construct-a-nondiagonal-2-times-2-matrix-that-is-diagonalizable-but-not-invertible

Question

Construct a nondiagonal $2 	imes 2$ matrix that is diagonalizable but not invertible.

Amy Jiang · Accepted 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.

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

Gennady Notowidigdo · 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

Griffin Johnston · Answer

suppose I had a square matrix. A So a is an in by in matric it has the same number of rows as column. And suppose also that the road space of a is equal to the null space of a no space of a And just as a quick reminder than no space of a is all the vectors x the set of all the vectors x such that a times the vector X is equal to zero. It&#x27;s all solutions to the homogeneous equation with this matrix. No, we want to prove in this video that a prove a the A cannot be convertible now, like I talked about in previous videos Ah, good first initial investigation for proof problems especially like this one. We&#x27;re trying to show that a or some some some mathematical object in this case AIDS and matrix. We&#x27;re trying to show that the A cannot have a certain property. In this case, we&#x27;re trying to show that they cannot be in vertebral, usually a good first initial approach to kind of get your hands dirty, and sometimes it ends up being the proof. It ends up turning into a foolproof, but regardless a good initial approaches to attempt to prove this by contradiction. So what is our goal? We try to prove something by contradiction. Well, we first suppose the contrary to suppose suppose a can be convertible can be convertible, then our goal, our goal when we suppose this sister derive a chain off implications. So if a driver change implications, what I mean by that is supposed taken, be in verbal without implies something. And that that&#x27;s something implies another something, and so on and so forth and to eventually we this the end of this chain of implications. We arrived at the end of his chain of implications. We arrive at some contradiction and maybe not the end, but one of these statements in here in this chain of contradiction, sometimes it&#x27;s we we stop when we arrive at a contradiction, so it ends up being the end of our chain. But it&#x27;s not necessarily the end of the Cheney. This infiltrated implications could go on forever, but regardless we eventually arrive at some contradiction. And when we arrive at a contradiction, that means that if we were supposed, if we were to suppose a can be in vertebral, then that would imply a contradiction. It would imply a statement that is absurd. That contradicts our original assumptions Are original assumptions that were given to us to be true for this particular problem. And thus this statement right here Supposing a can be in vertebral that can&#x27;t possibly drew a cannot be in vertebral if you were to arrive at a contradiction. Therefore, A Like I says, I said a cannot be convertible and we&#x27;ve proven a statement. So think for a minute what would happen if a is in verbal and make sure we want to leverage the fact that the road space of a in the north face of a are equal. So what would happen if a is and vertebral and we also have that the roast base of a is equal to the no space of a So I&#x27;m assuming you&#x27;ve had a go at it. So when a is in vertebral by the in vertebral matrix here, Um, all right, the in vertebral matrix here, um, there&#x27;s two things that we know. One of the things that we know if that if an matrix is in verbal, then the Onley solution x to the homogeneous equation A time of the vector X is equal to the zero vector. The only solution is the trivial solution. So X equals zero Vector is the Onley solution to this equation the Onley solution to this equation the Onley solution to this equation. But if X is equal to this year vector that that&#x27;s the only solution to this equation. Remember the set of all acts such that access solution to this equation is the knoll space of a So that means that the north face of a the north face of a is equal to the set containing zero vector and that&#x27;s it. That&#x27;s the only vector in the north place of a now something else that thean vertebral matrix theorem says is that if I&#x27;m matrixes in vertebral, which we are supposing a to b in vertical, remember by our by our our attempted a proof of contradiction proof by contradiction, we&#x27;re supposing that a is in vertebral. So if they were to be in vertebral then this will be true and also by the in vertical Matrix syndrome A is row equivalent to the end by in identity matrix And check out what we know we know that if two matrices are row equivalent, then that means they&#x27;re row spaces are equal. So the road space of a is equal, since a is row equivalent to the identity matrix because a convertible by the in vertebral matrix serum again, we&#x27;re supposing it to be in vertebral here, just just to see what happens to see if we arrive at a contradiction. So since we&#x27;re supposing it to be in vertebral, we know by the invariable major theme it&#x27;s row equivalent to the identity matrix. And we know there&#x27;s a theory that says that if two matrices aero equivalent, then that means that there rose spaces are the same. So we can say that the Roadway survey is the same as the rose base of the identity matrix, but the road space of the identity matrix, the in by an identity matrix. I&#x27;ll just write a kind of a scratch example of the identity matrix, so each vector looks a little bit like this, and finally we get to the last vector. In this I didn t matrix the identity matrix, the road space of the identity makers. It&#x27;s the span of all the row vectors now the span of all these row vectors is all of our in and namely this family&#x27;s row vectors for sure has the vector 1000 and so on, all the way to in or in minus one number of zeros. So the span of all these row vectors is definitely not equal to the set containing Onley, the viewer vector. Moreover, the span of all these row vectors actually equals all of our in So So let me let me emphasize this year. So So this span of all these reflectors is all in your combinations of these row vectors. So, for instance, one vector that would be in the row space off a would be the vector 100 all the way up two in minus one zeros bluff 01 all the way up two in minus one zeros, including this first year right here. And this vector is equal to 11 and then zeros for the rest of the entries. And this vector right here the fact that this factor is in the road space of a because any linear combination of the row vectors off the identity matrix, any linear combination of the row vectors. Identity matrix will lie in the span of the row vectors, a k a in the rose base off the identity matrix, which is the same as the space of a So therefore, the fact that this vector the linear combination of the other of the row vectors in this matrix, these two vectors plus zero times all the other vectors this vector lies in the road space of a. But remember, we suppose that the road space of a is equal to the no space of a. That means if the roads base of a sequel to the North based pay we were given this to be true. That means that every vector on the road space of Asia is in the north face of a and every vector in the north face of a is also in the rose face of a. But check this out. I found ah, vector in the road space of a That is definitely not in the north face of a. So the fact that a is in vertebral implies a contradiction, namely, the fact that we found a vector in the road space of a that is not in the north face of a implying that the road space of A is not equal to the North Bay survey which contradicts what we were given to be true. So if we were to suppose that a can be in vertebral, we would arrive at a contradiction at a contradiction. So therefore a cannot be in vertebral and we&#x27;re done.

Gennady Notowidigdo · Answer

Okay, so this is a kind of a proof question. So we&#x27;re going to show that is that if a score so some matrix A If it&#x27;s square is the zero matrix, then the urn, the Eiken value of a is zero. Okay, So to do this, we&#x27;ll start with a sort of a more very fundamental statement. So if ah, so if Lander and X is on aiken value, I convict the pairing for a matrix A. Then up a X is equal to Lambda X. Okay, so we start with that statement, then if we multiply both sides on the left, So remember of matrix R multiplication? It&#x27;s not community. So it&#x27;s important to state which side you&#x27;re multiplying on this case. We&#x27;re going to multiply both sides on the left, so we&#x27;re going to put on a so probably wrong you sprung with color. I&#x27;m gonna put a matrix a on the left hand side. Yes, sir. Back to blue. They&#x27;re more buy bread side on the left by a So what we get So you&#x27;ve got a squared X is equal to lander times a times X. So what I&#x27;ve done is sort of you have probably take one step back for a Times Land X, which then changes to sew because landers a scaler, you can always take the scale of outs. So we landed times a X, so it&#x27;s sort of using this sort of compatibility slash communicative ity slash associative ity type operation on major season victims. Okay, now a squared. He is 20 matrix. Okay, but this become zero times X, which obviously becomes the zero vector that has to become so remember a to land. The X is an icon value I convict apparent. So this ends up just being lambda squared. Times x good are this leads to two possibilities. So either land a squared equals zero, which implies Lambda equals zero. Oh, sir, Or be implied that X equal to the sirah. Victor. Okay, now, because Latin X and I again Victor by death, by sort of specifications or definition, this cannot be true. There are probably used a different college. Just cross this out, that this cannot be true, since I convict er&#x27;s are non zero. So all that your left with is possibility that Lambda has to beezer on it. It&#x27;s the case that land equals zero, then has to be the only Aiken value for a so this is obviously a relatively detail proof older. I guess I could fleshed it out a little more, but it&#x27;s definitely better than a sort of a simple show that without many steps. So ultimately, we sure now that if a squared is zero, so maybe up always say this. If a sweat 00 matrix, then this becomes that, and then that reduces the possibilities. So I&#x27;ll leave it there. Thank you very much.

Brianna Orr · Answer

we&#x27;re given the matrix A which is equal to zero negative to negative picture, native to zero and negative too negative to negative. Two year Onda were actually timing If it&#x27;s diagonal Izabal and if so, find a major eggs s such that as Invirase times A hives s is equal to on diagonal matrix that has the that I can values of a along the diagonal that you find Oh, it is diagonal Izabal we user Agon bodies give you get three distinct argon values for a then a is indeed diagonal Izabal So the determinant but a minus lambda times I is he hold you zero with almost a song for Agon buddies So the left hand side of recreation becomes native Lundell minus two minus two Negative too negative longer. I need you to nature to negative to jeer And you eat with that too Jiro on the right. Contrary finding dy determinant I&#x27;m going top I cool factor expansion along the first Your that uses negative Londell times negative The determinant off negative Lunda, I should see negative too native to negative Landau us to times He determinant off negative to negative two negative to negative Landau minus two times. He determinant off negative to Native Lavenda. Negative to negative two and we equally that 20 It&#x27;s funding users Negative. Lambda Times, Lunda squared minus four. This to times do Lunda minus four banished you times boy minus two. Your lambda is equal to zero. It&#x27;s finding that gives us negative Lum No cubed plus 12. Landau minus 16 is equal to zero using the remainder theorem not staged are if p of C he&#x27;s up only no meal is equal to zero when she is plugged into the polynomial. Then C is a zero off the polynomial Pete on. Ah, the fact a would be of the form Lambda minus c in this case, or the sea is equal to Chu because when truth plugged into the binomial, we get zero. Therefore, to is as you&#x27;re off p on door factor come Britain as Lambda minus two to find we&#x27;re all there possible argon values. So that&#x27;s right. Um, number one has been equal to do to find a other possible argon bodies you can divide or polynomial by. We&#x27;re factor here. London minus two. I&#x27;m going to use in digit the region where did the right side to the left sign. I should say we place Oh, I see which is to on then the coefficients off or probably normal taking into consideration that or Lundell Square term is missing. So we&#x27;re going to have to add a zero to seem that divide synthetically properly. So to the right of do we have negative one? Do you? 12. 92 16 on and applying Team Duchy Division we carry or negative one dumb two times negative one gives us native to adding negative to t zero gives us negative too. Multiply now you do too, but you argues is negative for adding negative. 40 Child gives us eight multiplying it by two. Gives us 16 1916 to negative. Two. Gives a syrup thio Uric ocean you of Lunda is equal to negative London squared minus two Lambda Less eight. We can equally that 20 on software worry zeroes off the polynomial the negative London to minus two Lundell plus eight is equal. Choose your something the quadratic. We end up with two factors negative Lundell pushed you on and Lambda Plus for So we have long dot being equal to cheer unloved of being equal to do you do for so London to has an allergy break Multiplicity off too, because he took her. It&#x27;s two times, as is your off the polynomial. Therefore, we only have two distinct Agon values. Sure, we can&#x27;t use the fact that three distinct Agon bodies would give us will tell us that is diagonal Izabal. We can use another approach to do so Using e minus London times The identity matrix times vector is equal to zero this all for or erectile or vector hair would be or agon vector of eight. Andi If with other vise of Lambda we get the total off. Three Agon vectors, then is diagonal Izabal. So where do you eat Lambda? One of his colleagues loved it too. We&#x27;re going to start talks with Lum. No one in this case Two a minus Lindell one, is he? You do too negative too, Major. June. It&#x27;s all 90 or two under constants. Are all series writing the augmented meat? You know that to solve these homogeneous system, you would have to are playing elementary rule operations. We get the re ruijin a lot for you to be 111 you I&#x27;m We get to ruse of zeros. You know, we see that the second and third column do not have piglets. Therefore, X two on X day are free variables on and you cannot sign A value to them Shall assign s two x two on and t two x tree writing over system off linear equations. Really? How? X one plus x two This X tree on that&#x27;s equal to zero and then we have X to being equal to us. Andi Extremely being equal to t solving for X one in the first equation, X one is equal to Thanks Negative X two minus x d You re writing or equations with the components being the subject? X one easy called you H two, my news. Thanks. Tree X two is equal to s on X trees equal to T being that substitution Excellent is there for equal to s minus t Well, negative. I came over getting the negative sighing that was placed here. It&#x27;s equal to negative as minus t onda as a vector Exxon extreme on eggs be is equal to you negative as negative. T S t andi within varieties in each Parametric for him. The fracturing the s we have native won 10 and then or a T thirsty. We have native one ju one. Therefore, we have to again vectors. Okay, native long 10 on the negative ones here one. So if we yet on additional Agon Victor from B, I didn&#x27;t value negative for then a would be diagonal Izabal working for you from the two is equal to negative four. We have a minus lambda times I and we expect to easy cool treaties. Your victor on Orlando is negative for sure. That e plus four times you had in teaching matrix times. The X factor is equal to this year. Victor, we&#x27;re augmented matrix becomes for negative to negative two Djourou on negative too for you do too native to nature. Jew for yours. Your using elementary rule operations to Yes, he&#x27;s in the reelection out for you were excellent for him. Is one native behalf negative or half, dear? Oh, you&#x27;re one negative. Aha, you&#x27;re on. Then we have a rule of jurors notice sucked away. They call him does not have people. Therefore X three is a free variable on. We can associate anybody with it? I&#x27;ll associate the value s so I&#x27;m solving for components on the vector X one x two and x d x one his equal Jew Ah Huh Thanks. Cheer those. Ah Huh. Thanks. Tree X two is equal to x three. Andi, hair over. X tree is equal to s giving back substitution on X two is also equal to s on X one. Is he called you thus? If away x one x two x three as a vector is equal to as as as factor in her s directors in the parliamentary crew we have one one run. So this is an additional Eigen vector of a on because we have three agon vectors. One here, 111 onto here. Negative 11 Jew and Eun Hee 101 We can confirm that A is because the number off Agon vectors he has easy call to any dimension of a so writing s s easy that has the agon vectors of a being. Its cover s negative, you know. Are you not? Then you can 11 year negative. 101 on 111 Andi s is such that he I called you the diagonal matrix stating or we&#x27;d have to two on. No, you do. For so notice in s the Agon vector corresponds to the Agon value. So Dragon Recto for the alien body to is negative. 110 needs, please. First in S I wise, it&#x27;s placed first in the diagonal matrix on door Second Eigen values to be occurs two times on the Eigen vector that correspond state is placed in the second column. That&#x27;s just be Kurt way too, right or s matrix and then related to the diagonal matrix. So the in very service, that&#x27;s 81. You won 111 multiplied by a zero. Negative, too. You didn&#x27;t you? Negative to jeer. Negative too negative to negative gear year multiplied by s negative. One negative. 11 one Jew one 011 That&#x27;s equal to the diagonal matrix with to be in place. Raced on the diet diagonal, then two, then negative for on. We have zeros everywhere else

Problem

[M] Diagonalize the matrices in Exercises $33-36 …

Problem 32 Hard Difficulty

Construct a nondiagonal $2 \times 2$ matrix that is diagonalizable but not invertible.

Answer

Thus $A$ has two distinct eigenvectors.
So it has two linearly independent eigenvectors.
Hence by diagonalization theorem the above matrix is diagonalizable but not invertible.

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

Linear Algebra and Its Applications

Lily A.

Heather Z.

Kristen K.

Michael J.

Vectors Intro

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Thus $A$ has two distinct eigenvectors.So it has two linearly independent eigenvectors.Hence by diagonalization theorem the above matrix is diagonalizable but not invertible.

Linear Algebra and Its Applications

Lily A.

Heather Z.

Kristen K.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Heather Z.

Kristen K.

Michael J.

Thus $A$ has two distinct eigenvectors.
So it has two linearly independent eigenvectors.
Hence by diagonalization theorem the above matrix is diagonalizable but not invertible.

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

Linear Algebra and Its Applications