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Define $T : \mathbb{P}_{2} \rightarrow \mathbb{R}^{2}$ by $T(\mathbf{p})=\left[\begin{array}{c}{\mathbf{p}(0)} \\ {\mathbf{p}(1)}\end{array}\right] .$ For instance, if $\mathbf{p}(t)=3+5 t+7 t^{2},$ then $T(\mathbf{p})=\left[\begin{array}{c}{3} \\ {15}\end{array}\right]$ a. Show that $T$ is a linear transformation. [Hint: For arbitrary polynomials $\mathbf{p}, \mathbf{q}$ in $\mathbb{P}_{2}$ , compute $T(\mathbf{p}+\mathbf{q})$ and $T(c \mathbf{p}) . ]$b. Find a polynomial $\mathbf{p}$ in $\mathbb{P}_{2}$ that spans the kernel of $T,$ and describe the range of $T$

a. $T$ is closed under vector addition.b. the range of $T$ is all of $R ^ { 2 }$

Calculus 3

Chapter 4

Vector Spaces

Section 2

Null Spaces, Column Spaces, and Linear Transformations

Vectors

Johns Hopkins University

Campbell University

Harvey Mudd College

University of Nottingham

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in this example, we have a transformation t and it's quite an unusual transformation that we're going to be investigating. What it does is it takes a degree to at most polynomial, they come from p two, and it maps it into our to. So imagine a function that takes a Paul no meal and gives you a point in the plane. That's effectively what we're doing. And so T R P is going to be p of zero for the first cornet and p of one for the second coordinate. Let's now do a full investigation of this transformation. T Well, the questions we might ask for such a transformation is, Is this transformation? T linear? It turns out that it is, but we need to show this. So let's start by showing that T is linear. Let's pull up the definition of linearity to guide us. So for the definition of linearity, we have two equations to show. Let's start with this equation here. So to that end, begin by saying, Let P and cue be in p two. We're using P and Q because that's suggestive of the typical elements from art domain. The set of Paul no meals. Then we need to consider the equation t of P plus Q, which is analogous to this left hand side. And our goal is to convert vector addition of the polynomial into vector addition of the vectors in our to So this is how we'll do it. First, use the definition of tea, which does the following for whatever a polynomial you have evaluated at zero and one. And that's your 1st 2 entries for the defector in our to, So we'll be writing P plus Q Evaluate at zero, then people Askew, which is evaluated at one. The next steps are function notation and function notation. If we have the sum of two functions evaluated at a 20.0 here, it becomes P of zero plus Q of zero, then p of one plus Q of one down below. Now that we've used up function notation, notice that we can break down this as this sum of two vectors will be p of zero p of one for the first factor, plus que of zero and Q of one for the second vector. Now it's a good time to pause and look at what we have from the first vector. This is really just TMP. So we can say that this is equal to T at P. The plus here copies to a plus here. And the second vector is nothing but a t at Kew. And that means we have broken down the vector addition on the inside to have the sum of vectors where we're looking at the images to have p anti of Q. So part one is complete For the second part. I'm going to continue to work with P, but we also need a scaler see. So let's start off by saying, Let see be in our. So now that we have a scaler, let's look at this left hand side of the equation. We can say then t at sea Times p will be equal to and before we get too far ahead, our goal is to do the following. We start with a C on the inside. If we can convert it to the outside, our work is complete so for the first step will take will use the definition of the transformation. T This is equal to see time's P evaluate at zero and c times P evaluate at one. So all I have done that this step is in place of the function p that we saw here and here. I'm using CPI. So they go here and here. So next will use function notation just as we did going here to here To say that multiple of a function evaluated zero is the same as C times PS zero. Similarly, the second entry ISI times p at one. So notice on the first equation, we had these elements here and we broke down into vector addition. Now we have these elements and we're going to break this down into a statement of scaler multiplication. We can put a C on the outside and placed p of zero p of one on the inside, but we know p of zero p. A one is the vector TMP. So this is now equal to C times T at P and this equation is complete notice, though we're not done until we stay our conclusion. So this is our conclusion. This shows the work above that t is linear. And now that we know that t is linear, there's a lot of wonderful properties we could look at, but we're going to cut our analysis a little bit short and just look at two more. Particular subspace is We want to look at the Colonel and the Range, starting with the colonel. So let's go to, ah, clean sheet and investigate the Colonel of the Transformation T. First, let's say what the colonel is. The kernel of tea is a set of all vectors. P such that t of p is the zero vector in our two. So let's see what that would mean. If t of P is equal to the zero vector, then that would imply P of zero is equal to zero and p of one is equal to zero. And this is because t. Of p by definition is PS zero and p of one. But we're setting this equal to zero vector. So I set both entries equal to zero at the same time. So if p of zero is equal to zero and p of one is equal to zero, then this would next imply the following Using ideas of algebra, we can say that p AT T is going to be equal to T minus zero where I've taken away the century times t minus one. Subtracting away this element here or to simplify p of t is equal to tee times T minus one. And this is for all t in our So this is our function and notice that we can do other things too. This function If we were to multiply by a number three then we still have t A P is equal to zero since changing a three here. If we put in a zero for tea or a one for t, we still get this equal to zero, no matter what scaler we put in front. So this has implicate implications for us. It tells us that the kernel of the transformation T is going to be equal to the span of tea times t minus one, the single vector. And this is because the span of this single vector is all in your combinations, which would just be constant multiples. Finally, the range of the transformation T is easier to look at notice that we set for T A. P. The first entry is PS zero, but that could be any real number whatsoever if we pick a generic polynomial p. Similarly, for P of one So this tells us that the range of tea is all of our two. So for this transformation, T we found that it's linear. The colonel is the span of this one polynomial, and the range is all of our two.

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