verify-the-statements-in-exercises-19-24-the-matrices-are-square-if-a-is-diagonalizable-and-b-is-sim

Question

Verify the statements in Exercises $19-24 .$ The matrices are square.
If $A$ is diagonalizable and $B$ is similar to $A,$ then $B$ is also diagonalizable.

Runpeng Li · Accepted Answer

Okay, so for problem 22 were first given Matrix K, which is diagnosed Technol Izabal. So that means by definition we can have a have any murderball metrics say ass times as he first time says is De were where these are diving a matrix. Excuse me. Now also, by the definition off similarity then were given that be similar to a So we also have, um t inverse times. Hey, times t she&#x27;s bi. Of course, T has to be a murderer matrix. Now we first soft a by multiplying I&#x27;m on applying t from the left hand side of right hand side so that mrs, uh, sorry. Second question. We must have a which is tee times be I&#x27;m stealing verse. Now we put the D. C. Based compression into our first question. No, we have asked members times t hey, t inverse times asked. Okay, Now here. Uh, yeah, I I missed the Dagenham matrix D on the right hand side. Now here, uh, let&#x27;s take you to be a scene first times t so that, um our yeah, so our you immerse will be t t inverse times has. Okay, So since these two matrices are all in murderball, so you has to be in veritable so we can rewrite our expression here. D to be you times B times Ewing burst. So we&#x27;ve shown that be, by definition, bees.

Runpeng Li · Answer

Okay, so for a problem, 20 were given the two to similar matrices. So by our definition of similar matrix, that means we There&#x27;s Derek. Six. There exists on the convertible matrix p such that a few times b times p m burst is a So this is our room assumption. Now we want to show that he squared is similar to be square now, To do that, we first consider a square to be a time saver. Now we substitute this expression of a So we have the PM&#x27;s be times P embers does be times. He&#x27;d have to be times p verse now because matrix multiplication is associative and we can calculate p first times peepers. So we have this and we kept with peace be times beavers then we who have we will have p times b squared times p yours. Now this shows that a squared is similar to

Peter Davis · Answer

So in order to see if this matrix a equals 13 minus 9 25 minus 17 is diagonal Izabal. We need to find that I can rallies of a But in order to do that, we need to find the determinant of a minus lambda times. The identity matrix said that equal to zero and sulfur lampa. So the determinant will be 13 minus lambda minus 9 25 minus 17 minus lander. So what set this equal to zero? So taking this determinant, we have 30 and minus lampa times minus 17 minus lambda minus negative. Nine times 25 equal standard. So multiplying this out, we have minus 2 21 minus 30 in Lambda Plus 17 Lambda plus Lambda Square plus 2 25 equals zero. So combining like terms, we have slammed a squared plus four lambda plus four equals zero. Now we factor this will get lambda plus two squared equals zero. Therefore, our Aiken value will be lambda equals negative to with the multiplicity of to now, in order to see if the matrix is diagonal sizable. What? We need to find a basis for the Eiken space for Lambda equals minus two and we can do that by taking the null space of the Matrix a minus minus two times the Identity Matron&#x27;s, which is just no a plus to ah, so we need to find an all space of the matrix 13 plus two minus nine 25 minus 17 plus two. So this will be 15 minus nine 25 minus 50. Now we divide the top of I 15 We&#x27;ll get one minus 3/5, 25 minus 15. And now we multiply the top row by 25 subtract this second row from it. We&#x27;ll get one minus 3/5 00 Now we&#x27;re called up the notes. Face of a matrix is the set of solutions to the equation. A X equals zero. So the north space of this matrix we&#x27;ll just be exit to times the vector 3/5 want. Now we see that the Eiken space of the matrix has a dimension of what? Since there&#x27;s only one factor, however, we found one I can value with the multiplicity of to That means that in order for the matrix to be diagonal Izabal, that Eigen space had to have a dimension of tube, which means that we needed to have two factors

Chris Trentman · Answer

we were asked to prove Ethereum about North Ogden only diagonal eyes Ball Mitrice&#x27;s. So we&#x27;re giving that A and B are orthogonal Lee diagonal Izabal. We were asked to prove that a B is also or thought finally diagonal sizable. We&#x27;re also told that a B is equal to B A. So they commute so first since A and B are thought only diagonal Izabal it follows that a and B symmetric major sees by here too. Therefore we have that baby transpose is equal to by the definition when a definition of the properties of transposition he transpose a transposed on the one hand and on the other hand we have This is equal to since and be commute be a transposed which by the properties of transposition is a transpose be transposed which is equal to since a is symmetric and beer symmetric a times B And so we&#x27;ve shown the A B transpose is equal to a B. So it follows that a b is symmetric and again we have that&#x27;s by theorem to a B is or Thorgan Aly diagonal Izabal

Brianna Orr · Answer

were given the Matrix eight, which is equal to negative 90 for native night. Andi were actually determine if it&#x27;s diagonal Izabal. And if you these were acts defying the matrix s okay on easy way for determining if or magics is diagonal realizable is to make sure that we&#x27;re matrix has you&#x27;re distinct. Agon values the distinct Agon values that it has that would make it hard. Diagonal Izabal matrix is dependent on the size of the matrix So because or matrixes are two by two matrix if it has two distinct Agon values than its diagonal Izabal. So finding the determinants off a minus lambda Times I Andy quitting that is you were going to solve who are ill and bodies for determinant would be a minus Lundell times I or identity matrix is one to you 01 And if we&#x27;re multiplying that by Lambda, then that becomes Lambda 00 Londell. Hence, if we&#x27;re subjecting a nominal from eight, we&#x27;re just going to subject Londell from the diagonals, Teoh or, uh, determinant would be negative nine minus lambda zero for minus nine minus lambda and that&#x27;s equal to zero. Finding or determine on. We have negative nine minus Lambda Times Negative nine, minus Lambeau minus here, which is equality? Is Europe expanding or bracket you this 81 plus 18 lambda? Because Lunda screwed and that&#x27;s equal to zero. Characterizing this expression gives us Lambda Plus nine describe his equals zero. Therefore, a lambda is equal to negative nine and it has an allergy break. Multiplicity off to because theory route is repeated because of the square we have the roots are carrying two times. Therefore, unfortunately, we only have one distinct agon value on. Therefore, we can&#x27;t used that to determine if four matrixes diagonal. Sizable. However, because we have the value off or because you have wagon value, we can set up or equation that would alot still. So for the no space which would give us the basis off or negative nine Agon space So we have a mine. Islam, The times I multiplied by the ex rector is equal to the zero vector, so that becomes a minus nine times. He and entity matrix times the ex rector, which is equal to is your director rewriting or equation in vector form, we have a minus nine times the identity becoming 00 for zero on We&#x27;re multiplying. You buy the ex rector, which is x one x two, and it&#x27;s equal to the zero meto solving or home a genesis t we can be right, its eyes on augmented matrix to do so where we&#x27;re solving the home a Genesis team show We are going to reduce it to the Rachel on form to solve or system stopping that rules one and two we get for zero Jew and years. Years year. Andi, then multiplying were first drew by 1/4. Gives us Ron Geo ju zero 00 So thanks to did he notice it&#x27;s, ah free variable because the column does not have our people. So because x two is ah free variable Oh, are negative nine wagons race for the Agon space for the argon value. Negative night is one dimensional. Hence the geometric multiplicity is one on because the geometric multiplicity is not equal to the algae break Multiplicity which is to or matrix a is not Diageo, not realizable

Pam Russell · Answer

So we&#x27;re gonna use direct multiplication to verify that the given to maitresse ease are in verses of one another. So if we have matrix A and we multiply it by a matrix, a inverse, if our result is the identity matrix that, in fact these two matrices are in verses of one another. So if we&#x27;re given Matrix A, which is a three by three matrix, and then we&#x27;re given a inverse three by three matrix If we multiply these two matrices together in our result is the three by three identity matrix that, in fact these two matrices are in verses of one another. So given a multiplied by a inverse, we&#x27;re gonna use matrix multiplication to verify that indeed, they are in verses of one another. So with a three by three, we&#x27;ll still use the dot product and we have three times eight plus five times negative five plus one times two three times negative 29 plus five times 19 plus one times negative eight and then we have three times three plus five times negative too, plus one times one one times a plus two times negative five plus one times two one times negative. 29. Close to times 19 plus one Times Negative eight one times three plus two times negative, too, plus one times one and then finally for row three. Two times a plus six times negative five plus seven times two two times Negative 29 plus six times 19 plus seven times Negative eight and then two times three plus six times negative, too, plus seven times one. Calculating all of these out in row one column one. We end up with a value of one in row One column to we get a zero and Row one column three. We also get a zero and wrote You column one. We get a zero in row two column to we get a one and in row two column three, you get a zero and finally in Row three column one. We get a zero and in row three column to we get a zero and in row three column three we get in one, and this is the three by three identity matrix. So we have verified that these two matrices are in verses of one another

Problem

Verify the statements in Exercises $19-24 .$ The …

Problem 22 Medium Difficulty

Verify the statements in Exercises $19-24 .$ The matrices are square.
If $A$ is diagonalizable and $B$ is similar to $A,$ then $B$ is also diagonalizable.

Answer

Therefore, the matrix $B$ is diagonalizable.
Therefore, the given statement is TRUE.

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

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Kristen K.

Michael J.

Vectors Intro

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Therefore, the matrix $B$ is diagonalizable.Therefore, the given statement is TRUE.

Linear Algebra and Its Applications

Heather Z.

Kayleah T.

Kristen K.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Heather Z.

Kayleah T.

Kristen K.

Michael J.

Therefore, the matrix $B$ is diagonalizable.
Therefore, the given statement is TRUE.

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

Linear Algebra and Its Applications