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In Exercises through delemmine whelher the indicaled sels |orn rng under Ihe indicated operations5.$ = {Aem(2,R) detA 0} , under matrix ;dklition ;ind multiplicatio...

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In Exercises through delemmine whelher the indicaled sels |orn rng under Ihe indicated operations5.$ = {Aem(2,R) detA 0} , under matrix ;dklition ;ind multiplication

In Exercises through delemmine whelher the indicaled sels |orn rng under Ihe indicated operations 5.$ = {Aem(2,R) detA 0} , under matrix ;dklition ;ind multiplication



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In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.
$\left[\begin{array}{ll}{0} & {I} \\ {I} & {0}\end{array}\right]\left[\begin{array}{ll}{W} & {X} \\ {Y} & {Z}\end{array}\right]$

Were given a matrix A and were asked toward fog only diagonal eyes. This matrix giving an orthogonal matrix diagram matrix were also given that the Eigen values for this Matrix or Lambda equals three and five. The Matrix A is 401004101104001 04 Noticed that a transposed says the Matrix has the same diagonal for four. For four. We exchanged zero with zero one with one zero with zero zero with zero one with one and zero with zero and recognize that this is the same as the matrix. A. So it follows. That's since a transfers equals a is a symmetric matrix. Therefore, it follows that a is north Ogden Aly Diagonal Izabal Your thought only diagonal eyes A when he defined complete set of orthe normal vectors. Eigen vectors for a So take lambda One to be the first Eigen value three. Then we have that. If the one is an Eigen vector then satisfies the equation. A minus three i v one equals zero vector The matrix a minus three I is the same as matrix except for before has become ones So we obtained the matrix. 10100101 10100101 in this simply reduces to the matrix. 10100101 0000 0000 from which we obtained a system of equations V 11 plus V one of three equals zero and V 12 plus view and four equals zero. This tells us that the 13 equals negative view on one and the 14 equals negative b 12 And so if we take V 11 to be the parameter s and he one to be the parameter t, we get that The general form for the I conductor V one is the vector s T negative s negative T which would be written as a linear combination s times 010 negative one or zero 01 My mistake s times 10 negative 10 plus t times +010 Negative one. So we see that this wagon spaces two dimensional. And if we take s to be one and t v zero, we obtained one Aiken Victor called V one is the Agon vector 10 negative one zero it would take Has to be zero. On TV one, we obtain a second Eigen Vector V two, which is linearly independent from V one, which is the connector 01 zero negative one. And so we have that set V one be to is a basis for this Eigen space. They also have that V one dotted with feet too is equal to zero plus zero plus zero plus zero, which is zero. And so we have that V one is or thought Donald to V two. And therefore we have this set view and be two of the north Ogle set norm of V one is the square root of one plus one, which is Route two in the normal. V two is the square root of one plus one, which is Route two. And so we have the unit vectors. You one, which is V one over. Norm of the one is equal to 1/2 zero negative. One of a route to zero. And the unit after you too, is the vector V two over the normal vector he to which is 0 1/2 zero negative 1/2. And we have that both. You want you to our Eigen vectors and are linearly independent in North Ogle. So that set you on you too. Isn't Ortho normal basis for this Eigen space? Next, Let claimed to be the second and last Eigen value five. Then if the three is an Eigen vector associated with this Eigen value, we have that A minus five I times feet three is equal to the zero vector matrix. A minus five. I is the sings matrix A Except for the diagonals now negative one. Instead of positive one, we obtain the matrix. Negative. 1010 ones. Zero negative. One 01 10 Negative. 10 and 010 Negative one. This could be reduced to the matrix with reduction 10 Negative. 10 010 Negative. One and 00000000 from which we can obtain these system of equations. The 31 minus V 33 equals zero and be three to minus fee. 34 equals zero. So we have that. These 31 is equal to be 33 and V 32 is equal 33 4 So if we take B 31 to be the parameter s and V 32 to be the parameter t yet the general form for the I conductor V three is S T yes, T is equal to the linear combination s times 1010 plus T times +01 zero one And if we take us to be won t v zero weekends the Eigen Vector V three, which is 10 10 And if we take s to be zero and t to be one, get second Eigen Victor, before which is 0101 and by construction we have that these two wagon vectors are linearly independent and thus form a basis for this wagon space. We also have that V three died with before is equal to zero, which implies that V three is orthogonal to before. So it's actually in orthogonal basis for this vector space. The norm of V three is equal to the square root of one plus one which is to in the normal before is the square root of one plus one which is equal to. And so we have the unit Vector you three, which is equal to V three over the norm of you three, Our normal V three. My mistake is a vector one over route to zero 1/2 zero. And the vector You four is a unit vector, which is V four over the norm of the four. This is the vector zero. Whatever too zero one of the route to and we have that both u three and U four our Eigen vectors for this wagon space and are valuable and are unit vectors. So it follows that set you three. You four is an Ortho Normal aces for this Eigen space. Since our Matrix A is symmetric, it follows that you won and you two are Ortho or thought Donald U three and U four. So it follows that to set u one u two you three you for is a complete Ortho normal set of Eigen vectors for this matrix. And we have that. If we take P to be the Matrix who's calling Victor's Air U one U two, you three and you four. This is the Matrix one of a right to zero negative one of a route to 01 of a route to zero negative 1/2 and whatever to zero Whatever. Route to zero and zero one of a route to 0 1/2 when we take matrix D to be a matrix whose entries our gagan values on the diagonal. So you have each Agnelli was multiplicity too. And so this is the matrix 3355 and then zeros at the worlds. And using these two matrices p and G, we can diet.

So they want us to orthogonal eyes this matrix. So first thing we need to do is actually find are arguing vectors so we could go ahead and try to normalize thumb. Eso will need to I'll call this a So we need to do a Maya slammed I, which is going to be one minus Lambda Negative five negative 51 minus lambda. And then we take the determinant of this. And remember, we're assuming that this is equal to zero. Yeah, so do the determined we do. The downs binds the ups. So that would be one minus lambda, uh, squared and then negative five times negative. Five is 25. That b minus 25 is equal to zero. So, yeah, we can actually go ahead and add that over. So one minus lambda square is you go to 25 take the square, root each side, and that's going to have one minus. Lambda is able to plus minus five and then we I'll add land over and I'll subtract The plus surmised fives that just as lambda isn't one plus minus five or, in other words, lambda is equal to six or land us all Lambda one is six and then lamp too is negative or so. Let's go out and find what are Eigen Vectors are going to be now and let me move this down here. So when Lambda is equal to negative for, I'll do negative for first, Um, a minus minus four I or, in other words, we're adding for should give us so would be five negative five negative 55 And then you can see that we could clear this to reduce it down to just one negative one 00 So then this tells us that X one minus x 20 or X one is equal to x two. So when we're creating our eye in vector over here, If we start with X one x two, then this is just going to be x two x two or x 211 Now we need to normalize this. So let's go ahead and, uh, find the length of it. So remember the length is just going to be both of those components squared square rooted. So the one squared plus one squared square rooted. So that's gonna be route to. So we multiply this by problems are. Divide this by route to and then that's going to give one over route to one over, route to And then this is going to be our Eigen vector. So this will be be too. And so this was associate with Lambda to equal to negative for all right. And now let's go ahead and do this with our other one. So this would be a plus. Um, What was it? Six were actually minus six, minus six I. So again, if we look at what a waas on you to skip this down, they want to get all that. So just yes, explain me this town and I'll speed this up. Yeah, eso we would do one minus six of being negative five negative five. And then here we have negative five negative five and then notice that this would reduce down to just 1100 So it's x one plus x two is equal to zero. Or, in other words, X one is even too negative X two and now over here we have x one next to and this should be equally so negative. X two x two Factor that out So the x to negative 11 and now we need to normalize this one as well. So same thing is before. So we find the length of this negative one squared plus one squared square. Root it, and that's going to give us a route to again. So we divide this by route to, and that's going to give us negative one over route to one over route to. And then this is going to be our first one when Lambda One was equal to six, right? So now to create are normalized vector, so we're going to have So let's first create the diagonal. So it's D. So we had negative 46 and then we have 00 and then for P. This is going to be wealthy one associated with negative for the Eigen vector. Let's come back appearance, see where it waas. There's one over route to one of her route to so we have one over route to one over route to, and then the one with six. We just found to be negative one over route to 1/2, which, if you want, you could pull out the one of the route to and just write this as one over route to and then one negative. One 11 Um, so that's one way you could do it. And do we need to find P embers? Yeah, we did. We just need to find this. So this is pretty much where we can leave it like this. So that's our diagonal ization. And then that would be our orthogonal matrix.

Were given a matrix. We were asked to or thought generally diagonal eyes. This matrix giving orthogonal matrix p in a diagonal matrix T matrix A is 3113 Notice that a transpose is equal to diagonals air still three and three. Then we exchange one with one. And so we see they transpose is equal day. This means that a is a symmetric matrix my fear him to From this chapter it follows that a is fourth orginally diagonal izabal meaning that such a meat sure sees P and G will exist. Dagnall eyes A. We'll need to find the Eigen values today, So we need the characteristic equation. This is the determined of a minus Lambda I equals zero. So we have the determinant of the matrix three minus lambda one 13 minus lambda equals zero. Or that green minus lander squared minus one equals zero. Which means that three minus slim two squared equals one or that three minus landed equals plus ra minus one so that Landau is equal to three plus or minus one. So we have to wagon values. You're distinct. Lambda one just to and planned it to, which is four Using these Eigen values, we can compute basis for the Eigen spaces. So you have that a minus two I. That's equal to times some Eigen vector Call it X one equals zero vector. This means that Matrix three minus two, which is 11193 minus two is one times the vector x 11 x 12 equals zero vector. So we obtain the system of equations x 11 plus X 12 equals zero and again x 11 plus X 12 equals zero He's a redundant and we obtained that x 12 equals negative x 11 So we have one degree of freedom. If we take x 11 to be equal to, say the parameter t then it follows that x 12 is negative t that the Eigen vector X one is equal to to negative t which is equal to tee times thieve vector one negative one And so, if we take tea to be one we obtain as an Eigen vector X one equals one negative one and to normalize x one, we calculate the norm of X one is equal to square root of one plus one, which is route to. And so we have that. Okay, Unit vector. You one is equal to x one over the norm of X one, which is equal to one over route to negative one of a route to and we have that you one is going to be enforced a normal basis for the I in space associated with Lambda one. Likewise, when the two equals four and we have a minus four I times the Eigen value X two equals zero vector. This implies that negative 111 negative one times x 21 x 22 equals 00 and be obtained The system of equations negative x 21 plus sex to two equals zero in the redundant equation Treating this to eliminate. And so we have that X 21 equals x 22 And so if we take X 21 to be equal to the parameter t, we obtain that in general the Eigen vector X two has formed to To which me written as t times the vector 11 If we take tea to be equal to one, we obtain the I conductor x two, which is 11 to normalize x two. The normal next to is the square root of one square which is one plus one, which is Route two. And so we have that a unit vector. You too is the big directs two divided by the normal, next to which is equal to one of a route to one of a route to. And so we have that you two is an Ortho normal basis for the Eigen space associated with when did too. We have bits since a is symmetric. It follows that both you wanting you tour struggle since they belonged to different I in spaces. Therefore, we have bit set you one. You too is an Ortho normal set since you wanting you to are unit vectors. And now take t to be the matrix. His columns are you one and you too written as one of a route to negative one of a route to and one of the route to one of two. And take a deed of your matrix is diagonal and has Eigen dies is its entry. So it has 2004 Then we have that key or thaw Guintoli they agonal eyes is a and so it follows that a equals PGP in verse since a equals p D. P in verse and PM versus equal to p transpose.

So they want us toe orthogonal e diagonal eyes. These, uh, matrix here. So since they don't give us the icon values and I get pictures for this, we have to go out and find those. So the first s. So I'm just gonna call this a So to a my Islam die, which is going to give six minus lambda minus two minus 29 minus lambda. And we're going to take the determinant of this now. And remember, this should be equal to zero. Um, so we'll end up with six finest lambda times, nine minus lambda and then negative two times. Second to its four. Uh, so that would still be just minus four. I was. Go ahead and expand this over here of that's 54 and then negative 15 lambda and then plus Lambda squared minus +460 Let's go ahead and combine those like terms. So we have. Why are not why Lambda Squared minus 15 lambda and then plus 50 is equal to zero. And is this factory herbal? Yeah, Could be goes We could dio Lambda minus five. Lambda minus 10 is equal to zero. So that tells us Lambda 05 Lambda is equal to 10. All right, so now we can go ahead and find what are Eigen vectors are. So let's take this Miss scoop this down. So looks Thio five first. So we have a minus five I So that's going to give It would be one negative too. Negative, Thio four. So you can see how this would get reduced down to just one negative 200 And so then this is going to imply that X one plus or plus minus two x two. Is he with zero? So x one is just too two x two. So when we look at our actual picture, actually, we expend screen will, um of x one, x two. This should be two x two x two or ex to let me oh clean out of So it would be X to To what Now we need to normalize this so we can go ahead and do that by finding the length of it. So it be two squared, plus one squared square rooted, so they'll be Route five. So we just need to multiply this by Route five, and then that's going to be our orthogonal I get better. That will be using for this. So, Pete to over Route five and then one over route five. So this will be the one on that goes with Lambda being equal to five. Now, we're going to repeat the same process, but with tin us. Let me Scoop was down. Yeah. Actually, it was 10 for the other one, right? Yeah. So now when Lambda is 10 so we have a minus 10 eyes, so that's gonna be negative. Four minus two and then negative to negative one. And so you can see how this is going toe reduce down to it. Looks like 2100 eso this is going to give x one is our x 12 x one plus X two is equal to zero. And so if we're going to go ahead and solve for X one So the X one is equal to so the negative one half x two So we can go ahead and love this end. So we have x one x two and then this is going to be, um, negative one half x two next to which is going to just be negatives. One half one times X two and we could go ahead and normalize this and doing that, we would get so negative one half squared plus one squared square rooted. So that is going to be 1/4 plus one, which is gonna be five. Force takes squared of that. So that's going to be, uh, Route 5/2. So we need to divide this by Route five over to and in doing that we will get negative 1/5. And then if we divide that by one, that would be too over Route five. As I think this is going to be be, too, when Lambda Two is 10. And now let's go ahead and step are diagonal eyes matrix. So D is going to be so 5. 10 along the diagonals and then our orthogonal matrix P is supposed to be so with five. Let me see what this was again. This was supposed to be to root file for one route five. So to route 5/1 5 and then or 10, this was negative. 15 over to Route five. The mentality that negative one, Route five over to route five. And then this isn't really necessarily. But I'm gonna pull out that one over Route five, since it's in everything just to make it look a little bit prettier. So 1/5. Um, then it's too negative. 112 So this is going to be our diagonal ization de along with that orthogonal matrix P.


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