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350 Chapter 5. Fundamentals of Systers of Differential Eqns {LHEHH 22 . Show that R2 is vector space: Show that R? is vector space; Show that the Bt of all matriCCS...

Question

350 Chapter 5. Fundamentals of Systers of Differential Eqns {LHEHH 22 . Show that R2 is vector space: Show that R? is vector space; Show that the Bt of all matriCCS with real entrier; vector SDacC Show that {['9] [e&] [%&] [e9] basis for R2x2 ([email protected] previ- question) Show that the plane (i.e: the =-v plane) is subspace of R: Show that the plane (i.e the V-z plane} subspace of R: 28. Show that any plane passing through the origin in R3 js subspace of R3. Show that the set of m matrices

350 Chapter 5. Fundamentals of Systers of Differential Eqns {LHEHH 22 . Show that R2 is vector space: Show that R? is vector space; Show that the Bt of all matriCCS with real entrier; vector SDacC Show that {['9] [e&] [%&] [e9] basis for R2x2 ([email protected] previ- question) Show that the plane (i.e: the =-v plane) is subspace of R: Show that the plane (i.e the V-z plane} subspace of R: 28. Show that any plane passing through the origin in R3 js subspace of R3. Show that the set of m matrices fOrms vector SPce with the previ- ously defined rules for addition and sclar multiplication. Show that the dimension of the vector space of HuErCAX six. Find bases for the column Shc in Problems 31 38 nullspace of the following matrices given Fio Fia Fi2



Answers

In each part, find a basis for the given subspace of $R^{3},$ and state its dimension. (a) The plane $3 x-2 y+5 z=0$ (b) The plane $x-y=0$ (c) The line $x=2 t, y=-t, z=4 t$ (d) All vectors of the form $(a, b, c),$ where $b=a+c$

Show that s is a subspace of or three. So Brown gives us s and sanitation retreats. That s is a set of all of actors X in or three such that the vectors take this form with the elements are and s being really value constants. And so s will be a subspace if one has is closed under addition and two s is closed under scaling multiplication. Remember that we knew. Quick check zero Victory does not exist in the suspect subset, as in that subset is not a subspace. But if zero vector does exist, then we still have to do I just have to show that one and two are valid. So we'll start with zero factory is your about your check so we can see that zero actor does exist If we just what are and as equals zero and, uh in the vector for here. And you can see if Oren s or zero than that vector will equal zero factory. So this director does exist now I can check close your under addition and we can let two vectors that are in this subset so you can say and save actor eggs which is equal to or honest us three r s this make that clear and the next sector will call why and that will have the same form when used t and you three team plus you and you snow These have to be That has to be close under addition. So if we add these two actors together, we need to get something that's in the saying that has the same components x plus Why will look like this look like or Blustein. But it's too. That's plus you three times are Blustein Waas that's plus you and thats plus you. And so now if we let se let w you will, that's plus you and say the you are close team. You may be right this as that addition looks like this V minus two w really be blessed over you on w which that has the same components as the subset in our civilization. So yes, it's closed under addition, I mean to say that close understanding, qualification. So let's ah multiply scaler K but a factory in the subset so we can well use. We use our why here we use this one almost like a scaler by that which will be it comes why? Which equals K times per victor Team owners to you. Green tea cost you. You That equals And both. Boy, that came through your canteen minus two K u three casing les que you on K u and you can see the you can make this. Where are eagles, Katie? And yes, equals, can't you? So then you can rewrite that for or But it's to us. Three are pleasures. Yes, and that has the same components. So that means it's closed under scaler. Multiplication So and that's what we need. So, yes, this is a subspace o r three for being as to show that the vectors and s Lyle claim, um, this with the form three eggs minus why plus seven z equals zero. So if the show is that the vectors of this form, if we plugged that into this equation, if it satisfies the equation than it does loud plane, so we'll go ahead and plug in, or a vector form, which or rewrite here, X or want us to us, we are blessed us. This so that'll give us three times are minus two s. Excuse gonna minus three are minus s. What can turn it? No. Plus seven s disease that can simplify down to you three are but its success. Once there you ar minus s plus seven ass equals zero and that is zero equals zero. So, yes, it was on that way.

So that the set of vectors and I'm going to write it down here, um, does not span three dimensional riel space. But it does span the subspace of three dimensional space consisting of all vectors lying in the plane with equation X minus two y plus Z equals zero. So I'm going to write down the vectors 345 and, um, it will spend this subset this plane. All right, So if we right that any vector is going to be a constant Times v one plus a constant Times V two plus a constant times V three, I should note that this is V one. This is V two, and this is V three, then a constant times 12 three. That's V one constant times 345 Let's v two and a constant times that which is V three. If I, um, right down the X coordinates, it would be one times see one plus three times C two plus four times C three is the X coordinate two times. See one plus four times C two plus five times C three is the y coordinate. And just to be clear, that's where I'm getting them from. And to be more clear, this equals X. Why Z any ah vector in three dimensional space. Okay, also three time See one plus five times C two plus six times C three is ze. Now we can write this as an augmented matrix. 13 four, 245 356 I got them from right here. 134 to 4 or five, 356 And then the solution. Now the solution. But the X y Z matrix is just x. Why Z the solution matrix would be C one C two C three. All right, we can ah reduce this matrix. Uhm, I'm going to take too negative too. Times the first rail and add it to the second row. Oh my goodness. That is not what it says. Okay, Two times the first run out of the second rule. That gives us zero negative. Two times three is negative. Six plus four is negative. Two negative two times four is negative. Eight plus five is negative. Three and ah, negative too. Times X is negative. Two x plus Why? Negative two x plus Why? All right now Gonna take negative three times the first row and added to the third row. So that's gonna give me zero negative. Three times three is negative. Nine plus five is negative for negative. Three times four is negative. 12 plus six is negative. Sex negative, three times x plus z Negative three X plus z. All right, let's keep working. I'm going to take negative, too. Times the second row and add it to the third row that's going to give me. Didn't do anything to the first row. Not doing anything to the second row Third row will be zero negative two times negative to positive. Four plus four is zero negative. Two times negative. Three is positive. Six plus negative six is zero and this is going to be negative two times negative. Two x, which is positive for X negative two times positive. Why is negative? Two. Why minus three acts less Z and from the last row we see that four x minus two. Why my eyes? Three acts plus Z equals zero. So for X minus three X is X. Don't do anything with the negative two y plus Z equals zero, and that's exactly what we wanted to show X minus two. Why plus Z equals zero. So that set of vectors does not spend the entirety of three dimensional riel space. But it does span a subspace and that subspace that its fans is the Plain X minus two y plus Z equals zero.

Okay, We've got a more in depth problem here. Um, so that these vectors negative Four comma, one comma, three, five comma, one comma, six and six comma. Zero comma to show that these do not span riel three dimensional space. But in fact, that they do spin a subspace of three dimensional space consisting of vectors lying only in the plane with the equation X plus 13. Why minus three Z equals zero. All right, so I'm going to write out the three vectors. Vector one is negative for one three. Make sure I don't make any mistakes either. Here, vector to is 516 Vector three is six zero to all. Right, So any vector that can be described would be a constant times Vector one plus a constant times vector to plus a constant times Vector three. So any vector which would be of the form X Why Z would be a constant times. Vector one just negative four one three plus a constant times vector to which is five 16 plus a constant, a different constant times. Vector three, which is six zero to Okay, So now I can write this in matrix form as, um, Wait a minute. I don't want it right in matrix form yet. That's school back a little bit. I want to write. This is negative for times. See, one plus five times C two plus six times C three equals X. And I got that from here. Time. See? One here. Time. See to hear time C three equals that the first coordinate. I can also write that one time. See? One plus one time see too. Plus zero C threes is why. And I can write the re C ones plus 60 twos. Plus to see threes equals Z. I could now rewrite this as an augmented matrix. Negative four five six. Making sure I didn't make any mistake Shut. Um, 110 and, um 362 x. Why Z Let's keep going. So I'm just gonna put one one zero. Why in the first row, Um, then, um, I'm gonna take that and multiply by four and add it to the old first row. So that's going to give us zero. So I'm doing this Times four. Okay, um so and adding it to the first row zero nine six um, X plus four Why? So I'm adding it. I've added it to the other row, which is verified that I did that correctly. Zero nine for six X plus four y. Okay, Now I'm going to take that second row, multiply it times negative three and added to the third row zero one times negative three is negative. Three plus six, his three to. And then this is going to give me Z minus three. Why? Let's keep going. Divide that second row by nine. Okay, Now I'm going to take this. I'm gonna take this, uh, second row and multiply it by negative 1/3 and add it to the third row. Um, okay, I'm gonna multiply it by negative 1/3 added to the third row. Negative 1/3 times. What am I thinking here? Negative 1/3 time. Six is negative. Two negative. 1/3 times. Z minus two. Why is negative? 1/3 Z minus. Think it's into by accident, but it's three. All right, so negative. 1/3 times e minus three. Why equals zero? All right. I'm noting a mistake in what I just did when I was, um, putting these Ah, line two and adding line to tow line three row to to row three. Um, accidentally wrote the wrong value here, So let's go back and try this again. Okay? I was doing negative 1/3 times the second row. That should have been negative 1/3 times the second row. And then I was adding that to the third row. I realized there was a mistake because the equation that I wrote before didn't have accident, and I thought that's weird, especially since the answer that were supposed to get has X in it. So let's just go back and make sure didn't misstate make a mistake. Negative. 3rd 1/3 Time's the second row plus the third row. So now negative 1/3 times X plus, or why plus Z minus three. Why equals zero? So now let's go ahead and simplify that. That's going to give us by multiplying by three on both sides of the equation, or even by negative three. That's going to give us X plus for why minus three Z plus nine. Why equals zero. And now we can combine like terms X plus 13. Why minus three Z equals zero. And that's exactly what we were intended to show that, uh, these three vectors do span this subspace of rial three dimensional space. Um, and that subspace is all the vectors lying in the plane of this equation. They spend that plane.

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


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