diagonalize-the-matrices-in-exercises-7-20-if-possible-the-eigenvalues-for-exercises-11-16-are-as--9

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 $(-2,1,2) .$
$\left[\begin{array}{rrr}{7} & {4} & {16} \ {2} & {5} & {8} \ {-2} & {-2} & {-5}\end{array}ight]$

Samuel Goyette · Accepted Answer

Yeah. Okay, so we have a three baby by three matrix eight. As we want to die, guys, this matrix has to bag and values so three and one for a to be diagonal sizable one of thes again values much must lead to an against place of dimension to And so we&#x27;re gonna figure out right now if that&#x27;s the case, by finding the egg and Victor&#x27;s off A on the basis of the egg in space, which is basically the same thing. So for lender One equals three. We saw this than your system of a minus three times identity times a vector X times of it with that times a minus threatening to tee times a vector X, which is equal to a vector full off zeros. That gives us a system that looks like this. So four for 16 and to to eight minus two minus two minus a times X one x two x three So the components of vector X and every team equals two 000 and basically all their line of this other lines of this matrix are the same. Uh, is there just multiple of each other So that first night is the second line, multiplied by two, and the third line is the second line multiplied by minus one. So we can only consider the second line if we won&#x27;t want to and divided by two cents everything in the Chinese. A multiple of two. And get this equations or the solution of this system were must satisfied his equation. So x one plus x two plus four x three equals zero. So the deck means that the basis of the against pace will look. Something&#x27;s this. So what? The X one of the bases will look like this minus x two, minus four x three, which means that X is equal to minus X two minus for expiry as the first component and x two ended x three as the 2nd and 3rd components. Therefore, X is equal to x two times, minus 110 at plus x three times minus +401 So those two factors here, minus 110 and minus force you want, are the basis for the egg again space. Associating with Brenda equals three. Therefore, it&#x27;s the Huygens space of dimension to so a is daikon advisable to complete and find two diagonal eyes form of A. We can compute another Hagen victor associated with Lambda one by solving just in their system. So a minus identity Times X equals zero. We can skip directly and write it as a linear system so six x a one plus four x two plus 16 x three equals zero two x one plus for X two, class eight x three equals zero and minus two x one plus not plus minus minus two x two minus six x three equals zero. So since we have four x two and 44 x two in the 2nd and 3rd line, I&#x27;m going to try have minus four x two in the third line. So if you find another way with more a quicker way to solve this system, it&#x27;s perfectly fine as long as you get to the right answer. So for me, I&#x27;m gonna multiply the third line by tools I&#x27;m gonna have minus four x one minus for extra minus 12 x three And which is of course, of course, to zero. And the second and third line will remain the same at the 2nd and 1st time were made the same. So two examples for ex troops plus eight x three equals zero and a six x one plus four x two plus 16 x three equals zero. Ah, Now where we doing? Um, I&#x27;m gonna have a line three. He called to 93 uh, plus line to and gonna have lying. One equals 91 minus nine to and that will give us the following equation. So four x one plus, uh, eight x three equals zero for, like, the first line that they&#x27;re lying We minus two x one minus four x three equals zero second and my sister I&#x27;m so two x one plus four x two plus eight x three equals zero. The 1st and 3rd line of this system are the same. So if we multiply their third line by minus two, we get the first line, so we don&#x27;t need to rewrite everything we were, and that will lead us to this equation. So X, uh, one equals minus two x three. So, basically, we just to the ext. Three to the right hand side of the equation and divided by four. And, um, the second nine will be will therefore become minus four x three plus four x two plus eight x three equals zero which becomes this so x two schools minus x three Therefore or effect er are Vector X will look something like this so x three times my next you&#x27;ll minus want one, and that&#x27;s the basis for the egg hiding space. Associate it with it. Lender equals one weekend. Therefore, write a equals so minus one once hero, first basis for first against face and minus four one. So did second places for the first time space and minus two minus 11 We write our agent values in so through pet 33 cents to egg and factors are so TT three and one with a bunch of zero outside of the diagonal and minus 110 minus four. CEO one minus two minus one poll minus one and a reality. We have to agonized A with P Times D, Times P minus one

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.

Amy Jiang · Answer

Okay, So what are your values? Are two and eight. So let&#x27;s start with a monitor speech. I that use me, you know, for 22 she didn&#x27;t give too weird for two to tune. The report came. This reduces to just 10 negative one. So there, one year of one&#x27;s own until zone. So case our general solution. Because sex three times 11 one. Our first alien space. It&#x27;s gotta be one is equal to 111 Now, where are you, dude? That&#x27;s a minus two. I give me this. 22222222 But I really sister this 11 one, Joe the rest of the girls. So we get General Solution is next to times. Think of 11 joke plus x three, which is never won one. Okay, so we&#x27;re gonna have to recluse here. It&#x27;s called a B two and B three. So are you. Explosion is gonna be Let&#x27;s do. This is we want me to be three. We have 111 negative. One 10 And they want so one and our dragon Army chicks think is gonna correspond to B one b two B three as eight. Gerald L 0 to 0

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

Samuel Goyette · Answer

Okay, so in this problem, we want to diet on ice. This matrix a three by three matrix. We know that five is one of its hag in values and that the vector minus 212 is an egg in vector off. So the first thing we&#x27;re gonna do to die agonizing Matrix is to find the Agon vectors for blander equals five. In order to do that, we&#x27;re going to solve the following linear system. So a minus five times identity matrix time vector X equals zero. So we solve this equation for X, we&#x27;re gonna get the following system. So minus 12 minus 16 and four. Ah, six eight minus two. 12. I raised a second line and we right there, just a bit more nicely. Eight. Okay, so they&#x27;re very light is 12. 16 and minus four times x one x two and x three, which is 00 Um, so all these lines are literally dependent. So we have We can see that line one is minus two times 92 and that 93 is two times 93 So the solution of the system will basically just look like mine, too. Which would which we will divide by 24 Simplicity shake and so ocean off The system will have to look something like this so three x one plus four x two minus X three equals hero. Therefore, x three is equal to three x one plus four x two So that means that are against pay the basis of our econ space will look something like this so x one x to three x one plus four x two as calor x one times 103 plus s collar x two times 01 That&#x27;s not 60 014 So these two vectors are the again the basis for the Agon space for Lambda equals five. So now we have three Eigen vectors since one was already getting in tow us now, as we must find his, we must find the egg and values associate ID with the Eggen Victor that was giving tow us so fine. The egg and value for the egg and vector Ah, minus tool one to eso No, we solve a new system. Except this time we saw for Lambda. So we have a minus land. Uh, times Identity times minus 212 equals zero zero&#x27;s and we need to solve four limbed that will give us minus seven minus and a minus 16. My enough. Minus four Just for, uh six 13 minus the Lambda and minus two, 12 16 and one Minus Glenda Times X one x two x three No, this is not x one x two x three We know these values already minus 212 equals toe a vector full of Xerox. I&#x27;m just going to write zero because I&#x27;m starting to elect space. Oh, we can rewrite our system. Ah, slightly differently. By multiplying I&#x27;m the 2nd and 3rd line. Uh, no. By multiplying by simply they computing this matrix vector products are. So now we just multiplied his vector by this matrix, which will give us this. So 14 plus two. Ah, then, uh, minus 16 plus a equals zero minus 12 to us on this. This is not in your system. This is a was the same thing. But this is a vector. You Well, much better. Minus 12 plus 13 minus and minus for and minus 24 plus 16 first to minus tool and equal. So that&#x27;s equal to a vector for the euro. So now we have a vector equal to another victor. And we get something like this when we simplify everything. So tool them. The plus six, minus lambda minus three and minus six minus tool and US equals 000 And all of these have one solution, which is Lambda equals minus three. So that&#x27;s our the remaining Hagen value. And therefore, we can write a like this. So a is equal to 10301 four. So those are the two alien values for them equals five plus the again Victor that was given. So minus +212 That&#x27;s our matrix. Pee wee writer died on matrix with the Hagan values. So 55 and not six, but minus three. So 55 minus three and a bunch of zero outside of diagonal and re simply inverse. Okay, zero for minus 212 and my sworn We know this matrix is convertible. Since all those pictures are literally independent

Samuel Goyette · 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

Problem

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

Problem 15 Hard Difficulty

Answer

$A=\left(\begin{array}{rrr}-1 & -4 & -2 \\ 1 & 0 & -1 \\ 0 & 1 & 1\end{array}\right)\left(\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{rrr}-1 & -4 & -2 \\ 1 & 0 & -1 \\ 0 & 1 & 1\end{array}\right)^{-1}$

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Calculus 3 Educators

Lily A.

Kayleah T.

Caleb E.

Michael J.

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.

Kayleah T.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$A=\left(\begin{array}{rrr}-1 & -4 & -2 \\ 1 & 0 & -1 \\ 0 & 1 & 1\end{array}\right)\left(\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{rrr}-1 & -4 & -2 \\ 1 & 0 & -1 \\ 0 & 1 & 1\end{array}\right)^{-1}$

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Caleb E.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Kayleah T.

Caleb E.

Michael J.

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

Linear Algebra and Its Applications