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V(atvxCousider thc lincar opcrators Tla +br) = (2a b) + 2br and Utr b) = (2a +6) + (a + 26)1 ou P(R).Find the cigenvalues of T anud U. [12 points] Find basis for th...

Question

V(atvxCousider thc lincar opcrators Tla +br) = (2a b) + 2br and Utr b) = (2a +6) + (a + 26)1 ou P(R).Find the cigenvalues of T anud U. [12 points] Find basis for the cigcnspaccs corresponding to thc eigenvalucs of T. [4 point] (c) Find basis for thc cigenspaccs corrcsponding to the cigenvalues of U_ [8 points] Is T diagonalizable? If $0, find basis 8 such that (Tla is & diagonal matrix_ and find [Tla. IF uot , cxplain why: [2 points] Is U diagonalizable? If s0, find basis 6 such that [Ul

V(atvx Cousider thc lincar opcrators Tla +br) = (2a b) + 2br and Utr b) = (2a +6) + (a + 26)1 ou P(R). Find the cigenvalues of T anud U. [12 points] Find basis for the cigcnspaccs corresponding to thc eigenvalucs of T. [4 point] (c) Find basis for thc cigenspaccs corrcsponding to the cigenvalues of U_ [8 points] Is T diagonalizable? If $0, find basis 8 such that (Tla is & diagonal matrix_ and find [Tla. IF uot , cxplain why: [2 points] Is U diagonalizable? If s0, find basis 6 such that [Ul: is & diagonal matrix, and find [Ule . If uot, explain why: [3 points]



Answers

Let $T: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}$ be defined by $T(x, y, z)=(2 x+y-2 z, \quad 2 x+3 y-4 z, \quad x+y-z) .$ Find all eigenvalues of $T,$ and find a basis of each eigcnspace. Is $T$ diagonalizable? If so, find the basis $S$ of $\mathbf{R}^{3}$ that diagonalizes $T,$ and find its diagonal representation $D$

The topic of this question is Eigen values and Eigen vectors. This question asks us to find a basis for each icons based for each value of this matrix. Um two find these bases. We first have to solve the characteristic equation to find the Eigen values and then uh real reduce or solve some linear systems to find some Eigen vectors and thus Okay spaces. So let's begin. What is our characteristic equation? It is the equation that says or that really finds the roots of the characteristic polynomial. Find where the characteristic polynomial is equal to zero and this characteristic polynomial is a polynomial function of the scalar lambda are defined by the determinant of our matrix minus lambda times the identity makes. Since this is what a minus lambda. I looks like this here is what the characteristic polynomial would look like. And so now we want to solve that but we can play the organ millions. So we have to evaluate a determinant determinants as you probably know or defined mathematically as being co factor expansions along any row or column of the matrix. Oh, co factor expansions are entry times go factor plus entry times go factor plus entry times go factor. Since uh because of all these editions of products we want as many zeros as possible in our choice of row or column. So we choose this room. This room has the determinant zero times the co factor of zero plus. Uh this term times let's go factor. Which we'll get to later Plus zero times of co factor which is zero. And so all we have is that's minus people slammed at times. Its cool factor. The expression of this determinant now what is the co factor? Well in this case it's just the determinant of the matrix we get by cancelling the determinant of the sub matrix we got by cancelling um, row and column containing our entry. So by the sub matrix, I mean the two by two matrix we get from the remaining terms six minutes, lamb done negative eight 1 -3- Line. Now, if we had so for any of these entries, that rule would be true. It's just You know, the co factor would just be the two x 2 determinant. We get by cancelling a ruin cohen in the specific location but any of the other terms would be the entry or sorry, the co factor of any of the other entries would be this determinant multiplied by a negative one. And so here is our three x 3 determinants. Bills Been multiplied by two x 2 determinants. And to be two determinants we can calculate more easily. Those are just these diagonal entries multiplied minus these diagonal entries multiplied. Yeah. And from there we get this fully factored form. So we know the Eigen values are negative too and positive. But now when we have our Eigen values we want to find the icon space. So the Eigen space of an Eigen value. Lam does something, something is the nail space of the matrix a minus that lend us something times the identity in the case of the Eigen value negative too. This will be the north space of this matrix which looks like this. Now what I mean by no space is the set of all vectors X, Y zed that satisfy this equation. This homogeneous equation has all zeros on the right side. Now the easiest way I see to solve this is to grow, reduce this matrix here april's two. I because it's real reduced echelon form, which is a lot simpler, We'll have the exact same north space. So if we find the roller, do a special on form of this and then take it small space, that will be our Eigen space. So I will reduce this. We can subtract eight of the third room from the top room. Or let me just do that. That is a valid road transformation. We're adding a scaler multiple of one room on to another room. The killer in this case is -8. Another legal row of transformation is rearranging rose. So I'm going to put The 3rd room on the top, The 2nd row on the bottom and the top row or the first row in the middle. And the third type of road transformation that has allowed is killing rose. And now I've scaled the middle rule by one third. And there is my role reduced echelon form which has uh leading ones, the first non zero entry. And each row is a one. If it has non zero entries at all, zero rows are on the bottom, the leading ones are the only non zeros in there are entries in their color and each leading one is to the right of the leading ones that are above it. So Since our role reduced Echelon four has the exact same null space as this matrix, we can say that the null space of this matrix and the no space of this matrix is given by Set of all linear combinations of this vector 1, 1 specifically, they are all the linear of the scalar multiples of this vector. Now we move on to, that's the basis of the Eigen space. So we're done for that icon and now we move on to the other Eigen value. As before we can really reduce the matrix find it's no space. If we add a negative of the bottom road to the top or yeah, we add negative of the bottom, roll to the top. We get this. We can also scale the rose. Actually, let's rearrange first like we did last time and put the bottom rope at the top with them. Keep the middle around the same this time. And but the top row now on the bottom and now uh well since we have two non zero entries in this column and we want our Non zero entries in the river. Good. Special on form To be the only non-0 entries in their column. What do you want to cancel this? one way to do that is at at 7/3 of this bottom room onto the middle room. Now we'll give you zero. And now I will assignment will do to transformations at once. I switched these two bottom rows and then scale The role with the three by one third. There is my river to special on form. It has a no space of 801, or has enough space. Who's basis Can be written as 801? And so the basis for the Eigen space Corresponding to the Icon Valium five is this.

Hi there in this occasion. We're going to work in the vector space of poor animals of utmost degree one. That means the space P one. We're going to consider to vacuum to basis for this space. The basis be that's going to be generated by the polynomial P. One and P two that you can find here. And the basis be prime. That is going to be composed by the point of Mills Q one and Q two that are defined below. We need to find first. The transition matrix from the basis be primed to be. What is the procedure? Will we need to consider next and extended matrix or first? We're going to put the new basis in this case the bases beat. And in the extended part we're going to put what corresponds to the old vases in this case the prime. Okay. The idea here is that This we need to write this point on those in the standard basis for P one. That means that in this case P one that is equal to six plus three X. Well corresponds to the picture 6, 3 In the standard basis for P one. Similarly, P two will be the victor. That is the .9 of 10 plus two X. Is the picture 10 to The victim. The point on your Q one in the more space here. Q one. Well, that is equal to two Is the vector just 2. 0. And cute too. That is a polynomial three plus two. X Will correspond to the Vector 3. 2. Okay. And we're going to use this back to representations in the basis. This coordinate factors to construct this extended matrix, that I define it here. So the new basis will corresponds to be. So we need to put here 6, 3 and 10 to in the extended part we need to put the vectors to zero on 32. Okay, so we have now or extended matrix. And what we need to do is reduce this matrix to the asian form in such a way that you obtain in the left part, the identity matrix. And in the extended part you will obtain the transition matrix from the old basis to the new basis in this case, from the prime to be. So that's what we're going to do. We're going to reduce this this matrix to the action form and to do that first we need to put a number one in the 11 entry of this matrix. That Can be achieved easily but multiplying the first row by 1/6. So one five thirds, one third on one half 3- zero and 2. Now we need to put a zero in the entry 1, 2 of these matrix. That means that we need to assign to the 2nd row. The second row minus 23 times the first row. So the first row will remain the same. And here we're going to have a zero, Then here -3 -1 and one Health. The next step will be put one in this position. That can be done easily. But my multiplying the second row by -1 3rd. So our matrix will become 110 here, five thirds, one third, one health one third And -16. Finally, we need to put zero in this position of the matrix and that can be done by multiplying the first row by taking the first road and obstruct him 5/3 times this second row. And now we have here on the left part, the identity matrix, that is what we wanted and in the right part we will obtain the transition matrix. That in this case is it was to to over nine, 7/9, 1/3 and -1 6th. So this is our transition matrix. We're going to regret this as from B prime to be. Yes. The transition matrix from Victory prime we're going to take Is a common factor. Why not? 19. So that we obtain here More writable matrix that these -2 seven three and I'm in here. Well, let me check yes minus three health's Or well, I think that this better 1/18. And in that way we will obtain the matrix here minus four, 14, 6 and -3. They think that this looks better. Okay, so this is our transition matrix from the basis. Be primed to be for the next part of this exercise we need to calculate now the transition matrix from B to B prime, but that can be obtained easily by taking the members of the previous matrix. That means the members from the transition matrix from from B prime to be will be this transition matrix from the basis B two B prime. So we just need to take the members of this matrix 1/18 times minus 46, 14 minus three. The members of this matrix, what's going to be that members will? The members is equal two, one over four times three, 14 six on four. And this will be the transition matrix from B to B prime. Then For the next part we're going to consider the following polynomial. The polynomial is equal to -4 plus x. We need to grate this in a coordinate factor in the basis beat. So what's going to be? Well, first, Well, to to to obtain this, we're going to write this first as a coordinate vector in the standard basis of the polynomial. That means the vector -4, 1. And we need to create this vector as a linear combination of the element basis in B. That means this is equals to all for one times the first element in the basis B. That is the vector 63 Plus α two times the vector 10 to when it to solve For Alpha one. Enough for two and we will obtain that of one is equal to one And that Alpha two is equals 2 -1. So the coordinate lecture of this point of melty in the basis B Will be the Victor 1 -1. And in order to write this this polynomial in the basis be prime. We need to multiply the previous vector by the maid transition matrix from B to b prime. Okay. Times deep coordinate lecture pete in the base is big. So these matrix was equal to 1 4th times three 14 64. And times in this case by the picture one minus one. That is the this polynomial in the basis be the coordinate vector of of the paranormal in the basement. Okay so we just need to multiply these pictures this picture by this matrix and we obtain that the polynomial the recording intellectual of the polynomial in the basis. Be prime as equals two. 1 4th times the vector minus 11 and two. And of course we need we can verify this directly. This is using the uh the transition matrix from the basis B two B prime. But we can do we can perform this directly. So let's regret. People was equal to minus four plus X. And the element basis in big prime. We're the we're the plan our meals too On 3-plus 2 x. Okay so again we can represent this as vectors to make the things easier we have here for one and we need to grate as a linear combination of the element of the element basis in the prime. That means you're all for one Times A Factory to zero. Let's put it be better. Better one plus B to to. And the victor here. 3, 2. And the solution here for B to one is minus 11/4 and forbidden to east one half. That implies that the the coordinate vector for P. In the basis be prime is the vector -11/4 And one half. Or what we obtain here, 1 4th Times The Vector -11. And here, too, just what we obtained in the previous part.

Were destroyed. So. Yeah. Whereas your people previous exercise 9 14 for a new matrix T. So we use the three x 3 matrix with entries 3 -1 1 7 -5 1. I mean I'm sure And 6 -6 2. No shit in part A wants the first time. The characteristic polynomial of B. To do this. We need a few things trace of our three by three. Matrix B is three minus five plus two or zero. Be determined of our matrix B 78. Well this is going to be wow negative 30 -6 -42. Yeah the 70 73. We talked plus 30. The 73-7 plus 18 plus 14. So yeah visit old caprices. We got headlights. This is a -16. We can see where you have a bubble bubble caprices like it's it's kind And we want to find the co factors be 1 1 which is the determinant of negative 51 negative 62 which is negative four 73- two. This is the co factor of 3162. Uh huh. Those are the cars those cars like which is zero worth. And the co factor B 33 is the determinant. Uh 3. -1 7 -5. But like which is not great. Get into the car. Therefore it follows that the sum of our co factors B II is negative 12. No, We know quickly by three. Nature sees our characteristic polynomial. Delta is given by t cubed minus the trace 12 times T plus the sum of the co factors negative 12. I'm sorry CQ minus the trace zero times T squared Plus some of the co factors negative 12 times two minus the determinant to be. Now you're 16 plus 16 and he ran equal. This is a cubic polynomial factor. This well notice that if delta has rational roofs, then it follows by the rational root theorem that has to be of the form plus or minus one. Plus or minus to plus or minus four plus or minus eight Or across your -16. If you're lucky you'll test two. You can sometimes division. So are coefficients are 10 negative 12 16, two times 1 is two, two times two is four negative eight seems negative, eight is negative 16 and zero. We have a remainder of 02 is a root of our characteristic polynomial. And therefore we can write a characteristic polynomial as delta t equals t minus two times the remainder T squared Quotient I should say plus two T -8. Which we can factor the quadratic and we get t minus two times t minus two to minus two squared times t plus four. See Yeah, now the zeros Lender one equals 2 And landed two equals -4. These are the Aydin values of our matrix B. I'd like to see, I'd like to see it. Dude chuck. Now in part B. Just as in the previous exercise. Mm hmm. We want find a maximum set s of linearly independent Eigen vectors of the Okay, this is in the previous exercise, we're trying to find the basis for the Eigen space of each item value of the It's working in the first Eigen space lend the one equals two. Mhm. Like, well, we're gonna subtract two down the diagonal of be. So our matrix M is b minus two. I I don't know. And this gives us the three by three. Matrix one negative 11 seven negative seven suck one and 6 -6 zero. That's a that's right. And this corresponds to the homogeneous system, X minus Y plus Z equals zero, seven, X -7. Wine Prince Plus z equals zero And six X -6 by Equal zero. And this simple system simplifies to X minus Y plus Z. Our first equation equals zero. And then the second equation is simply Z equals zero. Only here it's a bit. So we already see that any solution has to have Z equals zero. So we only have one independent solution. For example, take X equal one in life. Then it follows that the doctor you With coordinates 1 10. This is a basis. Siebert For the idea in space of λ one equals 2. Right? Yeah, comments are funnier than me and most Yeah. Now consider the other icon space for lambda two equals negative four. So we subtract negative four down the diagnosed B. To obtain M. So M. Is the matrix B plus for I. This gives us the three x 3. Matrix 7 -1 1 7 -1 1 she's And 6 90 of 6.6. She kept coming around after he found open. Mm This corresponds to the homogeneous system. Seven X minus Y plus Z equals zero. Seven, X minus Y plus Z equals zero and six X minus six. Y plus six Z equals zero. This system reduces to the two equations X porn site. All right minus Y plus Z equals zero. You divide the third equation by six and You only like the fucker for like 30 seconds. six Y -6. Z equals zero. So you do this. I just most pain. Yes. This one looks like eliminated the 2nd and 3rd equation right through my body. Yeah. Yeah right possible. And we see that the system actually has only one independent solution. So take Z to be one. You see that? Why must be one And the X must be zero. So we get the solution Z. Which is 011. Send somebody down here. And since the only independent solution this forms a basis for the Eigen space of land of two, which is negative. four garden hose in his ass. Therefore it follows the process the set S with vectors U. And V. You just Or 110 And 0 1. 1 elegant. This this is a maximum set like shows both P. T. Barnum of linearly independent audience. I've been vectors for B. one of them was where the what is it? Finally in part C. Just as in the previous exercise we're asking he is diagonal, Izabal. And if so to find a matrix P. Such that the diagonal make the matrix D, which is p inverse ap is diagonal. Okay, these are the current PM for cps diagonal on me. Sorry, some time. Well, we see that B has at most two linearly independent Eigen vectors. It's you and the. Yeah. Right. And so like you think like man people needs to be fucked up that they thought that was like a hand. Therefore it follows that our matrix B is not similar to a diagonal matrix, which is the same as a definition that B is not diagonal. Izabal War. Yeah. It's like, oh yeah, nobody fell for them. They all knew like this is just a retarded.

In this example, we have a two by two matrix say that's provided medical. Here's to find its Aiken values and corresponding. I get vectors to start off for the first part. To find Eigen values, we take the determinant of a minus lamb die or will be solving for the very bold Lambda as a determinant will have negative Lambda Negative eight in the first column than one end for my is Lambda in the second column, Then, to take this to determine it will first have the main diagonal multiplied, which is negative. Lambda Times four Mice Lambda and subtract the product of the off diagonal, which is negative. Eight. So we'll have a positive eight so far. Then, if we simplify this, we obtain Lambda Squared minus four. Lambda Plus eight equals zero, and this is our characteristic equation. The solutions are Lambda One equals two to minus two. Lambda or excuse me to I and Lambda Two is that's constant to with a plus two I. So we confined that using the quadratic formula, and the next thing we're going to do is determine the corresponding Eigen vectors for each Eigen value. So let's focus on Lambda one To start out, we can solve the system a minus lambda one, which is two minus two I times X equals zero vector. So let's augment and say Our matrix is going to be distributing the negative sign in. In the first calm will have negative two plus two I and the negative eight below. In the second column, we have one. Then take four and at together the result of taking negative two plus two. I now we'll have altogether two plus two. I they were augmenting with a zero vector, which I'll place here. Now it's due to operations as we ro reduce. I want to go to the second row, divide it by negative eight, then make it become the first row will have one. Then negative 0.25 plus 0.25 I and a zero. Then the second row is going to be the old first row, which is negative. Two plus two I one and a zero. Now we have a pivot that's here and uncle is to eliminate this entry. So let's start by copying the first row. It is a one negative points to five plus 50.25 I and a zero. And what we'll be doing is multiplying Row one by two minus two i, and adding the results to the second row will obtain zero. And if we take negative points to five plus 50.25 I and multiply it by Tu minus two, I we obtain exactly negative one, and that combines with this entry producing a zero here. Then we have a zero for the last column where we augmented. So in the row one of our Matrix, which is in row reduced echelon form, we find that if we saw for the very bill, X one will be equal to the opposite of 0.25 plus 0.25 i times x two and X two is a free variable Salt. Just write X two equals X two Oh, it will have a negative here that was neglected to be copied from there. So now we need to describe what our first vector, Aiken vector V one, is that corresponds to Lambda One and we have some choices. If we decide to not deal with these decimal values we see here, we can allow X two to be any real number except zero since zero is never Nike in Vector. So is pick X two to be, say, a four. If we do that, then all together will get a one minus I. That's a result of placing for here distributing as well as distributing the negative sign. Now we have to re piece repeat this row reduction to find the second I get vector, but there is a wonderful shortcut. That shortcut says that V two is the congregate of the one and the one conjugated would be one plus I and the contra gave a real number is itself. So what we've done is we found dykan values V one, V two and the corresponding Eigen vectors V one and V two. So these V one V two also form a basis for their corresponding Eigen spaces.


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2. (5 points) Give the thermodynamic enolale and the kinetic enolate forCHg
2. (5 points) Give the thermodynamic enolale and the kinetic enolate for CHg...

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