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(4 marks) Let F1 denote the real vector space of polynomials of degree not greater than with the inner product{p, 9}1, pl)ql) &xfor p,q € P_ Show that the...

Question

(4 marks) Let F1 denote the real vector space of polynomials of degree not greater than with the inner product{p, 9}1, pl)ql) &xfor p,q € P_ Show that the two polynomialsPi(z) = 1 P(T) =1-rbasis for Pi_ Use the Gram-Schmidt process thonormal hasis for Pe from the hasis (P1, Pz) .consttucn

(4 marks) Let F1 denote the real vector space of polynomials of degree not greater than with the inner product {p, 9} 1, pl)ql) &x for p,q € P_ Show that the two polynomials Pi(z) = 1 P(T) =1-r basis for Pi_ Use the Gram-Schmidt process thonormal hasis for Pe from the hasis (P1, Pz) . consttucn



Answers

Let $V$ be the vector space of polynomials over $\mathbf{R}$ of degree $\leq 2$ with inner product defined by $\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t .$ Find a basis of the subspace $W$ orthogonal to $h(t)=2 t+1$.

Hello there. Okay. For this exercise, we are going to consider the space defined, um, over the continuous functions on the interval, minus 11 with the inner product equipped when an inner product defined as the interval between minus one and one off the multiplication of the functions Mhm Andi were in particular. In this exercise, we need to find some orthogonal basis. So we need to find another tonal vases the span by the following polynomial one t n t square. Okay, so how to do this? Well, we're we need Thio. Think of it not as functions, but as a vectors. Okay, so we're going to use the Grinch meet procedure to obtain or through no basis. So let's start to see what happened with these two. Paulino knows here one anti So there are tonal. Well, let's see, by computing the inner product. So let us start by considering the inner product within one anti. This is just the integral within minus one and one off t. Sorry. Here is steep. T v t. Okay, So he we got on this interval minus 11 on T is an off function. So this is equal to zero, or you can just integrate us the usual way and you will find that is equal to do. Okay, this is due to the symmetry. You can think off. This also geometrically t is just a straight line. So the integral in this symmetric interval minus 11 you can see that the area here in the area here are the same. So they're going to cancel between them. There's going to be zero. Yeah. Okay. So this happened with odd functions in general. Or you can just integrate this and you will see that you will. Let's make it explicitly. Just this case. This will be t squared, divided to evaluate. On the interval one minus one, this will be one half minus one half and this is equal to this year. So this inner product is zero. So this means that so we got that won t is equal to zero. So that implies. Actually, that means that there is an equivalence that say that one and t r or phono. Okay, so we have the first part off or thrown of basis. So so far we got that foreign of places. Let's say Earth, this is until now will be a span by one t. And there is some other function here, but we don't know yet. Okay. Off course is need to have going to be a polynomial, the guy that have a t square on it. Okay, so now let's do that. Okay. So we got to also know elements in our basis so far, so we can use DeGrange mitt procedure here. Okay, So you can think this geometrically as like the usual vectors. As we got some vector here, it's going to be one. And here there is another vector t. So these two vectors here spun the space, don't you? Okay, so we're going to find W as the span off one anti. Okay, so we got this space here. Now we're going to consider DTs Square, okay? We don't know if they if this T square is orthogonal to one anti, Okay, But we without knowing that we can convert it Thio or Thorne. Well, you will see the actually Tsk. Square is not tonal toe. These two. We need need to add something. Okay. So what the Grant Smith procedures say is that we take the projection here on the space of you. I'm going to create this ASUs. It's purple. So this that you're here will be the projection off the square on the span off one. Until that is, these super space W Okay, so this is the the the projection on this soft space w on we can subtract. So what happened if we take T Square and we subtract the projection? Well, we're going to obtain. We're going to eliminate this part off this vector here on. We're going toe end with a North Onal function polynomial. Uh, that is our tonal to this super space dog. You Okay, so that is the Grand Schmidt procedure in equations. That means that this GT that is going to be or next function is defined a steel square minus the projection off T square on the subspace off you. Where did this Super space W is defined as the spine off one empty. Okay, So what we need to do now is compute that projection. So the projection off T square on the subspace w is defined as inner product off the square with one. And here the in the product off one with one times one plus the product off the square tee. Times t divided the inner product off Tiv ity with itself. Okay, so we need to compute this. So let's consider each in their product separately. So let's start with this one. So the inner product off to square one is the integral within minus one and one off square DT again. Here we got on, um symmetric interval on. In this case, the square is an even function, so we can just consider the half off the interval on multiplied by two. So we're going to consider the interval between zero on one times two times the interval between zero and one off square ditty on this is just two thirds. Okay, so the inner product between two square and one is to felt. Then we need thio Compute the inner product off one with 10 this is trivial. His minus 11 off won t t This is just too Okay now let's see what happened with the other inner product that is between t square here indeed with itself. Okay, so the inner product between t square t is equals to the interval between minus 11 off the Cube deity. So we have seen before that if we got a symmetric interval on Dhere, we got on what function then? This integral is opposed to zero. And again, you can think off this geometrically if you like, do the plot off. This this is more or less in this way. Here is defined minus one and here is defined one. So the integral here will cancel with this part off integral. So at the end is just zero or you can just integrate in the usual way and you will obtain that is equals to zero. But it is always good to know that if you got a symmetric interval like this one on what you have insight is a not function immediately. There's all will be easier. Okay, so we can just eliminate this part here because these inner product is equal to zero. So we all this term is going to Syria. So the projection. So the projection off T square on the subspace W is equals to is equal to two thirds divided too. But this is just one third. Okay, so this is the projection. Okay, So with this information, we can get the rial value off GT, that is or for vernal function to the subspace W is to find a sti square minus the projection off the square over the subspace W where w is the span. Great. This below where this w is the span. Just remember of one T, Okay? And we got here. The value of the projection is just one third. So at the end, GT is just T square minus one third. Okay, so from this or author onal basis is a spine. Bye. The functions one d and D square, minus one third. This is or phone all basis.

Yeah, in this trouble were given a set up. So we'll start solution by writing the organ directors for each polynomial. So for people in the quarter, Inspector Ho. But like this 101 said, the 1st 1 is the question. But these year old 2nd 1 is to one of you two. So using this Pete, too apartment that too perfect. You could be rich. Zero. What being? And for P three, that could be, um, that would be one one negative three. So then the augmented matrix phoned my bees. Corn factors would be 101 01 negative three and 11 Negative three. Now let's write this over mental midgets in traditional echo form. And that is he could do warn. 00010001 is you can see. We have three columns and each column has a few bits elements. So three columns, three events elements at each column. So it means that these vectors Spector's spend our fee. It means that the given Colonials no meals spin he two. Now in part, we were given 1/4 collector for a loan Jew, and that is next to 112 And whereas to find this bologna to using this So we know that cube of native one times people plus one p two plus thio three So Dunn will be negative one times ones You're one plus one times zero wounding three waas Two times One wonder three And we find this to be one the next 10. And using this, uh, using the information that these numbers represent corruption tease. We can ride the roller milk you is one plus B t minus 10 t spring.

Work, even Pelino meal overseas. One, then one minus t then to minus 40 plus T square. The last one is six minus 18 T. That's nine. T squared us. Sorry. Minus cube. Okay, so in terms of the Matrix we have of four by four matrix. So it is the first efforts column represented a constant term. Off off. Pretty nami second column represent our one and so on and so forth. We have off one are off to and our off three. So for the person in question, we only have one and other things will be zero. The second equation. We have one constant term and, um, negative one for that t and also right. There are two zeros and the third column. We have two for constant term and negative war or tea, one for t squared. And for the last column, we have six accounts in term negative 18 for teeth and nine for t squared. And next one or T. Cute. Okay, so it's no hard to observe that they are four people. Two columns. So this four is exactly, um excuse me. So these four people columns, uh, exactly equal to the number off. Go to the number. Oh, pretty no meals given complete onions given in our meals. Nam Joo's. So that means these columns for columns are columns. All right, the nearly independent. Okay, so the answer is yes. This is four. Really? Nah. Mills will spend the whole p three. So? So these four? Well, you know, meals lean on mules. It's been the three, so we're

The question States proved that if the vector space is pollen, no meals of any degree with riel coefficients and a subspace is polynomial zwah 12 up two k That is a set of actors each of different degree. So these are different degrees p one p two dot, dot dot PK are different degrees. We are of different degrees polynomial of different degrees, then s is linearly independent. So let's do what the hint says Assume without loss of generality that the polynomial czar ordered in descending degree so such that the degree of polynomial one is greater than the degree of polynomial to which is greater than we're just gonna put them in descending order. Now, um, if they are linearly independent, then some constant times polynomial one plus some constant times pollen a meal to plus dot, dot, dot plus some constant time. Polly new Me. Oh, okay. Missed equals zero. So, um, looks like we're proving this bite contradiction. Let p one PTO peace PK be dependent. And if they are dependent than a constant times, each of them, the some of that is going to be zero. But if that were true, then um since C one is the only polynomial not see one p one is the only polynomial that has the highest degree term in it. Well, then, um, I see one must be zero. And now if P two has the second highest term, but P one has already been multiplied by zero. So so if it had the same degree term as P to what was already multiplied by zero. So that's gone. So if P two has the next highest degree term, then si two has to be zero. And we can continue this because p three it's not even if P three has the next highest degree term. But since C one and C two or zero than all the terms in C one and C two are all the terms in P one and P two are eliminated. And so now p three is the only remaining polynomial of its degree. Or that has the term of that That degree and so see to ah, C three has to be zero. We can go all the way to, um P K minus one, So C k minus one has to be zero. So let's think of the last one. So now all we're left with is CK P of lowest degree is zero in the end. And if a constant times that polynomial is zero, then either the constant zero or the polynomial a zero. If the polynomial was zero, then that would be the zero vector making this dependent. And if the constant is zero, I mean, otherwise the constant is zero. So all the constants are zero, which is the trivial solution and therefore trying to scroll up. And therefore, um, p one p two on all of them that set would be independent. So we ran into a contradiction that this does equals zero. I mean, does not equal zero unless all the constants or zero, which would be the trivial solution. And so, um, we can come to the conclusion that p one p two. Yeah, that that he que is independent. I mean, the nearly independent


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