construct-an-example-of-a-2-times-2-matrix-with-only-one-distinct-eigenvalue

Question

Construct an example of a $2 	imes 2$ matrix with only one distinct eigenvalue.

Gennady Notowidigdo · Accepted Answer

Okay, So for this particular question, we want to find a two by two matrix with only one distinct the Eden Valley. Well, just said construct an example. Okay, Well, I could construct about two or three examples, but your identity, my tricks, obviously has early one distinct. I value we know that because the characteristic polynomial, we&#x27;ll just be modest. I land a modest one or square, which is a perfect square. Okay. And obviously the salute when you solve this equal to zero, it has only one solution. So that&#x27;s one, um, another example. So and then, in fact, that&#x27;s probably point out if we said this would be I, um other ways lander here, but any multiple of I sew anymore OK, off. I also satisfies this condition too. Okay, Uh, the second example probably ended here. Is any so any back any matrix two by two. Which sort of have the same, um, same Daigle entries. But let&#x27;s say if I vary it say one and zero. So as long as one of the entries zero one of the off diagonal a treat, this will also sort of fall under the first case here, where the characters, although if also a perfect square, and I&#x27;m probably gonna leave it, that because it is pretty simple

Tim Strang · Answer

everyone. So, uh, today we&#x27;re looking at a problem that asks us just to give an example of a two by two matrix A. So it&#x27;s just trying out a generic A rules. A the C D. It&#x27;s bad beef anyways, and we want to get this matrix to have a generalized liken vector that is not a regular Eigen vector. And in other words, we do not want this matrix to have two linearly independent Eigen vectors. So we wanted to be defective because defective matrices of the only type of major sees that will have Jordan canonical forms that are not diagonal and thus will have generalized Dagon vectors as opposed to having only regular AC in pictures. So how we&#x27;re going to do that? Um, honestly, you can just play around with this, But if you&#x27;re watching this video, it probably means that you would prefer a little knowledge in the correct direction. I certainly did need to take a little bit of a formulaic approach. So, um, we think about our criteria that it needs to be defective. We can say that we need Well, first of all, there can&#x27;t be two different Eigen values because we know that different Eigen values have, um, linearly independent Aydin vectors. So the first thing we need toe find is, um, some way to make sure that a only has one Eigen vector and it will have a multiplicity of to So the way we do that So the way we find Eigen vectors as we take a minus slammed Ah, I and that equals a minus. Slammed. Ah, the see. Do you mind us? Lambda? Right. Okay. And we need to take the determinant of that. So debt a minus. When did I get that right? Is going to be. And we take this times this minus this times this right, So that&#x27;s gonna be a minus. Slammed. Ah d minus slammed. Ah, minus C be perfect. And to find the Eigen values uh, lambda, we need to set this. The determinant needs to be zero. So we need to have this all equals zero right now. If we expand this out, we will get a quadratic equation and sparing you some algebra. You get that? Ah, And this squared plus negatives to you minus a When&#x27;d ah, plus a D minus C B. That&#x27;s all gonna equals zero. That&#x27;s what we want. Right? So this is just a regular quadratic equation. You can check that it comes from this, um, that we know how to solve. So one of the solutions to this using the quadratic formula X equals negative B, which is gonna be d minus a so di for us. A closer minus the square root of B squared. So the square root of negative one is I mean, this this sorry b squared, So be is negative. One times d plus a soak in a straight D plus a squared deep, plus a squared minus four a C minus four is one. So a D c b all over too. OK, so that&#x27;s our equation. Now, when will this have on Lee one of value? Well, it gets its two values from this plus or minus. So we need the plus to be the same as the minus the only number that that is true for a zero. So, really, our criteria boils down to this term right here in the square root being zero, so we can rewrite this as D plus a squared minus four times a D minus. C B equals zero. And this is our equation that we need to fulfill in order to have a repeated Eigen value. So anything that fills this equation will have one Eigen value, and it will be repeated. That being said, that doesn&#x27;t necessarily mean that it is, um, defective. If it has one Eigen value, it could have won Eigen value with two wagon vectors. Right? But, um, there&#x27;s no real point to try and go into the super nitty gritty and, like, objectively empirically solved for the specific exact formula that will always give you a defective matrix. So, basically, this should just be a guide lying for you. It was for me. And then I guessed some numbers that fit that. So, for example, I guess 01 negative one to where again? I used to find like that. And it turns out if we check the Eigen vectors air. Sorry, Eigen. Values of this, um, we do get a repeated number. We get lambda equals one with the multiplicity of two. And then if we check the Eigen vectors that we checked the egg in vectors, will you have to solve zero minus 11 negative one tu minus one sold for perfection will make that equals zero if you multiply it so we can do an augmented matrix 00 That goes to two to minus 11 minus one one 00 So, in other words, row echelon form. This would look like this. Right? So that means the Eigen vector that solves this mm is going to be of the form, so V and we&#x27;ll call it V one with just mark this land a one. We&#x27;ll see. What the second, um so V one equals Alfa, where Alfa is just some number one negative one. And that is our only Eigen vector, right? You can&#x27;t have an Alfa that makes a different linearly independent Eigen vector. That&#x27;s just counter to the definition of linear independence. So I there was the first few numbers I guessed I didn&#x27;t use any formula to get this other than the one we wrote up here in blue. Um, just play around with that. It&#x27;s not too hard. You can use my example. Um can find another one. Obviously, if you multiply this by any constant, it would also have ah similar properties, but yeah, this this blue answer just gets us in the ballpark gets us, uh, guessing the right answers. Okay, so we know that this this a so a equals this we know that that is a matrix. It&#x27;s an example of a two by two matrix that has generalized Eigen Vector that is linearly independent from it&#x27;s regular Aiken Factors because it only has one regular Eigen vector. So we know that it needs to have another generalized Eigen vector in order to fulfill this Ah, again space to make it essentially non defect. It&#x27;s defective because there&#x27;s a generalized one. But it has a multiplicity of two and only when I can vectors, so that needs to be a general. So how do we calculate a generalized Eigen vector? Well, we know that a minus lambda I is this in an Eigen vectors when that equals zero. But a generalized Eigen vector is when they might have slammed the I to the end equals this. So let&#x27;s just see a minus slammed the I to the to sorry, generalized Eigen vector. I honestly don&#x27;t even remember what I said, but it would be the General Times. This would equal zero if n and it&#x27;s not one for generalized. Okay, So a minus land I too. That&#x27;s just a natural place to start is negative one negative. 111 squared Which you can check me on this. I&#x27;m gonna save you The Matrix Alger books. Not super important. That just equals the zero matrix zero to right. So that means that any vector we multiply will satisfy this equation for n equals two. So, in other words, any vector is a generalized Eigen vector for an equals two. So we can literally right down the general if each and is just going to be, I don&#x27;t know, like your favorite factor. That&#x27;s linearly independent from the V one. Uh, let&#x27;s just go 01 Perfect. That&#x27;s our answer. They&#x27;re infinitely many possible answers and also different A&#x27;s you can write, write. This was not unique, but yeah, that&#x27;s how you generally go about it, and you&#x27;ll see if you watch the video on the next problem. That ah, doing larger matrices actually is much easier than this because you can build on having a two by two defective matrix, uh, embedded within your larger dimensional matrix. So it&#x27;s actually this is the most amount of work you&#x27;ll probably need to do for any questions like this. Anyways. Hope you guys learn something. Hope this is helpful and good luck.

Fuzail Shakir · Answer

All right, So these matrix called militant manger ces and they are, like 001000 You square this, you&#x27;re gonna notice that this is just the roof complete zeros.

Samuel Goyette · Answer

Okay, So for this question, we want to find a such that a minus one exists so that I e. A is in vertebral but that there&#x27;s no p no matrix piano matrix D such net a equals p times d times the inverse of people. No, For a question like that, you can try to retro engineer something. So let&#x27;s just try a couple matrix or at least one matrix for So let&#x27;s try the identity matrix. So a that says the identity matrix. The identity matrix, of course, is convertible because we know that a times a would be identity. So a times a minus one equals a minus one time stay, which is the court? A which is I didn&#x27;t. So it&#x27;s a convertible matrix saddlery. It is diagonal sizable because you can simply write a equals identity, times, identity, times identity. And you have I mean day and he again, which makes this it it said because we want something without and diagonal ization. So you can with the small size matrix. I think trial and error is the best way to find a matrix with the properties you want. Uh, I&#x27;m gonna give you one Now you can find some other. So let&#x27;s try matrix a close 1101 Uh, let&#x27;s say that let&#x27;s have be be closed toe one minus 101 So a times B is equal to the identity matrix and B times A is equal to also equal to the identity matrix. So we can say that a minus one is equal to be so a is an vertebral, so we&#x27;re happy we can draw a small, tangy smile is a diagonal sizable. Let&#x27;s figure it out. So that&#x27;s so for Lambda in this equation, uh, equal zero. So we try to find the pagan values. So my right small find the A and values of a that gives us this, then in Paris is this equation years of the determinant off one minus lambda, 101 minus when the equal zero. So that&#x27;s a good idea to take a diagonal matrix. Therefore the determinant will be, and the egg and values are easy to compute showed. Determine it. I don&#x27;t need to write dead anymore. We only have one minus not grand. A minus lander. We have one minus and square equals zero. Therefore, one is the only Eggen value with hopefully a multiplicity of to for a so land equals one is the only Hagen value of Hey, now I say the only could be a multiple, but it&#x27;s the only alien value of a only relevant. So we have and I equals one. Now let&#x27;s try to find the Eggen vectors which corresponds to them the equals want. So we have a minus identity times x equal zero, which will give us a system that looks like this on matrix 40 except for the upper right corner, which is one 00 Therefore, what the system tells us is that X two must be equal to zero. Otherwise this a swan time zero plus X two equals hero. Well, we satisfy. So okay, any Vector X, which is of the form X one times 10 would be a egg and victor, So we&#x27;re gonna just take one minus zero his basis of pagan space. So a only has one A as only has one Onley has one again, Victor, what has an infinite number of big in Victor, but as only one, uh, basis for its agon space. So I&#x27;m going to write. Maybe I can write it before he so it&#x27;s very clear as one basis for it&#x27;s hi in space one vector in the basis for attack in space. Therefore, since one doesn&#x27;t equals two and A is a two by two matrix A is and diagonal sizable. So in order for a to be diagonal sizable a diagonal I&#x27;s book in order for it to be diagonal sizable, we would have needed a to have two basis and it&#x27;s Agon space, but it only has one. So is in vertebral because we found the inverse of A, which is was one minus 101 But while we try to complete the egg in Victor, we only find out that it was only one basis. Therefore, a isn&#x27;t diagonal visible. So here&#x27;s an example convertible matrix that isn&#x27;t thank unrealizable

Amy Jiang · Answer

okay, That&#x27;s not a equal 11 serves up. So based on this, if you take a term in it, uh, a minus, You know that I get the city Could set of Terminix of one minus Lambda one. So and then the lambda that has been negative. Landa Times one minus lambda. Is it good? So we have to again, Really? That&#x27;s lying is he could still win one. And the Matrix is diagonal Izabal based on their seven, but this is clearly not inevitable.

Alexandra Embry · Answer

one example of a matrix that is its own inverse. Is the Matrix negative? 100 negative one? That&#x27;s because if we multiply that matrix by itself, we get the at we get 1001 And so if this is our matrix A and it&#x27;s also our matrix a inverse, then we know that a times a inverse equals the identity and a inverse times A is equal to the identity.

Problem

Let $\lambda$ be an eigenvalue of an invertible m…

Problem 24 Easy Difficulty

Construct an example of a $2 \times 2$ matrix with only one distinct eigenvalue.

Answer

$\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]$

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

Linear Algebra and Its Applications

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Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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$\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]$

Linear Algebra and Its Applications

Lily A.

Kayleah T.

Kristen K.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Caleb E.

Samuel H.

Michael J.

Joseph L.

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

Linear Algebra and Its Applications