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Let D(As) = {c() eC'(o,L;C):40) = 0} and for %6) e D(As) As[pk)](x) = 0(x)-206x) [4.3.6] Show that the operators A, is dissapative: i,e., (Aspo),6)) < 0 for...

Question

Let D(As) = {c() eC'(o,L;C):40) = 0} and for %6) e D(As) As[pk)](x) = 0(x)-206x) [4.3.6] Show that the operators A, is dissapative: i,e., (Aspo),6)) < 0 for all 0) e D(As) [4.3.7] Show that the operator A, is not symmetric[4.3.8] Show that all the eignevalues Y of A, must satisfy Y <-2.[4.3.9] Can you find an eigenvalue of A6

Let D(As) = {c() eC'(o,L;C):40) = 0} and for %6) e D(As) As[pk)](x) = 0(x)-206x) [4.3.6] Show that the operators A, is dissapative: i,e., (Aspo),6)) < 0 for all 0) e D(As) [4.3.7] Show that the operator A, is not symmetric [4.3.8] Show that all the eignevalues Y of A, must satisfy Y <-2. [4.3.9] Can you find an eigenvalue of A6



Answers

Let $\mathbf{u}$ be a unit vector in $\mathbb{R}^{n},$ and let $B=\mathbf{u} \mathbf{u}^{T}$
a. Given any $\mathbf{x}$ in $\mathbb{R}^{n}$ , compute $B \mathbf{x}$ and show that $B \mathbf{x}$ is the orthogonal projection of $\mathbf{x}$ onto $\mathbf{u},$ as described in
Section $6.2 .$
b. Show that $B$ is a symmetric matrix and $B^{2}=B$ .
c. Show that $B$ is an eigenvector of $B .$ What is the corresponding eigenvalue?

Were given a matrix, a vector V and Eigen value of day and we were asked to show That's five is an Italian value of A and V is an associate Eigen vector. And then we were asked to or fog Nelly Diagne eyes a check the see if five is an I in value of a we know that and Eigen value satisfies the characteristic equation of a that is you want to check to see what the determinant a minus five I is. This gives us the determines of the Matrix. Same is okay, except for now, the diagonals or negative 14 So negative one negative one negative one negative one negative one negative one negative one negative one negative one and you can see this determined. It is clearly going to be zero since we have duplicate rose and therefore it follows that lambda is an Eigen violet value of a check to see if V is an Eigen vector in this wagon space. What to see if these satisfies the equation? A minus five i times V equals zero or alternatively, we could compute eight times V to see if it's equal to five. I times V yeah, that a Times V is equal to for a negative one. Negative one negative 14 Negative one. What we have. We have four times. One is four minus one minus ones of two negative one times one is negative, 14 times one is four negative times. One is negative, one again to and finally negative one times one is negative, one negative one times one is negative, 14 times when it's four. So again, too, we have that. This is equal to two times I times. Well, still doing that. We calculate a minus five i. V. We'll get something a little bit different since equation implies that a minus five I'd be is equal to this is matrix that negative ones times V. So we have negative one negative one negative one so negative three all the way down. So it's clear that V is not an Eigen vector actually associate with this Eigen value. But we have shown that a V is equal to 222 which is equal to two times V, and so B is an Eigen vector of a associate ID with and I didn't value Lambda equals two. Now I noticed that transpose of a is equal to the Agnos stays the same four for four in exchange negative with negative one negative one with negative one negative with negative one. We recognize this is the same as a So we see that A is a symmetric matrix Since a transpose equals a and therefore follows that a is orthe orginally the ag analyze herbal tour following Lee diagonal eyes a You to find a complete set of more than normal lighting victors. We have one Eigen vector. He is Thea connector associate with the item value landed too. We don't know. However, the multiplicity of fighting value landed too, which we could calculate. You can also calculate the multiplicity of backing value Linda five and use that to find multiplicity of Viking value. Lambert to Let's do that. We have it a minus saying Linda one is equal to five Eigen value must satisfy a minus five by V one equals zero. So we have that this coefficient matrix negative one negative One negative One negative. One negative. One negative One negative one negative with a negative one Reduces to the matrix 111 000000 So we have two zeros which tells us that despite in space has mentioned to and we obtain the system of equations V 11 plus V 12 plus view 13 equals zero. And so we have that view 13 is equal to negative. The 11 minus B 12 And if we take V 11 to be the perimeter s and V 12 to be the parameter t in the general form for the I connector V one is yes t negative s minus t which can be rewritten as the linear combination s times 10 negative one plus tee times 01 negative one. So we obtain the Eigen vector V one if we take has to be one in TV zero She's 10 negative one and we obtained Eigen Vector B two if s zero and t is were one which is the vector 01 negative one. By construction we have that V one and V two or when nearly independent and Stan the Agon space. So it follows that you want to be too are a basis for this Agon space. The basis of Eigen vectors and we have the norm of V. One is equal to the square root of one plus one, which is to route to Norma. V. Two is the square root of one plus one, which is rare too. And so the unit vector you one is the vector V one, divided by the enormous V one, which is equal to one of a route to zero negative one word to and unit vector. You too is the vector V two, divided by three normal V two, which is the Vector zero. Whatever route to negative one over route to. And so because these are unit vectors and they're still Eigen vectors we have that you wanted you to is in before I do this to check something. So I went ahead and I calculated thes unit vectors not checking something. I need to check if they were orthogonal who really wants You know what the normal set from this Eigen space. So we have that V one died with me too is equal to zero plus zero plus 1 90 quarter zero. So it follows that the one is not to put a dog in all to be to and so we're going to have to use Gram Schmidt or thought generalization procedure to obtain a second orthogonal member of the basis. So we'll take you want to be director V one? Let you too be the vector V two minus the projection of the two which is on the view one which is v two dot v one over v one dot you one sometimes the vector V one. This is equal to you Have defector V two is 01 negative one minus and then v two dot over the V one we determined was one and then v one diver to be one Mrs to times fy one is 10 negative one. This gives us zero minus 1/2 is negative on half one minus zero is one and negative one plus one have is negative 1/2 instead of you here She probably had made these w one and w two And now my Gram Schmidt we have that These vectors w one and w two not only are still Eigen vectors because that's a linear combination of my conductors, but are and orthogonal said we have the norm of w one is going to be the normal V one, which is the square root of one plus one which is written to And we have that the norm of to and likewise we have that to w two is also and orthogonal specter to w one normal to w two. This is the norm of negative one to negative one which is the square root of one plus four plus one which is Route six. And so we have the unit vector you one which is w one over. The norm of W one is equal to 10 Negative one divine favor too. We're one of her 20 negative. Whatever route to and the unit vector, you too is the doctor to w two divided by the norm of defector to W two. This is equal to you know you two is the vector negative 11 negative one or negative one to negative one. So negative one of a route six two of a route six negative 1/6 evidence and now we have that you want you to is a north a normal basis for this wagon space. Returning to previous results showed that v was in Eigen Vector. So she with the Eigen value, landed two, which could be the only other Eigen value. And it must have a multiplicity Get one since a is a three by three matrix. So we have the already The itself is a basis of Eigen vectors in the Eid in space. And so the normal V is equal to square root of one plus one plus one, which is three. And we have the unit vector you three, which is be over. The normal V is the Vector 1/3, 1/3, 1/3. And since a is symmetric, it follows that you three is orthogonal to both you wanting you to and therefore we have that the set you one you too. You three is a complete Ortho normal set of I Connectors of the Matrix A And so if we take P to be the Matrix with Colin vectors u one you too And you three this is the matrix one of around 20 negative one or two. You too is negative whenever Route six to a very six native wonder over route six and then you three is one of a route three 1/3, 1/3 and we'll take matrix D to be the diagram Matrixes entries of the Eigen values. So this is going to be five five to the rest of the entries are zeros and using these to maitresse ease you can are thought Nelly diagonal eyes.

Okay, so this is a kind of a proof question. So we're going to show that is that if a score so some matrix A If it's square is the zero matrix, then the urn, the Eiken value of a is zero. Okay, So to do this, we'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's not community. So it's important to state which side you're multiplying on this case. We're going to multiply both sides on the left, so we're going to put on a so probably wrong you sprung with color. I'm gonna put a matrix a on the left hand side. Yes, sir. Back to blue. They're more buy bread side on the left by a So what we get So you've got a squared X is equal to lander times a times X. So what I'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'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's are non zero. So all that your left with is possibility that Lambda has to beezer on it. It'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'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'll leave it there. Thank you very much.


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