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If Bis the standard basis of the space Pa polyncmials then let B={1,t,4,+} Use coordinate vectors test the linear independence of the set of polynomials below Expla...

Question

If Bis the standard basis of the space Pa polyncmials then let B={1,t,4,+} Use coordinate vectors test the linear independence of the set of polynomials below Explain your work1+812 _ 0,t+20 , 1+t+ 812Write Ihe coordinate vecior for Ihe polynomial 822 _ ? ODDO Write Ihe coordinate vector for the polynomial t + 2t? ODDD Write the coordinate vector for the polynomial ~t+8t2To test the linear independence of the set of polynomials, row reduce the matnx which formied Dy making each coorainate vector

If Bis the standard basis of the space Pa polyncmials then let B={1,t,4,+} Use coordinate vectors test the linear independence of the set of polynomials below Explain your work 1+812 _ 0,t+20 , 1+t+ 812 Write Ihe coordinate vecior for Ihe polynomial 822 _ ? ODDO Write Ihe coordinate vector for the polynomial t + 2t? ODDD Write the coordinate vector for the polynomial ~t+8t2 To test the linear independence of the set of polynomials, row reduce the matnx which formied Dy making each coorainate vector column of the matrix possible vrte the matrix in reduced echelon form; Are the polynomials linearty independent? Since the matrix does not have pivot in each column, its columns (and thus the given polynomials) are not linearly independent Since the matrix has pivot in each column;its columns (and thus the given polyncmials) are not linearly independent Since the matrix has _ pivot in each column its columns (and thus tne given polyncmials) are linearly independent 0 D. Since the matrix does not have pivot in each column; its columns (and thus the given polynomials) are linearly independent



Answers

Find two matrices $A$ and $B$ such that $A B=B A$

X. Well do we has at most one solution but E X. You could be hedge and minus R plus one linear independent solutions. No. Aah am into an metrics. This implied number of columns in a is equal to and and our been restoring given n minus R plus one is equal to one. So this implies N -R is equal to 1 -1, equal to zero. So this in black hand. Uh huh. Equal to our this implied rank of A. Each and so this imply and columns of E. Uh Mhm linear fully independent because yeah, rank A. Mhm. Equal to roaring equal to cool. Um Rank Mhm. Mhm. Of A. So this impact columns are literally independently. So columns. Yeah, yeah, linearly independent and script.

To check this subspace. What we need to do is we need to verify three exams. The first axiom is zero, vector should be belonging to a subspace. W. I mean the set W If you won't, you too belongs to W. Then we should verify that you want to see. You too also belongs to W. What is the closure property? Third, if you belongs to W then land up. You should also belongs to W Where lambda. Is it number from the field? Okay, if all these three exams satisfied, then we call W is a subspace of the factor space week. No, these set off all environ mattresses over the field kit. All right now W is the set of all symmetric matrices. So what do you mean by a symmetric matrix? The matrix was transposed the same as it's A So that's called a symmetric matrix. All right. Now if is symmetric then that's null. Matrix belongs to W. Yes, of course. Because the non matrix transpose his narrative so yes, it's true. Second axiom some of two symmetric matters is A and B. Supposed to be a symmetric B is also symmetric. Then a place be whole transpose is equal to the property of transport. It is a transport transport. But he transposes A. Because asymmetric transpose is B. Because B symmetric diet implies a place to be. He's also symmetric so that implies equals B belonged to W. So second exam is very fine. Now coming to Torrance if is symmetric lambda E. We'll transport the property of transport is lambda and a transport is equal to the lambda. A. Because the symmetric it is lambda lambda. He will transpose islam day. So that means lambda is also symmetric so it belongs to doug. So all the three exams are verified. So yes. Set of all symmetric matrices is a subspace. The second one upper triangular mattresses. So what do you mean by an upper triangular matics? All the elements below the main diagnosed should be zeroes, for example A B C 000 123 So this is an upper triangular matrix below this. All the elements are zeros of is a pet Wrangler. He's a pet Wrangler and he's also a strangler. Then it let's be is it a veranda? Of course, yes. Because these zeros get started with the other metrics in the same place. For example 123 4045006 If you add with 178 089 00 10. So what do you guys? 000000000 So some of the upper triangular mattresses. Also an upper triangular markets. So this is very frank. Null matrix is not trying biometrics because null metrics by default, all the elements below the main diagnosis anyway, zeros. In fact all the elements are real zeros. So it belongs to going mathematics, The Time magazine. If you want to play any number with the matrix scale are multiple, zero into any number of zero. So these zeros will still be retained. So lambda you Orlando A is also an upper triangular matrix. So yes, the set of a particular mattress is a subspace. The third diagonal matrix, basically, we can just individually say that the diagonal matrices from the subspace because lower triangular mattresses also from subspace. Because what do you mean by lower strangler? All the elements about domain that loves you with the same reasoning as a crime. And what is the diagonal matrix? It is both the lower triangle ring up strangler. Since both our Wrangler and Wrangler mattresses from subspace set of diagonal mattresses. Also from subsidies. Simple reason. Right. And skill harm metrics is a special case of diagonal matrix, scalar matrix means all the diagnosed elements should be zero, sorry, All the diagnostic elements should be same and other diagnosed elements other elements should be basically zeros. So this is a special case of diarrhea magics. And since diagnosed mattresses from subspace, the set of the subset of it also forms the suspects. Alright, so this is a prison. So

Okay, we want to find if these three vectors are linearly independent. So usually when you have to find out whether vectors are linearly independent or not, and uh you basically are given the same number of equations as the number of vectors. But here you see uh sorry, as the dimensionality of victor's. But here you have four dimensional vectors and you have three victors. So um there are many ways which we can use to see whether they are linearly independent or not. We can use the basic definition of linear independence or we could um uh okay in this case we'll just use the basic definition of linear independence. Let's say that let's assume that they are linearly independent. Um Sorry, let's assume that they are linearly dependent. And let's see if it works out or if there is a contradiction. So what is meant by linear independence? It means that if this vector is a, this victory is B. Sorry, let's let's just call them what we have been given. We've been given this as we won. Me too. This is V. Three. Right then. Uh these uh like we can say that um one of like these vectors do not form, these vectors are linearly dependent if for some A. And be we have eight times we one plus three times with two equal to B. Three. So this is the basic definition of linear dependence. Now, let us use this definition here. Will calculate eight times V. One and B times we do. And we'll set it equal to B. Three. So then we have the equations A to a minus four. B. I'm taking the first term of every um director to A minus for me equals two and right. Not the second term. Uh A minus to B. My sorry, A place to be. It's plus. I don't know. I got a minus here. Uh please to be equals nine. Now we can solve these two equations. Women be able to solve these two equations uniquely for A and B. Let's just multiply the second question by two. Then we get to a place for B equals 18. This gives us um Now let us add this equation with uh let us add this and this equation we have four A equals 20 so A equals five and be equals. So I'm using this equation now uh B equals nine minus five by two. This is four by two. It is equal to two, so equals five and equal to do. Now let us see if this is consistent with the next term. So we should also have three A. From here. Uh Plus three B. Three A place three B equals 33 21 here uh equals 21 but we already have found A. And B. Values. So three A. Is five multiplied battery. 15 and three B. Is six. Uh It's three multiplied by 26 Okay so this works out correct. 21. Now let's um now let's go with the last question. Sorry? Last victor element. For A. Plus four four A plus B equals 22. Four A. Please be equals 22. So um Yes. Uh for A. S. 20 and B is too, so indeed 20 plus two is 22. So we find that if A equals five and B equals two then uh five B. One plus two. We two is indeed equal to eat three. That means that V three is um not independent. Like these two elements are not independent to each other. So for so our answer is they are not independent or they are dependent linearly dependent. So this is the answer for the eighth problem there. Okay. I'm not sure if I should be during the other two, but okay. Since we are at it. Let me just um do the other two problems. Also ninth one 123 215 Okay. In this case we have four victors and we have and there are three dimensional vectors. So like if you have a plane, if you have a piece of paper, this is this is my piece of paper here. And for example I have three victors. Can three victors, it will be linearly independent. Um No because no matter how I don't three victors, I can't find some combination of other two vectors which will give me the third vector. So basically uh for example, if you want me to find uh some combination of these two vectors. Um Okay. Some combination of these two victors which gives me a toad victor. So I can always stretch one victor, then I can stretch the other victor by multiplying them with constants, and then I can take the minus of this victor, so I can go backwards. So in that case I'll have this victor backwards, and then I can add uh these two vectors, finally in such a way that they end up giving me any victor, which I want. What I'm trying to say is that if we have four victims in a three dimensional space, there's no way that they can be linearly independent. For a piece of paper is a two dimensional space. So three vectors cannot be linearly uh independent. Uh Here we are given four vectors which are three dimensional, so they cannot be linearly independent. You need to have at least four dimensions, like if this was 123 and had another term and all of these had another term, then there was a possibility that they could be linearly independent. But right now there is no possibility you can just see that there are four terms and three dimensional terms. So you know already that the answer to ninth oneness linearly dependent. Just like the eighth one. Okay, 10th one. Finally um three full one. Okay. Do five and minus 10 Three minus two. One and 24 minus three. Zero. Do. Okay, so let's do the same thing. This is X one, this is X two, This is X three. We are going to multiply this by A. We're going to multiply this by B. And then we're trying to go and see if there exists any N. B. For which everything works out. So for which A X one plus B. X two equals x three basically. So let's let's do it. So three A minus B equals two. And for a this is 00 right? Yeah zero for a place zero equals four. So from here we immediately at equals one. So three A minus B equal to do which it means three A's 123 minus me equal to two. So we have B also equal to one. Let me check this. 343 minus one. Yeah. And two. Okay. And four. Yes. Zero. Okay. Four. Okay, so three A minus B equals two and for a minus zero equals four equal to 13 minus B equal to two. So three A three minus me equal to do. So B is equal to 123 minus one equals two. Okay, so both A and B. A. One. Now let us see if this is consistent with everything else that's over here. Let's take the toward victor. A multiplied by one plus be multiplied battery A plus three B equals a plus three. B equals um minus three told victor Yes minus three. But is one and B is also one, so four is equal to three which is four equals minus three. Which is wrong. This means that it is not possible to have an A. And B. Such that Exxon and bx two will give you X. Three. So this proves that these three vectors are linearly independent. So the 10th one linearly independent. So these are the answers. Um Eight and nine are linearly dependent and 10th is linearly independent. Thank you. My name is hank it and I hope you like my um I hope you like my radio.

We need to find two matrices and be such that a B equals B. So out off the two matrices, let a be a unit matrix that is 100 va on Let us consider be Toby Quito. 1234 So the left inside that is a B will become equal toe. 1001 multiplied with one don't 34 which is equal to one. Do three and four on the right. Inside B A will be equal to 1234 multiplied with 1001 which is equal to one YOU'LL three and four. So therefore, a B is equal toe B A If is ones it'll 01 on B is equal to 1234


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