 

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