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S is nol a subspace of Mzz since I' %e sthe above is true...

Question

S is nol a subspace of Mzz since I' %e sthe above is true

S is nol a subspace of Mzz since I' %e s the above is true



Answers

If x and y are linearly independent vectors in Rn , then {O,X,Y} is linearly idependent

So who will show that? That's it s given by the order there A and two to the poor. A such that is really number. Ah, this It is subspace off the vector space from previous problems. 13 which is given by a one A to such that it too is positive. Okay, so we assume we want to check the proprieties. Ah, closed under addition and clothes and the scale of multiplication. So we assume we have to ah, points in the city as and ah, that like the set of skills here are is the rial number. So we assume the scaler. Okay, Azriel number and ah. Then you will be written as, let's say one to to the Bori one and the will be written as they're too Tools to the boy do. And now we want to check the closeness under addition first. So we dig you love the and this operation defined in ah problem 13 as follows. So ah, we would have a one to a one. Plus a two to borrow a to And this will give us Ah a one plus a two and then Ah, cooperation between the tool to the poor everyone and to the forehead. Tooth is going to be, uh, defined as this. So through the border, one glass, a Jew. And this is the new order pair. And as we see here, these two numbers Ah, like the summation of these two numbers is going to be a real number. And ah, this quantity here as positive. So ah, that tells us that this order there ah, lies and Victor Space V So that's Ah, first propriety. Now, the 2nd 1 we're gonna check the scale of multiplication, so take a and multiply by you Ah, on this world, people's according to the oppression, defined and problem 13. So this will be ah que times a one and ah, we raised the two to the power Anyone toe the scaler k so that will be true to the K A one and again this order pair. If we look at the first argument, this is ah Reum number. And if we look at the second argument, this quantity is positive. So again, this order there lies envy. So we have proved that ah, sit as a subspace off the victor space V

Wow, Alright, in this question, were considering a subspace U of V. So I kind of give you a sample problem here because I think it's easiest to understand if you also have a visual image. So, for example, if we have the space V, so that's outlined by this black rectangle that contains all of these letters A through J. So we have a set V that contains a number of elements, and then you is a subset of those elements. So you contains only a subset of the elements of the A, B and C. So if we define you, you is a subset of V containing elements a be n c. And now remember, whenever we're defining a subset or just a set in general, it's defined by its elements. So even if we have, for example, V and V, we had two A's, those A's are still just listed once. So when we take you, plus you this set A B C plus itself, the outcome would still Onley contain elements a B and see. So because it's on Lee, these same elements we can say that you, plus you, equals you. So what's your solution to this problem. Same thing would happen with numbers, shapes, anything that you want to consider equations, right? It's the same elements being added to itself. So the solution is just that same set of elements.

This question we have been asked which of the given statements are true. Now the statement is S is equal to two Sine Square X. to cost Square X three is linearly dependent. Now the set will be linearly dependent. If there exist constants K one, K two and K three says that K one times two. Science where X plus K two times Yeah, two car square x plus K three times three. This should be equal to zero. So we should have A set of constants. K one, K 2, K three says that not all of them are zero so that we can obtain this sum to be equal to zero. Now for this purpose let us assume that K one is three three times 2 sides. Where X let us consider K 22, B three. So we have three times two car square X. Let us assume play three to be minus two. So we have minus two times three. So we have six sine square X plus six, cost square x -6 is equal to zero. Now we can factor out six over here, we'll get sine square X plus call square X. And we can leave the last term as is no science or expose cost where X is known to be equal to one. Using the pythagorean identity. So this is what we end up with. So six times one is six minus six. And so we have zero equals to zero, which is a true statement. So that means that if we consider K 12, B three, K 22, B three and K three to be minus two, then we can see that this linear combination is equal to zero, hence this set is linearly dependent, and does the given statement is true.

So we have to affect us X and Y. Which we assume a linearly independent in RN. And the question is is the set zero? Xy nearly independent in our own. Okay, so remember a collection of vectors V one, V two VK is linearly independent if the following is true. So the only way the definition of linearly independent is that a one, V 1 plus a two V two plus dot dot dot plus a K V K. This is equal to zero implies that a one equals a two equals dot dot dot equals a K equals zero. So we can only write a linear combination of these vectors equal to zero. The only way we can do this is if all the coefficients zero. Okay, so we want to write too right? These vectors as a linear combination and set them equal to zero. So let's say eight times the zero vector plus B, times x plus c, times y equal to zero. But notice that a times zero is always zero. Um eight times 0 is always zero. So this implies that B x plus C, Y is equal to zero. But since I am since X and Y linearly independent, this implies that B equals C equals zero. But here A could have been anything we wanted. Right? And the equation is still satisfied. So for example, if A is equal to one, then this equation is satisfied. A times zero plus B X. Let's see Y equals zero. So we find three coefficients A equals one. For example. And then B and C both zero. Such that this linear combination is zero, therefore it's not linearly in the.


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