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Suppose that a four by four matrix A has vectors va, vb, vc, vdthat satisfy the following properties Ava =vc, Avb = −2vb,Avc = −va, Avd =−2vd + 3v...

Question

Suppose that a four by four matrix A has vectors va, vb, vc, vdthat satisfy the following properties Ava =vc, Avb = −2vb,Avc = −va, Avd =−2vd + 3vb. What is the Jordan canonicalform of A and what is the transformation matrixP?

Suppose that a four by four matrix A has vectors va, vb, vc, vd that satisfy the following properties Ava = vc, Avb = −2vb, Avc = −va, Avd = −2vd + 3vb. What is the Jordan canonical form of A and what is the transformation matrix P?



Answers

Find the Jordan canonical form $J$ for the matrix $A_{1}$ and determine an invertible matrix $S$ such that $S^{-1} A S=J$. $A=\left[\begin{array}{rr}4 & 4 \\ -4 & 12\end{array}\right]$

Heavy one. So the question we're looking at today gives us this matrix A and, ah, the first thing it asks us to do that we'll just jump right in its toe, solve this and, uh, for its Jordan canonical form. So in order to do that, first we need to find the Eigen values. So we need to find the solutions to the equation. Determined of a minus slammed. I equals zero and we're solving for Lambda. So we're going to write that, Remember? Well, first off, what's right? A mighty slammed I seven minus slammed, uh, for minus lambda or minus land. Uh, and it's the same right. In the determinant of that, I remember how to take the determinant of three by three matrix ease. It is Ah, it builds off your knowledge of determinative to buy two years. So it's going to be this element times the determinant of this matrix plus this element times the determinant of this matrix. So you're basically per muted it and then plus this times the determinant of this. So let's write out all our different components here. So you have seven minus swim. The times four minus land squared minus minus one. That's plus one. Then we have negative to just negative, too, because it's negative. Two tons, uh, negative one squared, and then the other term is gonna be a zero. Okay. And then we have 2000 0 minus negative for post slammed us. That's just two times four minus Lander. We need to some these All right, so let's just right that all out. And I'm going to skip over some algebra in simplifying this. You guys don't need to watch that. You have trouble on the algebra. Then you can go to an online calculator with, uh, the steps out. There's a very nice ones out there, but that's not really the crux of this problem. So I'm not going to address it. So we get finally because we want this all to be zero. We get zero equals Lambda cubed minus 15. Lambda squared. It was 75 Lambda minus 1 25 Perfect. So that's a characteristic polynomial. We're setting it equal to zero, and basically, you just need to fiddle around again. I'm gonna skip the algebra or, uh, your teachers said it's OK. You can use, um, some numerical methods or online solvers to do this, uh, cubic equation. But it actually all shakes out after Cem work and educated guesses into Lambda minus five. Que perfect. And that's equal to zero. And, uh, you could just check. Sort of that, uh, this actually just calculated out Lambda minus five. Cute if you want, and you can sort of reverse engineer how you get that from this. And so what this means is that there is only one I can value, so we don't need a subscript when two equals five, right? And it has a multiplicity of three. And that's gonna be important. So we need to find how many distinct linearly distinct. So linearly. Independent Eigen vectors there are for this matrix, which is the same is finding the dimension of just this Eigen space since there's only one and we need to find that because that will give us the number of Jordan blocks that were working with in our Jordan canonical form. So had we find Eigen vectors, but we set a minus That was a terrible minus heavy, minus one and I times are Eigen. Vector is going to be easier, right? That's how we do it. It's a definition of Eigen vector in your solving for V. So we can set up a augmented matrix. And I'm subbing in from Ah, this a minus Lambda I right here on Southern in land equals five. So seven minus five. That's two to to zero. Negative one negative one. But if one 14 perfect. And we want that to equal zero. So we're gonna get this into a reduced echelon form, right? So you can divide the first row by two. Because it's gonna be zero. So one negative 11 We see that that is the negative of this bottom row. So we can just add them together. And this bottom round was gonna go to zero. Perfect. We have negative one. Negative one. Okay, so then we can add these two together, actually going to subtract them. So here we can divide these out my negative one. Because that's not important that we're going to get 10 to 011000 Yeah. Uh huh. Sweet. So that is our role reduced echelon form, as you can see. So we are ready to write down our vector V so V equals the last component needs to be negative two times the first component and negative one times the middle components. So we can write our first component. That's going to be Let's just put it as one, and then the last needs to be negative to write, and it needs to be to for the middle. And you can check that that all works. And of course, we can multiply this by any real number A. And that will still work because, well, we just arbitrarily chose one. It could be any multiple and still satisfy this. It just matters the, uh, components relative to each other. Okay, so that is our Eigen vector, and you will see that there is only one. There's only one option we could have had up to three. Because the multiplicity of this I can values three with the dimension of the Agon space is just gonna be one. Which means that a is a defective matrix. So it's only going to have one Jordan block. So what that means is that we have to sort of glued together here. Let's write out Air Jordan economical for so we know there's gonna be fogs along the diagonals zeroes over here, and we're basically just gonna glue these all together into one big Jordan block. But big ones up there. There we go. That's our only option. We only have one blocks work with. So that is going to be the answer to the first part of our questions. However, Clifton also asks us to find in Matrix find a matrix s satisfying Jay equals s inverse times A s. So weaken, basically construct J from a with this. But this ah matrix were about to find. It's the equivalent of a diagonal ization matrix for a Jordan canonical form. And we need to find the generalized Eigen vectors, right, So we need there to be three of them, so we need the length of the cycle of generalized Aydin vectors. So, um, we have a theorem that the size of the Jordan block equals thelancet of the cycle for the generalized Eigen vectors. So what we need is we have two criteria, so the length is three. So the terminating power is gonna be three. So we need a minus five I the third to be equal to zero, and we need a minus five I Sorry. Time. Some vector will call it fee one, because we're too used to be up there. No v one squared to not equal zero. Okay. And if a minus five I squared times V one doesn't equals here, then obviously a minus five I times V one won't equal zero. Okay, so let's go about this. This is just gonna take some rather unpleasant matrix multiplication. But we have our A minus five I right here. So we can just copy that town okay. And squared. That's gonna be equal to again. We'll see if you guys the algebra. Alright, so I've done out the calculations, so we have a minus five. I squared is going to equal this. And then if you look here, you'll see that a minus five I cubes. This is a pretty simple check for you guys is going to just equal, uh, Big three by three matrix full of zeros. So you can sort of see from the symmetries in these major cities if you want to do it out, but yes. Oh, anything Times zero is obviously gonna equals zero. So actually, RV, one can be anything that is linearly independent from this Eigen vector that we've found because we want them to be generalized. Eigen vectors. Right? So, um well, what did we find? Let's see, if 100 works, that's always a nice one. So we just need to check that a minus five times V one does not equal zero. So, a minus five I squared times ones here, a zero that's going to be equal to Well, it's just gonna be to zero negative. Perfect. All right, that's not zero clearly. So we are set. We're ready. So we need our cycle. Are cycle is going to be a minus five. I squared V one a minus five. I fy one the one right. That's just the definition of our cycles that we're doing here for generalized Eigen vectors. And it's a blank three. That's perfect. And basically, RS is just going to be This is gonna be the first column that's gonna be the second, and this is going to be the third, and it's very important that you keep them in those orders. Um, not necessarily for this problem, because it has only one wagon value, right, and it's all sort of the same it's all sort of symmetric that I'm It's very important that you keep those in general. So we pretty calculated a minus five. I squared times V one. What's a minus five pi times V one. So that's just going to be, well, the same things there, too. Negative one. I was sorry to. This is incorrect. This is supposed to be sorry, guys. I looked at the wrong things firstly to one because a ticket of this matrix, it's supposed to be over here and this is the one in the middle. So but a minus five i v is the vector I accidentally calculated earlier. And that's going to be 20 negative one. And of course, our V one 100 We've just chosen Ah, somewhat arbitrarily, but not quite. Okay, so now we are ready to write r s s equals. Well, it's to one negative. One, 20 negative one and 100 Perfect. And that is only need feel free to check that, but this is for all intents and purposes are answer right. It's just like diagonal ization. But the slight camp yacht that it's not actually diagonal, But I'm Yeah, So this is pretty standard problem. It's very good for you guys to, uh, really get familiar with it. I hope this is helpful. And, um, everyone, Good luck.

Sorry about that. The problem we're looking at today gives us a matrix. Give says gives us a equals 5110 for zero. Negative. One needed. One three. Perfect. Gives us this three by three matrix A. And first off, it asks us to find the Jordan canonical form. So let's do that real quick. You know what to do that you need to find the Eigen values and then determine how many blocks there are. So the Eigen values take the determinants of a minus. Lambda. I set that to zero and sold for a equals zero. Sorry. Sulfur lambda. So that's gonna be debt. Ah, five minus land. Do four minus land, uh, three minus slammed. Perfect. And so, if you remember for three by three matrix, that's going to be this times the determinant of this matrix. Then we can ignore the zero term, because that's going to multiply 20 plus this times the determinant of this matrix. Okay, so that's going to be equal to it's right in two lines because we're running out of space. But I want to keep it all together. Five minus landed times for minus slammed it times three minus lambda zero. So that's good. Just close off these parentheses and then summing so minus one times the determinant of the blue. So let's just keep this here. This here minus one times one time 00 minus four minus slammed us so that slander minus four. And so this some of the green and the blues is going to be easier. Perfect. So basically, we have written all out and then we can just change this to a plus for my Miss Landau. So you see, we have to common terms so we can sort of factor these in here to make it easier on ourselves. Right? So we see off the bat, Here's one Eigen value lamb. People's 45 vehicles for this whole thing will equal zero. So write Lambda one equals four, and then we can have up to two more hiding values here. So we have to solve that. Let's say that dotted green thing equals zero instead of four minus Landau so that factoring doing foil right 15 minus three lambda minus five Lambda plus slammed us squared. And then we have that plus one term. So we're sitting that equal to zero. So squared, minus it. Lambda plus 16 is gonna equals zero. Then you can either complete the square. I'm just gonna quickly do the quadratic equation so and equals negative B plus or minus square root have eight squared minus four times 16. Well, that's just gonna be eight squared. So this is gonna cancel out, and that's gonna be over to a So Lamda equals four. And that's going to be of multiplicity to write because it is repeated. Um, remember, if if this equals zero, then the plus or minus is just going to be at eight plus 0/2 and eight minus here over two or the two other Eigen values, so they're gonna be the same. So that's going to be Lambda Two equals four. And that's going to be a multiplicity where m two equals two. Because it's repeated twice the case. We have her Eigen values. Now we need to determine how many, um, Jordan blocks there are. So the amount of Jordan blocks for each Eigen value is gonna be the amount of linearly and Eigen vectors for that wagon value. And oh, wow. Well, I just noticed that all of our, uh, Linda's. Therefore, so is actually not going to be a lambda. One Orlando to it's gonna be land equals four. And the multiplicity is going to be three. So you gotta watch out for that. If you solve them separately, make sure you remember it would have cut myself later. But, uh, Sam equals three. So, um right, So we need to find the Agon vectors for this to see if it's defective or non defective, or how many? How many Eigen vectors there? Okay, so how do we find Eigen vectors? Well, we take our A We subtract Lambda Times, I So that's going to be five minus 44 minus four and three, minus four in the diagonals. That's 111000 negative one negative on negative ward. And we can do it augmented matrix because we need to find what multiply did this equals zero, right? Because it's a minus slammed I times. The Eigen vector equals zero 000 You could see pretty clearly this is going to go to Inro echelon form 10 negative one off. The roads are the same. So it's just one copy of one of them. And so we have to free variables. So that's going to give us two different Eigen vectors. So the one is going to be, Well, let's just write this. I like this method more generally, so V is going to be of the form. So there's no restriction on this column right on the middle column. So that's going to mean this chemical anything right here. The second component, the first component, plus the last component. Um, first component, minus last components gonna be zero. So they have to be the same. So we can just write b e and then weaken separate that out into right. So these are gonna be R V one and V two at two different Eigen vectors. But are multiplicity was three. So this still is defective, right? Um so the number of linearly independent Eigen vectors is too. So we have a theory that says that's gonna be the number of Jordan blocks that we have as well. So without loss of generality, weaken right, Jay equals, Will we know that it's gonna have four along the diagonals, right? And the, um, tradition is the standard form is to write the largest Jordan block on top and then do the smaller ones. So we have two different Jordan blocks in three different numbers. So we need to combine two numbers into one Jordan blocks that's gonna be there and then the other one is already one. So that's to Jordan blocks. That's the only way we can do this. We don't need to look into it any further, right? Just fill in all the other zeros. And so that's gonna be RJ. That's the first part of the question. Now we need to find a s such that we can get this and Shea from a by doing this operation. So it's essentially diagonal izing, but for just Jordan canonical form, Right? So, uh, normally for diagonal ization, our columns are Eigen vectors corresponding to the Eigen values. And that's gonna be true. And it's it's ah adorned bulk of one because that's analogous. So this last column of yes is going to be the Eigen vector is going to be one of these Aiken vectors, and then we need to find the generalized Eigen vector cycle for this right for the 1st 2 So let's find the generalized Aiken value cycle because that's gonna be the most difficult to find. Really found the to Eigen vectors. So how do we find generalized Eigen values? Well, it's a block of size two, so you know, it's gonna have cycle of two, right? That's a fear. Um, So what is the cycle? We need to find a vector. Such that a minus four. I sets landau v comma the That's gonna be our cycle. Those are gonna be are too left most columns. And, um because this is a cycle, we know that a minus four I squared times v needs equals zero. And these both need to be don zero. Right? So let's calculate a minus four. I squared. Well, that is Did we put for a minus four? I yeah. All ones. All zeros, all negative ones. Perfect. And just do out the multiplication from the A three by three. That's gonna be zero is gonna be zero. It's gonna be zero in, since all these are the same. We see that this is just the zero matrix, and that is a minus for I aa minus four. I squared. So any vector we put their times a minus four I squared. Any V is going equals zero. So we just need to choose one that is linearly independent from the Eigen Vector, which using here, right, Because we need them to be linearly independent. So let's choose the simplest Eigen vector There. She's V two. All right, so it needs to be linearly independent from V two. And it actually also needs to be linearly independent for B one because we have the condition that this is non zero, right? You know, we solved for V one by making you do the sums actor that makes that zero so linearly independent from both of these, actually. So we can just Teoh this. There are many options you could do here, but let's do V equals to zero negative one. Why not find one that you can just quick and easy clearly see, is linearly independent for both is okay. Now we need to calculate this cycle out fully completely. So we have V and we need a minus limbed all u times v. So that's negative one, right? No sorry. Negatives code in the last column 111 to Syria three, three, three. Perfect. Okay, so we have our four sorry three columns ever gonna use. Right? I said, we're going to do this. And that has to be our last one. This corresponds to blue, and then we need to do these in this order. So first, it's going to be if and then it's going to be this for SS. And it's just the columns are the Eigen vectors Didn't generalized. I connectors. So first is 333 and we have was a 20 negative one that we have 01 zero and that's it. That's our s. Um, yeah, it's not unique because we chose just to random one that wasn't in the, um I wasn't in the span of these two, but this These problems are very easy to check. Just take the inverse there. Lots of online calculators. If you don't feel like doing it yourself. It's not too hard with a three by three. But take the inverse, make sure it satisfies the relationship we started with. And, um, if it does, then you're golden. Uh, yeah, that's that's the question. I hope this was informative. And ah, good luck.

You're going to sort problem Number 80 from the section matrices. This problem we had to find the Jordan canonical for as I love the inaudible Madrid's Yeah. Here, the inevitable Metis is given us minus one cereal, cereal minus. Do we know that U minus lambda I in the X equals zero? Let lambda would be. It goes to minus one for Lambda One Indo I will be minus lor zero Siddle minus one, then a minus. Lambda I into its will be a minus. Lambda I Indo eggs it would be and close to it will look like fearful peril zero minus one Indo. Excellent. Excellent. Thanks to we just equals zero in this little get X two equals zero. Excellent Can have any value. So you invited that one. So the veterans will be X one fiddle. But if one come on, Phil Similarly, then Lambda Tau equals minus two a minus. Lambda Tau Tau I into X equals zero. This will be getting us a man. Islam backed away. So I will be one Filo Filo zero Indo It's learned X two. This is a considerable. We'll get X Article zero extra can help any value So designers be see, you know, comma extra clinics to be zero comma. One we will get. Yes, it was x 10 x 10 Then zero a stool. Then we can write as in worst. Yeah. Yes. But if Because as in West Aliases one zero theater the U. S. Minus one. Serial Sierra minus two. Then again as one Cyril Cyril, off market bodies to reserve begetting us minus 100 minus two. India this Look one. Cyril, Cyril, this really get research Pass minus one. Seal zero minus two. That standoff question. Thank you.

Everyone. So the problem were given. Today starts us off of the matrix A equals to negative to 14 three negative seven to planets and upper triangular matrix. Because everything below the diagonal a zero and first things first. It wants us to find the Jordan canonical form. So we're going to know that the diagonal of the Jordan canonical form is gonna be the Agon values. Let's find the item values. So to do that, we set the determinants of a minus, slammed I to zero and solve that characteristic polynomial for lambda. So a minus lander, I that's going to be tu minus land. Uh, negative to 14. That's a 10 14 0 three minus Landau. They have 700 to minus lambda. Okay, we need to take the determinant of that and set it to zero. Well, what's the determinant of three by three matrix? It's this number times the determinant of this matrix, right? Plus this number tons the determinant of this matrix. But because it has two zeros here negative, sometimes zero minus two money slammed. A times here is gonna be zero. So that counts for nothing. And similarly, this sum is going to go to zero because the determinant of this matrix and blue times 14 the determinant with matrix and blue is just gonna be zero right? Three of its elements are zero, and they're all multiplied at least one other and zero time 00 So the only thing we have to worry about when this determinant is this first green part. All right, so what's that? That's the ah to minus Lambda Times three, minus lambda Times two minus lambda Because it's times this minus this products. But that products gonna be zero. So it's just that it's just down the diagonals that actually works for any upper triangular matrix. So just a quick trick. I did it out just to show you why it works because, uh, I'm not sure if the theorems in your book, but, um, it probably is somewhere, but sometimes it's just helpful toe realize why you have what you do, and it is pretty easy to show. So that's gonna be zero. So we see that Lambda one equals two isn't Eigen value, and that has multiplicity of two because it's twice in this characteristic polynomial, so just write em one equals two and Lambda too cools three. And that just has until vehicles one multiple cities have to add the three right, because it's a three by three matrix. Okay, so the next step is we need to know how many join blocks there are, because we do have one Eigen value with multiple Steve to use that could either have one or two Jordan blocks in it. So we need to find the number of linearly independent Eigen vectors in the Eigen space or window one equals two because that's going to be the dimension of the argon space is gonna be be number of Jordan blocks for the ag value. So the way we find Eigen vectors is we do nt minus two. I times 50 will call V one because you know that Lambda two has an Eigen vector associated with an equal Sterno. Okay, so a minus 22 I, that's going to be 010 on the diagonals. It's your 10000 scroll up here. That's negative to 14 and negative seven, and we need to solve this for 000 right? Just set up in augmented matrix. Now it's easy to see that this row is negative two times this row. So they just cancel out and we actually get the very simple matrix. Um, 01 if seven. All right, quick and simple. And that means that r V one, we solve it. So the second component, minus seven of the third component, needs equal zero, so we can make the second component seven times the third component just like that, then our first component can be anything we want. So we're actually gonna We can multiply these each by any number, right? Say times a and then the the first component can be anything. It doesn't have to be a otherwise would be one times a rights we can write B and that actually, we can split into two vectors. So let's call him. He won a and he won't be. So. You wanna a equals a time 071 and b one b equals be times 100 Right. And if you have these together than you get few one. So this is there to Eigen vectors for this Eigen value of two. So that means it is actually non defective. Um, it has the dimension of two and a multiplicity of to. So that's perfect. So we can just write out our, um, Jordan canonical form. And because it's not effective, it's just going to be a diagonal matrix with the, uh, Eigen values down the middle. I'm perfect. That's it. Um, there you go. We have a joint economical form. But then the question asks you to find the transformation Matrix s such that we can get J from a by doing this right. And so you know how to do this? This is the last section. This is diagonal ization. So because of the Jordan canonical form is just the diagonal form, this s is just going to be a Each column is going to be the Eigen vector for the Eigen value that it lines up with. So we know the 1st 2 Collins, they're gonna be V one and V one b. So we need to find the Eigen vector for Lambda equals three. So, again, that's a minus three I. So that's going to negative 10 and negative one on the diagonals. So native born zero negative one zeros. Just make sure I got it right. Let's scroll back up negative to 14 and snacking seven. Perfect. And this is gonna go to zero, and we need to solve this, and we'll call it V two. So, uh, we need to solve this and find the vector. So that is going to be, well, these air just linear multiples of each other. So they're gonna cancel out, Um, negative one negative. 20001000 And we can erase these negatives just because we don't need him because we can multiply the whole thing by negative one, right? And then that's that's just it. That's in real reduced form. So we can write the two equals. Well, the third element has to be zero, so you can just put that in there. The first element has to be negative two times the second element. So we can just put one and negative, too, and then that can you can multiply that by anything so we can call it. See, it just has to be of that form, all right, because zero times sees just gonna be zero if we plug it into our A minus Lambda High times V two equals zero. So that's our third again Vector. And now we have to put them in the order in which we listed are Eigen values in the Jordan canonical forms. That's 22 and three. So we need to do that to their that their the first the 1 81 b and then B two so as equals, if you want a 1071 and then 100 right, and then V two is negative two 10 Perfect. So that's our s. And that's all we need to know. We don't need to calculate the inverse, although I highly recommend it just to check your work if you're doing it yourself. Obviously, I've given you the answer here, but for future reference, and you can check my answer. If you multiply a out as such, you should get back this this J and that's if you've done that. It's hard to have done anything else around. So a good litmus test at the end. But yeah, that's the problem. I hope you guys learned something. I hope this is helpful


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Moeorh L-nta rolnis] DETAILS IkestAT1T 8,6.,037. MY MOTES Ask YOVe TEACHIR Woea Kmo hundrrdecr catonl n Cavlae arag Maon Io neln 14Ainchi: Find tne 904 canldence teo Jnu |Opulst 7n ncin Icnain Lower Lmnil pouLoLon Ma naald doatica 6 2 9 ncnul; (Giye Fouf Bnbror cortect to (42 @ecunai Brcot) Uppor L...

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