diagonalize-the-matrices-in-exercises-7-20-if-possible-the-eigenvalues-for-exercises-11-16-are-as--3

Question

Diagonalize the matrices in Exercises $7-20,$ if possible. The eigenvalues for Exercises $11-16$ are as follows: $(11) \lambda=1,2,3$ (12) $\lambda=2,8 ;(13) \lambda=5,1 ;(14) \lambda=5,4 ;(15) \lambda=3,1 ;(16)$ $\lambda=2,1 .$ For Exercise $18,$ one eigenvalue is $\lambda=5$ and one eigenvector is $(-2,1,2) .$
$\left[\begin{array}{ll}{3} & {-1} \ {1} & {5}\end{array}ight]$

Samuel Goyette · Accepted Answer

So we have a matrix A that we want to Dad, analyze. Let&#x27;s get started. So how first step to diagonals matrix is to find the egg in Victor&#x27;s at the age and values we started again by finding again values? Well, Ford it determinant of a minus them. The times identity equals zero. So we solve this equation for Lambda. We have three minus then, uh, minus 115 minus than the court zero. That gives us a nice in your non linear equation of the form three minus than the times five minus lambda, minus one times mine. This one equals zero. We want to solve this equation. We have 15 minus three Ramdas minus five. Lamb does purse rammed the square. Thus one call zero. We rewrite this equation by simplifying everything&#x27;s we get them to square minus eight. Them does 1st 16 he court zero Oh, we get than the minus four square. Its course is you. So you if you can see it directly, good. Otherwise you can use the formula to solve. Ah, second degree polynomial. The important thing is that you figure out that render one is the same as them. The to which is four. So we have one alien values with a multiplicity, too. And now we&#x27;re gonna find So Step two, we&#x27;re gonna find the Eggen vectors of them. Duh equals for so how to find? Okay, Hagen Victors, we solve this system so a minus four times and then to d. C for being Oregon value, times X equals zero. So zero is a vector for zero and X is it, uh, what we want to solve for? It&#x27;s a vector. So we have this linear system, these linear equation, and we&#x27;ll have something that looks like this. So minus one minus one 11 kinds x one x two equal 00 So that&#x27;s the system for this problem. What does it mean to solve that systems? It means that minus x one minus x two equals zero and x one plus X two equals zero. And that means that x one equal minus X two. Because those to a question are the same, you can simply multiply. Either of those two is equation by minus one and you&#x27;ll get the other one. So that means that X, which looks probably theirs so x two times not equal X two times minus 11 is our Eggen vector. So any multiple of minus 11 is our again Victor, So minus want one? Is is the Onley Eggen sector? Uh huh. Hey, so that means we can not agonize A So we can&#x27;t agonize a dia gone No lights, Okay, Because a means a needs and linearly independent Angan vectors to be diagonal Izabal dia Gone lies and it obviously doesn&#x27;t have And linearly independent again vectors. It only has one. And in our case, fan equals two because a is a two by two square matrix. So this is an example of a two by two matrix that we can not bag on arise because it has Onley one Eggen vector So we can maybe have it little conclusion. Therefore, a isn&#x27;t died otherwise and there you have it. So we found we find we found are a invent values by looking at the determinant of a minus lambda times identity and refined that we found that there&#x27;s only one pagan value for a Eggen vector for a Therefore it&#x27;s not final Diagonal Izabal

Amy Jiang · Answer

were given 5105 Okay, we know, eh? It&#x27;s causing a is triangular. It&#x27;s our only idea. Values five and five. Okay, so you have another five minutes to anyone is five I. That gives us there. A one goes out which said steak that X two equal to zero and we have that X one. It&#x27;s free. So are generous solution Compass. It&#x27;s one time ones out.

Amy Jiang · Answer

Okay, What is a minus? Lambda? I That&#x27;s two more is Lambda Three for one month, Islam, that now let&#x27;s take exterminated that to get the characteristic Well, you know me up. That gives me No! Do you like London? 11 is long. The and minus three times four that simplifies to land the squared minus three Landa minus 10. Excuse me. Land them when it&#x27;s five times on the plus two. So our alien values are *** two and five. Okay, let&#x27;s start with Monday is equal to five. So you met in a minus five? I Yeah. This is equal to negative. 334 Negative. For so seeing this is the stewardess multiple to shutter, right? They seem to be a factor. Or the values 11 multiplied by three. Or a negative 11 What? Deployed by three. And then multiplied by negative four. So we get that on. General solution is to 11 So Orion space is just 11 Okay. Now for that landed, you could too. Negative too. It&#x27;s all for that. We got a plus two I that&#x27;s just equal to 4343 We have another repeat. So we have that little boys do for three year old Joe. So we get that X one is equal to They got three over four. Next to an excuse free. So our general solution becomes X two times negative. Three over four one. So are again space for this one is negative. Three over 41 What? But simplify that. Or make that it&#x27;s a whole numbers. So we get this in the other three times four. Okay, Now it&#x27;s construct to again, um, spaces. So that&#x27;s this. What did we have? 11 in the 34 and our that. I got a delectable matrix 5002 where five has this matrix or has that again factor. And you go to your house this

Samuel Goyette · Answer

Okay, so we have a matrix A and want toe. Dagon. Arise. So what do we need to do and retired? Analyze a matrix? We need to find e Hagen value. So how to find Hagen values? We look for the determinant of a minus Lambda times Identity matrix equals zero. So we look for Lambda, which saw this equation. In our case, we will get the determinant of one minus lambda 00 minus one minus than, uh equals zero. And the determinant of the two by two matrix is pretty easy. So we&#x27;re multiplied to diagram components. And since there&#x27;s a zero right here, the sixth is not important. So we have one minus than the times minus what happened here? Times minus one minus them the equal zero. All right. Which means that and then the equals one so that no, once he corn and render two equals minus one are the Eggen values. Hey, now that we have ta of those of a what do we do? We find it a n vectors of so number. Step to fine again vectors. Oh, okay. How to find the Eggen vectors? Well, we so well in their system which will look like this before. All right. What&#x27;s from with my pen for then? Don&#x27;t want equals one. We have a minus the identity matrix time surgeon Victor X, which is equal to zero. So we want to solve for X and that equation X will be our egg and vector. That system will look like this. So 006 minus two times x one x two 00 So remember that when you write this, this is a bit of a notation. Zero denotes a vector of the appropriate size four zeros. Now we other than your equation. So sick six x one minus two x two equals zero. Which is the same as saying that X two is equal to three x one. Uh, so x looks a bit like this, so you will have X one. Oh, no equal. What did I do? X equals X one time one times three eso. That&#x27;s a solution for the vector one and three is a solution for this system and any multiple of that. Victor is also solution. But we just kind of take a more simple for impossible. So Ah, the first Eggen vector is one minus three is the egg and victor for Randa, one equals one, not minus one, but one. And we&#x27;re gonna figure we&#x27;re going to do the same thing for them, uh, equals minus one. So for them, they pulls mind this one. Now we solve that in air system A plus identity matrix, Times X equal zero once again to sell in their system we want to solve for X. Uh, so that system is this same as writing to 060 times X one x two equals two factor 00 Which basically, what does that mean? It means that we need a two x one. He holds zero and six x one equals zero. So that&#x27;s pretty simple to infer that X one will be equal to zero and x two can be any value because it will, in that linear system, extra will always be multiplied by zero. So the victor 01 we take the most simple form it could be. Any multiple of that vector is the egg and victor for Randa cools minus one. And there you have it. Now we have a in vectors Diegan values, and it&#x27;s pretty easy to write down the diagonal ization. So we have 130 one. So we have first Hagen Vector, Second A in Victor. The Matrix D will be one 00 minus one. So we have first Hagen value in second and value and choose. Now we have the interest of peace of one&#x27;s 031 minus one. And there you have it. We have our Daggett in that factory ization diagonal ization of a with the proper alien values and Eggen.

Bobby Barnes · Answer

right. So they want us to diagonal eyes, this matrix if we can. Um So let&#x27;s just start to try the diagonal eyes. This And for some reason, we can&#x27;t think we&#x27;ll just stop there. Now, The first thing. Um, since they actually give us these Aiken values, we just go ahead and set up our Eigen vectors Oregon matrices, and then we can try toe go from there. So remember, what we&#x27;re doing is it&#x27;s a minus. Lambda I and I&#x27;m just gonna be calling this matrix A So when lambda is equal to, um, let&#x27;s do four first. Help me. This is going to be a minus, or I, which should be 00 negative 2 to 1400 one. And if we went ahead and we&#x27;ll reduce this, um, we should get one one. Half 0001 000 Yeah. So this will now tell us that we have X one plus one. Half x two is equal to zero, and then x three is equal to zero. Just remember, this is supposed to be equal to zero for everything, because technically, we have this equation here like that. So, by reducing this. We just ended up getting that. This is equal to zero vector. And now we can say Okay, well, x is supposed to be actually, we need toe first. Solve this top one here because we would subtract x two overs. I guess X one is equal to negative one half x two So we&#x27;re gonna use that over here. Do you have x one x two x three And so x one is supposed to be negative one half x three I&#x27;m sorry x two x two was just x two and x three they say is zero. So we just plot x two. This is X two that we have negative one half, um, 10 And at this point, I don&#x27;t necessarily like having the fraction in this. So what I&#x27;m going to do, you just multiply everything by two on doing that won&#x27;t necessarily change anything. It just kind of changes that input. Uh, So what I&#x27;m going to use for the Eigen vector here is if I multiply that by two, it will be negative 1 to 0 and doing this just like I said, No fractions involved, but it doesn&#x27;t really matter. You can use this other one and still good. A good answer. Yes. So this is gonna be our Eigen vector, though, for Lambda is equal before that will use. Now we&#x27;re going to repeat the same thing, but with a box. Eso let me scoot this down here. So now if we had Lambda equals five a minus five, I is going to be equal to eso would be negative 10 negative too. 2525 to 04 000 And then if we go ahead and roll produced this, we should get one zero to and then everything else is a zero. So, um, this equation or the first equation tells us that x one plus two x three is a go to zero. Um, X two is free, and x three is free. So x 263 just free. Then that&#x27;s all we really get out of this. So if we go ahead, do the same thing we did before her action. First, we need to solve for X one. So x one is going to be able to negative two x three and now we&#x27;ll get negative. We&#x27;re actually, uh vs Rio x one x two x three So x one is going to be negative. Two x three Next to is just x two and x three is just x three and how we can write this So next to well, there is no excuse in the first or third and we just have one in the second and then plus X three. Um, there&#x27;s negative to in the first, None in the second and one in the third. And so then these two are going to be our two Eigen vectors that we use for, um, this one or ah, for what was it? Lambda is equal to five. So let&#x27;s go ahead and set up our diagonal ization of the actually mean. See how I set this up the first time when I did it. Okay, so I did four or 55 s, so it doesn&#x27;t necessarily matter the order you do this in. But I just went ahead and already wrote all of these down. So first, our diagonal ization early. So when I&#x27;m using is gonna be 455 along the diagonals And now to get p what we do remember, we line up our Eigen vectors with our Eigen values. So for Lambda being four, ours was supposed to be what was sick. And so is this negative 1 to 0. So you have negative 1 to 0. And then we could place, um, either of our Eigen vectors here, um, into either of these columns. It doesn&#x27;t really matter. But when I did this earlier, I put 010 in the negative 201 But again, you could flip these and get a valid answer as well. So now the one thing we need to look Teoh is fine. P inverse. You could do that however you want. Um, but the way I did it was I just set up a pay tricks, um, augmented matrix equal to the identity. And I just wrote produced it. So let me go ahead and do that really quickly. So negative. 10 negative. 210001 And then just 111 And then everything else zeros. And so, if we were to go ahead and produce this, um, this should give us. So now we have the identity on the left side and on the right will be negative 10 Negative too. 214 001 Okay. Yeah. And remember, this is supposed to be p adverse. So let&#x27;s go ahead and write out What are diagonal ization is now. So we have a is supposed to be so p D. P in verse. P was negative. 10 negative, too. 210001 are diagnosed. Realization D is 455 along the diagonals. Everything else heroes and then Pete in verse, we got to be negative. 10 negative 2 to 14001 So now, again, would you make it might be slightly different. Um, depending on how you set up your diet, you&#x27;re diagonal eyes major steep on. Also, since we have a repeated vector depending on where you are goingto repeated Eigen value depending on where you put those two factors that correspond to it, um, it will also give you a different P and verse as well as Pete. But as long as you like, multiply everything else and it gives you the same thing and they end, it really doesn&#x27;t matter. Eso At least this is what I got when I set it up

Bobby Barnes · Answer

so they want us to go ahead and diagonal eyes this matrix and they already give us our Eigen values. So let&#x27;s first go ahead and check out what we get when we do. Um, a minus. So it&#x27;s supposed to be Lambda I and so I&#x27;ll do first where Land A is equal to two. So in this case, we get negative to negative two, three, and then everything else stays the same for this would be negative. Four negative six negative one negative three 12 And if you were to reduce this, um, this should give us a 123 And then the rest of these were just zeros, because, I mean, this is actually not too hard to do by inspection. You could see here 123 that cancels out with second row and then multiply that by negative, or just do two times rotary onto that. That also council&#x27;s up. So even if you don&#x27;t have a calculator, this one&#x27;s not too bad to see. So let&#x27;s go ahead and get our Eigen vectors out of this. Remember, this should technically be equal. Uh, 20 matrix. Um, So what? This is going to give us now is the expression x one plus two x two plus the three x three is a go to zero and then that means X one is gonna be negative two x two minus three x three And so if we were to take some random variable Yeah, so x And then we had x one x two x three Well, we come over here and plug that in for X ones that be negative two x two minus x three and the just x two x three And then remember, we can break this up into just the excuse in the X threes, which is going to be so the X twos. So we have one in column are in rogue one. So that being negative two, we have one in Rome to which is just one and then not in row three. So zero and then plus X three times eso We have negative three x three in that first room Second nothing. And the first roar Third very. We have one. So so far we have these two hi conductors on those who look pretty independent. So no problems there. Then we&#x27;re going to repeat this, but with lamb day equal to what? So, actually, let me move this down. So over here. Yeah, we did. A minus slammed I with Lambda equals one that is going to give a negative one. Negative four negative six Negative. One zero Negative. Three, 1 to 5. And if you were to go ahead and reduce this one, uh, this should give us 102 011000 And so from this, we end up with two different one. So the first one being, um it&#x27;s all right to sell it here, x one plus two, x three physical to zero, and then we have x two plus X three is equal to zero. And so if we were to solve for X one and x two, that tells us X one is negative. Two x three x two is equal to negative x three And then if we were to do that same thing, set up that little vector. So, um, X is equal to ex wine up next to next one next two and x three So x one is supposed to be negative two x three x two is supposed to be negative X three and then x three is just x three. So we can pull that X three out when we get negative to negative 11 And so this is going to be our third Eigen Vector eso Let&#x27;s go ahead on pull both of these down here. I was just actually picked this up. Scoot it down. Okay, so our Eigen vectors with So this is Lambda is equal toe one and then these are Lambda is equal to two. So when we go about constructing P now, so we have p is going to be actually let&#x27;s say what D is first before we do that. So d I&#x27;ll say is one, 22 Because remember, it will just be our Eigen values along the diagonal and now key. We just need the what is called the Eigen value to match up with the Eigen factor. So for one, we&#x27;re going tohave negative to negative 11 and then for two we can use either of these Doesn&#x27;t matter, so I&#x27;ll do negative to one zero and then negative 301 So this is P and now you go about finding p inverse. However, you want. Um, the way I normally do it, though, is I set up a augmented matrix equal to the identity. And then I just wrote, reduce it. So I went ahead and actually already did that. So let&#x27;s go ahead and set that up really quickly. So we do. Negative two. Negative to negative three. Negative. 110101 And then I want this to be 10 not 10010001 And now we&#x27;re going to reduce this, and, uh, let me see where I wrote this down. Eso This should give so 111 So we have the identity on the left now and then. Over here, we get negative one negative to negative. Three negative one negative. One negative. Three. 124 like that. And we can go ahead and use this over here as p in verse. And now to get our actual diagonal ization. Remember, we&#x27;re saying it&#x27;s supposed to be a is equal to P. D. P. In verse. We just go ahead and write this out. So P was negative. Two negative to negative. Three Negative. 110101 de was 1 to 2 along the diagonals everywhere else. Zero and then p inverse is negative. One negative to negative. Three negative one negative. One negative. Three, 12 Right, That but better. 124 And so this is going to be our diagonal ization of that matrix that we started with. And again, This is not a unique diagonal ization. You could get something different. Uh, if you put your Eigen values in a different order. Um, since we have one Eigen vector that goes with or to begin vectors that go with one Eigen value, you could switch those two, and you should still get an equivalent one. Um, you just need to make sure that whatever makes whatever way you set it up, you&#x27;re just being pretty exact. But again, it doesn&#x27;t really matter how you have it set up, because again, there is multiple ways to write this Diagon localization. And as long as you plug this into your calculator and it spits out the same answer, that&#x27;s all that really matters.

Problem

Diagonalize the matrices in Exercises $7-20,$ if …

Problem 9 Easy Difficulty

Answer

diagonalizable

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Anna Marie V.

Heather Z.

Kayleah T.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 5

Video Transcript

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

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Heather Z.

Kayleah T.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

diagonalizable

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Heather Z.

Kayleah T.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Anna Marie V.

Heather Z.

Kayleah T.

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

Linear Algebra and Its Applications