Is the polynomial vector f(t) = t^2 – t in the span of P = {t, t^2, t+t^2}?

Is a Polynomial Vector in the Span of a Set?

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How to Determine if a Polynomial Vector Is in the Span of a Set

This lesson walks through a detailed linear algebra problem involving polynomial vector spaces, spans, augmented matrices, parametric solutions, and verification of the final answer.

The question investigated is:

$f(t)=t^2-t$

and whether this polynomial vector belongs to the span of the set:

$P=\{t,\ t^2,\ t+t^2\}$

The lesson explains that span problems are fundamentally questions about:

Solution systems.

If a consistent solution system exists, then the vector lies in the span.

Translating English Into Mathematics

A major theme throughout the lesson is learning how to translate the wording of the question into formal mathematical structure.

The statement:

“Is $f(t)$ in the span of $P$ ?”

means:

Can the vector be written as a linear combination of the basis elements?

Thus the span expression is written as:

$x_1(t)+x_2(t^2)+x_3(t+t^2)=t^2-t$

where:

$x_1,\ x_2,\ x_3$

are unknown constants.

Expanding the Polynomial Structure

The lesson carefully rewrites each polynomial vector into coefficient form:

$t=0t^2+1t+0$

$t^2=1t^2+0t+0$

$t+t^2=1t^2+1t+0$

This allows the polynomial vectors to be rewritten inside matrix form.

Jonathan refers to this process as creating:

“The Bubble Matrix.”

Constructing the Matrix System

After expanding the vectors, the problem becomes a standard augmented matrix system:

$\begin{bmatrix} 0&1&1\\ 1&0&1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix}$

The lesson explains how the polynomial space is converted into a standard vector space so that ordinary linear algebra techniques may be applied.

The matrix is then reduced using row operations and solved using a general parametric solution.

Infinite Solution Systems

The lesson demonstrates that the system produces infinitely many solutions because one variable becomes free.

Jonathan replaces:

$x_3$

with:

$s$

to avoid notation conflicts and writes the resulting parametric solution.

The final solution vector becomes:

$\begin{bmatrix} -1-s\\ 1-s\\ s \end{bmatrix}$

Since a consistent solution exists, the vector:

$f(t)=t^2-t$

is indeed inside the span of the set.

Verifying the Solution

A major emphasis throughout the lesson is verification.

The solution vector is substituted back into the original span expression:

$(-1-s)t+(1-s)t^2+s(t+t^2)$

Combining like terms yields:

$t^2-t$

which matches the original polynomial vector exactly.

Thus the solution is verified to be correct.

“When you are solving mathematically based problems, you are given the instructions within the question. You simply follow the instructions.”

The Importance of Mathematical Communication

The lesson repeatedly emphasizes that linear algebra is not only about arithmetic.

Students are expected to:

Translate definitions into equations.
Communicate solution structures clearly.
Understand basis notation.
Organize matrix systems properly.
Verify results rigorously.
Use professional mathematical formatting.

Jonathan explains that many students become overwhelmed because they try to memorize isolated procedures without understanding the definitions driving the problem.

Research Challenge

An additional challenge presented during the lesson involves investigating whether the factorization process could be rearranged differently by distributing matrices to the right side instead of the left side.

The lesson discusses how changing the order of matrix multiplication changes the dimensions and output structure of the system.

Students are encouraged to explore the dimensional consequences themselves as a research exercise in matrix multiplication and vector space structure.

“This is a pretty heavy duty question for anybody in linear algebra.”

Final Message

The lesson concludes by emphasizing that high level linear algebra requires:

Patience
Notation fluency
Matrix organization
Verification habits
Understanding of definitions
Repeated structured practice

Students are encouraged to focus not only on obtaining answers, but also on communicating solutions clearly, elegantly, and professionally.

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Transcript reference: :contentReference[oaicite:0]{index=0}

Original Transcript

All right, kiddos. We’re going to take a look at if the polomial vector f of t = t ^2 minus t is in the span of the set p is equal to t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t t ^2 T + T ^2. Now this this particular book, this particular problem is uh not in from any book. Um this this question is let me put it like back in here. This that’s let me get to the primary file

standalone question.

I didn’t take this from a book. I just made it up. And um I used I used slaying techniques. See there’s a difference between working from the book when you’re in your class and then just applying techniques outside of that. So let me uh let me show you this. So is the polomial vector. This is a vector in the span of P equals this set here. When you are solving mathematically based problems, you are given the instructions within the question. You simply follow the instructions. In this case, the instructions come from relating the English to the math. As one sees the forest through the trees, one sees the math through the physics. Coming from the definition of span and solution sets in this case is f ofx in the span means to the set. I used f of t. So let me change that to t and okay I might let me change that to t

is f of t in the span means it means to set the span equal to f of t and see if there are unique infinite or no solutions in general. All right. From the question, the solution lies within the keywords vector and in the span. In the textbook, these words would point to a definition that would point to the following solution. Thus, according to the span theorem definition, uh I set span of the set equal to f of t, which then I write span of the set equal to the function of t. Just like that, just notation. And then uh the span is the product of each element with an unknown constant. In this case, I use x1 and x2 and x3. And then that’s equal to the function of t. I rewrite the function of t as a linear combination here. So boom boom plus boom boom plus 0 * 1. That’s exactly what this is right here. Okay. So then I have this piece here now equal to this piece. I undo the product there. And then this one here, I do the same thing. I undo the product. Can you guys see this right here? that this let me do this. This right here is exactly this. Can you guys see that? T ^2 * 0 + t * 1 + 1 * 0 is equal to t. I will I will do another line in here to show you guys how this works in case you’re not seeing it. This is this is going to be t ^2 * 0 + t * 1 + 1 * 0.

Okay. So that that’s exactly what this is now. So let me change this here to blue so that you can see the blue the blue and the blue. All right. That that’s how that that’s how that operates right here. This piece here gets stretched out to this piece which is then this piece. Okay. So then now this piece and this piece and this piece is in every term including the right side of the equation. So I can put it in what I coined as the bubble matrix. I call this the bubble matrix because it you’re taking this you’re taking this polomial space and you’re stretching it out and now you’re allowed now you’re able to just do standard vector operations. This is an equation right here and you can extract that into its own augmented system. Uh and then I wrote some things here. Note you cannot move this to the right because the dimensions would conflict meaning you can’t you can’t factor this out to the right side. Uh let me uh

this right here would be wrong.

Do you understand? It has to be on the left side because you don’t do a 3x one by one by three. I mean you could I guess you know but uh this is a 1×3 and this matrix is a 3×3 and this is a 3x one. So I guess I guess you could uh maybe you could pull it out on the other side. I guess it wouldn’t matter really. What am I thinking about here? I guess, you know, it’s in every single term and it’s a 1×3, but if I put it on the other side here, then I get a bunch of matrices. It you could you might you might be able to you you you may you may be able to do a workaround with the following.

Um, but what it what it but it would it would output matrices instead of vectors.

because this would be a 3x one by a 3×3 which would be a a three. So let me do this. So that that is a 3 * 1 * a 1 * 3 damn it

is equal to a 3 * 3. So if I pulled it out this way, I’d have a bunch of a bunch of matrices instead of vectors. You see this? If I pulled it out this way, the product would output matrices.

So for you guys, so so research research challenge research challenge. Let’s let’s let’s uh add that to the key to the key here.

Book key.

I have to organize these by alphabet at some point. Research challenge.

All right. So, for you you kiddos out there that are hardcore math people like me for the re research challenge. Research challenge. See if you can solve this by factoring to by distributing

to the to the right.

So I’ll do this wrong method maybe. So you guys that are hardcore students, this is a research wrong method. Maybe this is a this is a let me put this in there. I want to I want this is a challenge for you guys. All right. I if we pull it this way, this would be a matrix because this would be a 3×1 by a 1×3 output of 3×3. It can you do that? you’d have to rewrite the whole entire thing. So, that’s a challenge for you guys to do it. Um, this that’s not going to be put on the website. I’m editing this. I have a bunch of problems I got to edit and get in my book and put new batches on the website. So, this is going in the book after I’m done. Uh, so then now this becomes this. And then we do the arithmetic. We do the augmented matrix. We reduce it. I plug it into Wolfra alpha like this to do the calculation. I’m not doing it by hand. That’s absurd to do this by hand this day and age. So, I go through and I have just a standard matrix. Now I solve it with the general solution. I find the solution. I undo the basis here. Pay attention to the basis right here. This basis notation is the same thing as this right here. Let me do a different color. I’ll do purple. That’s the same. This right here means the same as this. It’s just a little less writing. But this placeholder here tells you it’s still being multiplied by this. We’re just doing all the work on the inside now. And I do that and I from the matrix we see that there are an infinite number of solutions. Thus f is in the span is in the span of p and the solution is the following. If the if the solution is correct then the solution should be able to balance the initial equation. So I go through and I I take the answer that I got with my general solution here. I just put it all together into a parametric solution and then I simplify it and I get f is equal to this. Okay. And I got f ofx here that should be a t. And so now f of t is equal to f. And this is the this is the this is the vector we found right here in the solution. I multiply it out by the set here the basis which is which is just multiplying this this t right here out. So let me um let me do this for you guys. So you can see this is uh

this is now going to be

this piece here.

Change that to be black. Actually you know I’ll make this one purple. So this is is um equivalent to this. And then this is equal to that. So I get this piece here which is which is uh what my function of t my vector function here. If we want to simplify our if we want to verify our work is correct. If it we want we want to verify our work is correct. If it is correct then the above answer should mean that the span equals the function. And um when we plug it in how did I do that? span of this I get x1 x2 x3 is t ^2 minus one. So I plug the vector in for x1 x2 and x3. This x3 is different than that x3 there. So what what I want to do is I actually want to um since x3 is free I’m going to let let x3 be s I’m going to let x x3 be s instead.

So then uh let me do this here

let x3 b s just so that it’s not conflicting with that notation. And then this here will be S S.

That way it doesn’t it doesn’t look so bad. And so f the function f is is is this which is um the same thing if you guys are new here to linear algebra notations here. This is a really cool problem. I I love this problem because it’s like you look at it you’re like there’s no way this is going to this is going to come out like this can’t be it can’t be it just can’t be it. And it is this this um so this this right here this function here is oh I did that on purpose because this this right here is exactly the same thing as this here. And so this right here is f of t. Oops.

This is our solution f of t not the original. So I’ll call this I’ll call it this g of t and I’ll call this g for our solution is g of t. So let me change this back to g. Let me change that to g because our answer our question our original question f of t is this. So that would be the statement would be incorrect to call that f of t as well. I need that to be g of t. And uh so then this so then this right here is g of t. That’s the answer. And if that is indeed the answer we want to verify our work is correct. If it is correct then let me get this here

then the span of this we span it out x1 x2 x3 and we uh need to I we need to make point here that this right here is x1. So this right here is X1 and then X2 is going to be I’ll do green so that you guys can see this. X2 is green and then this one here is X3 and I’ll do that one as red right there. Uh sorry that should be X3.

All right. So x1 x2 x3 is x1 x2 x3. You plug those in there and you should get t ^2 -1. Let me get this all back to s here.

Oh, I I did that already. I’ll just leave it like that then.

So I went back. Okay, I had done that already. So, you should read the whole paper before you edit it. You might have already done what you want to do. Happens all the time. Uh, so here, so I plug in for x1, x2, and x3. I plug in negative 1 – s 1 – s and s in there. I plug those in to there. And um, let me um, let me keep the color coding here so that you guys can see what I’m doing here. Oops. For that. And then X1 is orange. So orange,

green,

and red

so that you guys can see that. And you go through and you do the arithmetic and voila, you got the answer. Therefore, f of t I kept doing f ofx there. f of t f of t and that should be g of t. Therefore g of therefore f of t is in the span of the provided set in the solution system of the equations is g of t is equal to that. All right. Um this will be on your test. This definitely is a question test. So test solution exam solution

would be

So in this area, if you’re new here, what I write is what you need for the exam. If you’re doing a question like this, what you need to write on the exam, you can go, you can go directly to this. I mean, you don’t need to do everything that I did. You can see exactly by looking at it. Oops. Hold up. I got more down here. I did not do what did I what did I miss here? I guess there’s more stuff here. Oh, I already did it. How to structure this exam and max credits. Okay, so I had already done that.

Okay, so exam solution

and span is f of t. f of t. I’m surprised nobody pointed that out in the editing.

The above is all you really need to write in your exam, but you may have other arithmetic based on your skills and time exam. Okay. All right. So, this one concludes this edit. Hopefully, I didn’t miss anything. And they the teacher might ask you to find the solution like this as well. So with an if you do this then you can see thus

g of t equals this is the answer which is the which is the solution. g is equal to that vector that’s the solution. All right these are this is this is a pretty heavyduty question for anybody in linear algebra. There’s not much out there that is like really streamlined when it comes to doing a question like this. This is about as good as you’re going to find anywhere. All right? And it took me a long freaking time, a long freaking time to put all of this together in in uh such a uniform sort of, you know, elegant fashion more or less. And uh you know, you should be grateful. You should be grateful that someone like me is willing to go through the effort to straighten this stuff out for someone like you. You know, this is not this nothing about this. For me, learning this was quick. And I didn’t have any online resources for this stuff. I had to figure all of this out on my own using books. And uh the internet would never have helped with this. I mean I the internet’s helping you now. You’re welcome. But uh have some respect for this. This was not easy for me to to learn how to do all this and type it. You guys understand to type it and I did this live in a live session on demand and I made the question up and I prove I made the question up. I solved it and I proved that it was correct. All typed. Have some respect for that. Okay. Took a long time for me to do this. Okay. This is part of the Pllem Academy research position.