# [Linear Algebra] Kernel and range

Tags:
1. Jul 23, 2016

### talrefae

1. The problem statement, all variables and given/known data
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.)

2. Relevant equations
$p(t) = a_0 + a_1t + a_2t^2$

3. The attempt at a solution

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.

2. Jul 23, 2016

### Staff: Mentor

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. Jul 24, 2016

### pasmith

Hint for part (2): What is T(p) for p(t) = at?

4. Jul 24, 2016

### vela

Staff Emeritus
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. Jul 24, 2016

### talrefae

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. Jul 24, 2016

### Staff: Mentor

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. Jul 24, 2016

### talrefae

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. Jul 24, 2016

### talrefae

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. Jul 24, 2016

### Staff: Mentor

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 $\mathbb{R}$.

10. Jul 24, 2016

### talrefae

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!