what-3-times-3-matrix-will-have-the-same-effect-on-homogeneous-coordinates-for-mathbbr2-that-the-she

Question

What $3 	imes 3$ matrix will have the same effect on homogeneous coordinates for $\mathbb{R}^{2}$ that the shear matrix $A$ has in Example 2$?$

Lu He · Accepted Answer

So if we want to buy the major city by dimensions there have sent if asked the one on December who we need to make sure the lost Borio upto chess move after transformations that also you can be job I mean, up before and after the transformation that values should be the Seine. Okay, so it s set a new majors called B and B Ys one you know, you know, one person by look Sorry. You should be one point who if I one year old one you know, you know? So this maitresse meant it tries the boy as why you it&#x27;s get the new majors after transformations and the better should be as prospect 1% and five I And why? And yet so you can see that the last world does not trench. Ah, that means the yet are you can be dropped. So if we drop the yet borough, we get the san transformation as the eight times ah s five better. So if we drop the yet you get this them I&#x27;m so after. Ah, a major is I mean, the measures in assemble one ties That&#x27;s why better So the answer we get it should be bay me is

Michael Jacobsen · Answer

in this video, we have a three by N Matrix. So that&#x27;s three rows and columns, and we&#x27;re going to assume that the columns of a are going to span all of the space are three. Next, let&#x27;s addict something extra to this picture. Let&#x27;s say assume de is a three or N by three matrix, for which eight times D equals I three. So what we want to determine here is what does this matrix de equal in particular? Well, let&#x27;s start with I three. We know that I three is the three by three identity matrix, so it is formed by columns E one E two e three, where e one is 100 E two is 010 anti three is 001 so altogether were then looking for a matrix D, such that when we multiply a times d, we obtain this three by three identity matrix we see displayed here. First, let&#x27;s also note that since the columns of a span or three each time we write a matrix equation eight times X equals B, where be is in are three. This matrix equation is always consistent. This is because knowing that the columns of a spin or three is logically equivalent to the statements that a X equals B is always a consistent system of equations. Let&#x27;s use that fact next to construct our matrix de here, we can say that the systems eight times X equals e one eight times X equals e to eight times X equals e three were e one e two e three are the columns of our identity matrix here. Each one of these systems are consistent, since we could have placed anything in these positions on the right hand sides and they&#x27;re still consistent. Let&#x27;s suppose then that let&#x27;s call this d one de to d three our solutions. So in other words, let&#x27;s say what we mean exactly were saying that since eight times X equals e one e two or three here is always consistent. It follows that a Times D one must be equal to the one. Since this is our first solution. Eight times D, too, is E to and likewise, eight times D three is E three. Now let&#x27;s then define the matrix D By columns D one do to d three, where these came from. The solutions that we had constructed. Then when we take a times D, this is what will produce by the definition of a matrix times another matrix. We take a Times first D one, which came from here. Then the second column of the product is a times D to and lastly, we have eight times d three for the third column of this product. So this is what eight times D looks like. But we also know that eight times d produces e one e two, any three so we could make a substitution here, here and here to get the following. We have the one e two and e three, but this is now the three by three identity matrix. So we&#x27;ve just shown how can to construct the particular matrix de such that eight times d equals I three.

David Barker · Answer

soon says, um, give an example of two maitresse. Ease that when you multiply them. Part of this product is a three by two main tricks. All right, now what they mean again, they mean when I&#x27;m not supply to major cities together, my answer is gonna be a three by two. That meaning I&#x27;ll have three rows 123 and two columns. So my answer is gonna look just like that. What are two matrices that will end up looking like that? Well, it&#x27;s unlimited a Sfar. My answers go. But remember, it&#x27;s the outside dimensions that determine what your products gonna end up being. So I could have a three by one times a one by two number. The no parts have to be the same or it&#x27;s not defined. So I could have a three by one that would look like this with three rows in one column times a one by two. All right, that&#x27;s one row and two columns. See you in a get this product would end up being a three by two. All right, I could do a long Is the outside numbers all three by to remember, that&#x27;s what I&#x27;m trying to end up with. So something outside numbers or three and two. I could have anything on the inside. As long as they&#x27;re the same. I could have three by 25 time to 25 by two Now. I would take hours to do, but you could do it. Remember? As long as the inner numbers of the same read anything. Three of my five times in 552 You&#x27;re the same number of the outside. It&#x27;s gonna be mine too Mentions.

Tim Strang · Answer

everyone. So the question we&#x27;re looking at today is I just asked you to find an example, just one example of a three by three matrix that has a generalized Eigen vector that is linearly independent. So essentially, it&#x27;s just asking you to find a, um, matrix that has a generalized Eigen vector that&#x27;s meaningfully different its regular Agon vectors. Because regular ibeacon vectors are sort of, ah, trivial subset of generalized Eigen vectors that you confined. So what that means, And I highly recommend you watch my previous video. It&#x27;s on the question right before this in the textbook that ask you the exact same question just for a two by two matrix. And if you&#x27;re wondering why there is a two by Two Matrix and a question about three by three matrices, it&#x27;s because B is Theo Example matrix that we found in the previous question. It&#x27;s not unique, clearly not the only one. Otherwise, we wouldn&#x27;t really studying generalized Aiken pictures, but, um, so please go back and watch that it&#x27;ll be pretty integral to the things we&#x27;re doing here, although not entirely necessary, will still proved that what we&#x27;re about to do with this still works just on its own merit. So let&#x27;s try and build our A. So I cheated. I already did this, and I checked. But I&#x27;m going to explain why this works. And I feel like that will be the best way to go about this problem particularly. So we&#x27;re just gonna put one 0000 Then we&#x27;re going to embed the matrix be down here, right? So this right here&#x27;s or two by two Matrix B and then we&#x27;re just adding a one along the diagonal. Perfect. Now, the reason why we&#x27;re going to show, um that this is ah, defective and therefore has generalized Eigen Value is because when you take the determining when you find that when you find the Eigen values well, let&#x27;s let&#x27;s do that. So a minus lambda I What&#x27;s that? That&#x27;s one minus slammed of 0000 negative land, uh, one negative one tu minus slammed up. And then the determinants of that. The determinant of a three by three matrix, if you remember, is just this number times the determinant of this number. And then you&#x27;re going to ad you&#x27;re basically going to commute this. So then it&#x27;ll be plus this number times the determinant of this matrix, right? And then yada yada with this number. But those two numbers are zero, so their terms actually cancel out. And it&#x27;s just the top left corner, which is which is why we have made all those other values. Zero. Because it works out quite nicely. And we get to preserve the structure actually, of this embedded be matrix in the sense that well, if you do this out, I will spare you the algebra. But remember that it&#x27;s cross cross cross minus cross. So on minus Lambda Times Negative Lambda Times two minus lambda minus minus one. So plus one. And that&#x27;s going to be equal to Lambda minus one. Cute. All right, so we can see that Lambda equals one as a multiplicity of three. It&#x27;s all all of the, um, Eigen values are Lambda equals one. They&#x27;re all the same, so that doesn&#x27;t necessarily mean it&#x27;s defective. But it&#x27;s It&#x27;s a good start because, I mean, if we had three different Eigen values and they don&#x27;t be heavily nearly independent Eigen vectors, so I&#x27;d be non defective than we can do anything, but basically we need to prove that There are less than three linearly independent Eigen vectors for this system, so we can actually just look, we don&#x27;t have to do the calculation. That&#x27;s necessarily so. We have our way up here. We can just see that if we multiply this by if he call it V one first. I didn vector by 100 right? Well, what&#x27;s that? It&#x27;s Rose Times column So it&#x27;s one times one plus 00 Anything multiplied by 00 and then zero times 100000 Right, So there&#x27;s on Lee going to be a the one that&#x27;s gonna equal V one So perfect. That&#x27;s in Eigen Vector with Eigen value of one. That&#x27;s one of our Eigen vectors. OK, so we can have a maximum of two more, and we don&#x27;t need to have any more, however, because this matrix is a thing that has not come into play yet. We will see that if we just define RV to and this is where the last video is going to come into play. V two is going to be Alfa. Where else was just any number time zero one negative one, and you will see that that is also in Eigen value. And obviously it&#x27;s linearly independent from that that is also in Sorry, Eigen Vector of Eigen value one for this matrix A and you can check that. And, um basically, what we&#x27;ve done is we&#x27;ve just If you remember from the last video the V two and the two by two matrix was I&#x27;ll foot times one minus one. So we&#x27;ve basically taken care of the Eigen vector for the ah third, the the added added, uh, spatial dimension. So it used to be our to announce our three that we&#x27;re operating in. So we&#x27;ve just taken care of that by making it super trivial, but is making it a one, and that allows us to just sort of focus on the interesting result that we already found. And you can generalize this to any number of dimensions. If someone asks you what&#x27;s ah, seven by seven major extent, you know, has a generalized Eigen vector. You can just go well but But but But but But the doctor talk, you know, and then put be at the very bottom all of these diagonals or ones, everything else zero. And that&#x27;s that&#x27;s all you need to Dio holds on iteration and cat. So if if you If you&#x27;re curious how we got this and how we got B then please, please, please go back to my other video. As I have said so we need to now find a generalized Dagon vector. So we we we&#x27;ve prove that these, um since this has a free variable that&#x27;s in the place of being a specific Sorry, it&#x27;s not true, but I&#x27;m Yeah, I did. I don&#x27;t know what I was saying there inside, huh? Yeah. So we can just take this away. We don&#x27;t really care about the AL Foods. Just still linear factor. Those aren&#x27;t super super important in linear algebra in general. So we now need to find a generalized Aiken vector because we have to Eigen vectors. There should be one more generalized Aiken vector. So how do you find that? Well, generalized Eigen vectors are of the form a minus slammed I to the end the equals you&#x27;re right. Where a regular Eigen vector is just n equals one. Okay, so we&#x27;re just gonna calculate a minus lambda I, um, to to to powers until we get something useful. So a minus land I squared is going to be, well, a minus Lambda, I What was that? That was Ah, this that we took the determinant off. I&#x27;m sorry. That was bad medication again. So that&#x27;s just going to be It&#x27;s right it out again. One money slammed a 00 00 and then be That&#x27;s negative. Planned? Uh, 11 tu minus lander. Okay, Perfect. And we&#x27;re going to square that, and I know I&#x27;m skipping a lot of the algebra here. Um, I trust you guys. If you really need to complete it into an online calculator that show great steps. I&#x27;m not really concerned about that. Um, made it this far in the class. That&#x27;s probably not the issue had on this. Um, conceptually. So I&#x27;m just gonna cut to the chase here. You&#x27;ll see that this equals 00000 then here we have negative one negative one. One more. And that&#x27;s for Lambda equals. I have already plugged in. Linda is gonna be one, right? So that&#x27;s perfect. That&#x27;s fine. Um, squared. So that&#x27;s squared is going to be well, that&#x27;s quite is actually just equal to you. Preserve the zeros and then you take the square of this matrix since we saw in the last one b squared is its That&#x27;s gonna be actually just one big zero. Feel free to check it out. It&#x27;s pretty easy to just see right there, that negative ones and the ones they work out very nicely. So any vector times this any vector v times a minus slammed I squared is going to the zero because it zero time zero. So we just need to choose a vector that&#x27;s linearly independent from these two guys. So take your pick. Um, my personal favorite is just always keeping it simple. The three is 010 That&#x27;s an example of one of infinitely many possible generalized Eigen vectors. So we go back to the question, What did we need to find? What we need to define this generalized Eigen vector. In addition to this, say that we wrote. And now you can see, because we found a generalized Eigen vector than, um, proof that a was defective that a did indeed, uh, satisfy the parameters of the problem. You can just in bed be into a Okay. Well, uh, I hope that straightens things that hope that all made sense. Um, everybody could look on your future mathematical endeavors

Patrick Vaughan · Answer

Hello. So today let&#x27;s do a more theoretical type problem. So let&#x27;s show that a arbitrary three by seven matrix of a, uh, has banality of four and it must have a column space. Let&#x27;s show that it must have a column space, um, where it&#x27;s in the, ah, third dimension and is the road space of a then also in the third dimension. So let&#x27;s review our rules. Rank ability zero. So our rank milady here, um, says that e rank of some arbitrary nature. A plus, the reality, the dimension of in all spaces day equals the number of columns. So in this three by seven matrix, we have three rounds, seven columns that we know the number of columns of seven. We know that the utility we&#x27;re assigning that to four and the rank we don&#x27;t know. Yeah, but using this rule, this tells us that the bank of a equals those seven columns minus the formality that were assigned, so that gives us a record three. So let&#x27;s build this arbitrary matrix. And let&#x27;s just assume that it&#x27;s already in produced garage one form just because that makes everything a little be easier. So we have one to you. Three rows. We have our leading one. Our next one. When I went after that, then we have seven columns. So 12 three for five, 67 And we&#x27;ll just fill battle up zeros to make this nice and easy. So this is our three by sudden Matrix. So and reduce little up. Throw up, Sean form The, uh we have the rank is three. So that&#x27;s the number of tipping point 12 three. And then that told us that the column space is the, uh in the reduced Raj lawn form is the Matrix Farm. But we&#x27;re just assuming about this matrix and the original Given. So not only is it released Rachael on form, it equals our original matrix. So our colons stays is actually going to be these three columns because that&#x27;s where a pivot points are. So the column space is the one. Oops, let&#x27;s undo that. See, one equals one 00 see, to equals euro 10 and C three equals euro 01 So we have three doctors in the column space, so this makes it such that the column space all day equals three dimensions because you could just arbitrarily say that this is you&#x27;re x dimension, or why dimension and your son dimension. So yeah, it must have a concept of that. So let&#x27;s done. Look at the road. Space doesn&#x27;t have to be three dimensions. Or is it three dimension? So the road space of our original matrix, But we just arbitrarily chose Do you already be in this reduced row? Actual on form 1234567 Okay, you&#x27;re 10 There&#x27;s one. So the roast space is the non zero rose. So we look at it. We have one hears looking on zero a non zero and a non zero. So the roast space in land such that. And let&#x27;s do this in blue or green. No one equals one 0 to 3 for 567 wrote to equals zero one 03456 seven And then our road three. I&#x27;m sure you can see where this is going. No. Is zero euro one, 4567 So, again, we can look at this and say that this is our excess. Why space? And once again we see that Yes, indeed, for this three by seven matrix that we arbitrarily chose. That the row space zooming back in. Now we don&#x27;t need to see that the row space, okay does indeed show up in three dimensions.

Runpeng Li · Answer

All right, so in problem 34 suppose A&#x27;s of three by three matrix and A&#x27;s B&#x27;s a vector in our three with the property. That Exodus p D has a unique solution. And we want to explain why the column Self Bay must spend our three. So Okay, so in this case, since we know the leaner system they X equals B is has a unique solution. So that means if we consider the or commended matrix and furthermore, if we considered the rule reduced form off this matrix, it must has has to ruin form like this 81 1 a two to something on the diagonal, diagnose and have some stuff there some stuff here. But it doesn&#x27;t matter. What are they? And we have zeros here, and I&#x27;m on the right hand side. We have B one b two B three as a constant. Okay, So this is the only impossible for this system to have a unique solution because only in this way we can solve a unique by the By the last row, we can solve the unique, unique x three, and furthermore that from the second road that X two will be determined by X three and x one would be determined by X two and x three. So this the only impossible way that we can have a unique solution for this system. So that means by observing this matrix, we have three people, two columns and three people. Columns means, uh, means axes. Um, yeah. So three people columns actually gives me, gives us the information that all the columns because A said three by three matrix And since every every single column is, is that people column So these columns are linearly independent, very independent, and furthermore, they must spend our three because in this case, we actually have three vectors and our three estimation three and those three Rector&#x27;s Arlene Early independent so thes three directors must spend our free so

Problem

Use matrix multiplication to find the image of th…

Problem 1 Easy Difficulty

What $3 \times 3$ matrix will have the same effect on homogeneous coordinates for $\mathbb{R}^{2}$ that the shear matrix $A$ has in Example 2$?$

Answer

$\left[\begin{array}{ccc}{1} & {.25} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right]$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Alayna H.

Kayleah T.

Caleb E.

Michael J.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 2

Video Transcript

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\left[\begin{array}{ccc}{1} & {.25} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1}\end{array}\right]$

Linear Algebra and Its Applications

Alayna H.

Kayleah T.

Caleb E.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Alayna H.

Kayleah T.

Caleb E.

Michael J.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications