orthogonally-diagonalize-the-matrices-in-exercises-13-22-giving-an-orthogonal-matrix-p-and-a-diago-4

Question

Orthogonally diagonalize the matrices in Exercises $13-22,$ giving an orthogonal matrix $P$ and a diagonal matrix $D .$ To save you time, the eigenvalues in Exercises $17-22$ are: $(17)-4,4,7 ;(18)$ $-3,-6,9 ;(19)-2,7 ;(20)-3,15 ;(21) 1,5,9 ;(22) 3,5$
$$
\left[\begin{array}{rr}{6} & {-2} \ {-2} & {9}\end{array}ight]
$$

Bobby Barnes · Accepted Answer

So they want us toe orthogonal e diagonal eyes. These, uh, matrix here. So since they don&#x27;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&#x27;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&#x27;re going to take the determinant of this now. And remember, this should be equal to zero. Um, so we&#x27;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&#x27;s 54 and then negative 15 lambda and then plus Lambda squared minus +460 Let&#x27;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&#x27;s take this Miss scoop this down. So looks Thio five first. So we have a minus five I So that&#x27;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&#x27;ll be Route five. So we just need to multiply this by Route five, and then that&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;t really necessarily. But I&#x27;m gonna pull out that one over Route five, since it&#x27;s in everything just to make it look a little bit prettier. So 1/5. Um, then it&#x27;s too negative. 112 So this is going to be our diagonal ization de along with that orthogonal matrix P.

Chris Trentman · Answer

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&#x27;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&#x27;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&#x27;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&#x27;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.

Bobby Barnes · Answer

so they want us to or thought analyzes. And since they already give us the Eigen vectors arming the Eigen values, we could just go ahead and find the Eigen vectors and then normalize them and then plug everything in from there. So let&#x27;s first do for when Lambda is equal to negative three. So we do a minus minus three. I. So that&#x27;s really just like we&#x27;re gonna add three along the diagonal. So this is going to be for negative 64 negative 65 negative to four negative 20 And if we go ahead and roll reduces, this will end up giving us one zero negative one half 01 negative one and then all zeros down here. And so what this implies is that x one minus one half x three is equal to zero. So that&#x27;s just x one is even told one half x three And then over here X two minus x 30 or X two is a good X three so we could go ahead and write. So we have x one next two x three here, so x one we said Waas one half x three x two is just x three and the next three is X three. So we go ahead and pull that out. So it be x 31 half 11 Now, before we actually normalize this, I don&#x27;t have toe square fraction. So I&#x27;m just gonna multiply this by two. So it won&#x27;t really change anything since when we normalize it. That&#x27;ll get accounted for. So just be 1 to 2. So now let&#x27;s normalize this one. So to do that, it would be one squared plus two squared plus two squared spur voted. So that would be, um Well, what I do to here, Right that to square. So that would be four plus 48 plus 19 So that&#x27;s actually just gonna be three. So we divide all of this by three. So that actually gives us when Lambda One is even to negative three. Our first Eigen vector is going to be one third, two thirds and two thirds. All right, now we&#x27;ll do the same thing. But with Lambda equal to negative six. Um, let me pick this up and scoop this down. Just we kind of have it with us. Eso would be a minus minus six eyes. So again, it&#x27;s gonna be like we just add six to all of this. So this is going to be equal to s 0785 and everything else stays the same. And now, if we were to go ahead and roll reduce this, we should end up with. Actually, this is in a five. Here. Um, there&#x27;s maybe three plus three are negative. Three plus six is three, right, but only very reduced. Now, this one that should give us 101 are 011 half and then 000 So then this implies that x one plus x three is equal to zero. Um, which is gonna be next one is equal to negative x three and then for the next one. This is saying x two plus one half x three is going to be zero or X to negative one half x three so that we can use that to build our Eigen vector. So we have x one x two x three So x one was negative x three x two is negative one half x three and x three is just x three So this is x three negative one negative one, House one. And just like before, I&#x27;m gonna multiply all this by two. Uh, just cause I don&#x27;t wanna have toe square a fraction. So this is going to be negative, too. Negative one to. So now we can normalize this one on doing that would get negative two squared plus negative one squared plus two squared, all square rooted. So that would be four plus 48 plus one. So, again, three once described it. So then that tells us when lambda, uh, two is equal to negative six. We get our second director being so we just divide all that by three. So negative two thirds negative. One third and to third. And then our last one. We&#x27;re going to do this with nine. Let&#x27;s go ahead and scoot this town and going to now do a minus nine I. So we just tracked nine along the diagnosis. That be negative. Eight negative seven and negative 12. And then everything else stays the same. And then when we rode use this, we should end up getting eso. It&#x27;s 10 negative too. 012 and then 000 So then this implies that x one minus two x three is equal to zero or x one is equal to x three. And in that second equation is saying. X two is x two plus two x three is equal to zero or x two is negative two x three and then we go ahead and plug that into x one x two x three. So then again, X one is, uh, it&#x27;s just a equal sign, uh, two x three x two is negative two x three and x three is just x three. Pull that out. So we have to negative 21 and then we&#x27;ll go ahead and normalize this one. And doing that, we should get it&#x27;s gonna be two squared plus negative two squared plus one square rooted. So again, that&#x27;s just going to be, uh, Route nine or three. So then that would give us when Lambda is equal to nine. Uh, our third Eigen vector is going to be too negative to one, or we need to divide all these by three first, so it would be two thirds negative two thirds and one third. So let&#x27;s go ahead and write out with our diagonal ization matrix system. This is gonna be D is equal to, um So we had negative six negative. Three and nine. Remember? Doesn&#x27;t remember the order you write these a song is when you write your or thought little matrix. You kind of line everything up. But I&#x27;m just used to draw writing it from smallest to largest like this. All right. And so now let&#x27;s write what are vectors are actually the one associate with nine here, So if we write, p is the one we just found. So this is that two thirds negative. Two thirds. One third now for negative three. Uh, this should be actually, let me just pick these up and scoop them all down. Then there will be easier for me to just plug it. So he&#x27;s down. Okay, so now for negative three, we have of one third, two thirds to third and then or negative six, we have negative two thirds negative. One third to third. And if you want, you could go ahead and pull out that one third just to make it look a little bit prettier. So this isn&#x27;t really needed, but it will make it look better s that we have negative too negative. 12 1 to 2. And then to negative. To what? So now these are going to be our diagonal ization, along with far orthogonal matrix P.

Bobby Barnes · Answer

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&#x27;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&#x27;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&#x27;s going to have one minus. Lambda is able to plus minus five and then we I&#x27;ll add land over and I&#x27;ll subtract The plus surmised fives that just as lambda isn&#x27;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&#x27;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&#x27;ll do negative for first, Um, a minus minus four I or, in other words, we&#x27;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&#x27;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&#x27;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&#x27;s gonna be route to. So we multiply this by problems are. Divide this by route to and then that&#x27;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&#x27;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&#x27;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&#x27;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&#x27;s going to give us a route to again. So we divide this by route to, and that&#x27;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&#x27;re going to have So let&#x27;s first create the diagonal. So it&#x27;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&#x27;s come back appearance, see where it waas. There&#x27;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&#x27;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&#x27;s our diagonal ization. And then that would be our orthogonal matrix.

Pagadala Kishore Reddy · Answer

Bye. Right. Minus minus My no land one my neck than that you might be then that he was then, Lambert. One that again. My one. And when Lambert truly minus then Wait. We cool One body. You When again That responding even. Really? My, my What? No, not planes again. Victor! Oh, you won my name. One. You all right? You too. One day, Jewel. And you be quiet, My baby! No! Minus one. Jeeva One label well, and one brake and know that I wanna be mine. Three You don&#x27;t. People, people, people jiggle, jiggle. Thank you for watching.

Chris Trentman · Answer

were given a matrix and were asked towards orginally diagonal eyes The matrix giving north ogle matrix P in a diagonal matrix t You&#x27;re also given that the Eigen values of this matrix are 15 and nine. The matrix is a which is 431134111143 in 1134 with likened I use lander equals 15 and nine. We have that A is a four by four matrix. So we know that one of these exactly one of these Eigen dies. Must have a multiplicity of to. So we&#x27;ll keep that in mind. I noticed that a transposed is equal to Dagnall stays the same 44 for four And then we&#x27;ll exchange three with three one with one one with one again one with one one with one and again three with three and we see that a transpose is the same as a So it follows that a is a symmetric matrix. Therefore it follows that a is orthogonal e day agonal izabal And so there do exist such matrices p and g such that a is equal to PDP inverse to find these matrices we need to find Eigen vectors for a So take Lambda One to be I can value one. We have bits. I connector V one satisfies the equation A minus I The one equals zero vector. And so we have that matrix a minus. I is. 3311 33 11 11 33 1133 and using real reduction to be contributes this to the matrix. 11 33 33 11 0000 0000 And so we can further reduce this matrix to the matrix. 1133 and then attracting two of row one from row three. He gets three minus 303 minutes. Three a zero one minus nine is negative. Eight and one minus nine is negative. Eight 0000 0000 We can actually divide this second room by negative eight. Simply get 11 And then one more time, we&#x27;ll subtract three of row two from row one to get 1100 0011 0000 0000 And so, with this maitresse matrix and reduced row echelon form. We obtain the system of equations. The one one plus the 22 equals zero and the 33 plus current. This is V 12 and V 13 plus the +14 equals zero. So we have that the 12 equals negative view on one. In that view on four equals negative V 13 So if we take the 11 to be the parameter Yes. And you won three of the parameter t We get that The general form for the Eiken victor, The one is vector Yes, negative ists t negative t which may be written as the linear combination s times one negative one 00 plus tee times 00 one negative one. And so, if we take s to be equal to one and TV quarter zero we obtain and I in vector V 11 which is the vector one negative one 00 And if we take as to be zero and t to be one, we get the I convert your V 12 which is 00 one negative one and by construction we have that he&#x27;s like in Victor&#x27;s V 11 and V 12 are a basis for this I in space. The 11 dotted with V 12 is equal to zero plus zero plus zero plus zero. And so it follows that this is equal to zero and that we also have that V 11 and V 12 are orthogonal. And so this isn&#x27;t orthogonal set as well you Finally we have that the norm of the 11 is equal to the square root of one plus one, which is route two. Likewise, the norm of the 12 is equal to the square root of one plus one, which is Route two. And so we have that the unit vector you won is equal to the 11 over the norm of the 11 which is equal to one of a route to negative one of route 200 And the unit vector U two is the vector V 12 over the norm of B 12 which is equal to 00 One of a route to negative one of her too. And we have it you want and you two are Ortho normal vectors, that former basis. So that&#x27;s you on You too. Is an Ortho normal basis for this Eigen space. Take land A to to be the Agon value five. Then we have that. The connector V two satisfies a minus five. I be to equals zero The matrix a minus five. I is the matrix. It&#x27;s native ones down the diagonal and then 311 11 three. And then you symmetry. Fill in the rest. Using row reduction. We can rewrite this as nobody arranged the rose 1st 11 negative. 13 113 Negative one negative. 1311 and three Negative. One 11 And here I will subtract three of real one from row for and add one of rue one to row three so that I get 11 Negative. 13 I&#x27;ll also subtract one of Rome one from row two. This is one minus 10 one minus 10 three minus negative. One is for and negative one minus three is negative for So we have one plus negative one and zero. One plus three is four 81 plus one is zero and three plus one is also for Finally, we have three minus 30 Negative one minus three is negative. Four one minus three times negative. One is one plus three, which is four and one minus nine is negative. Eight. We can actually rewrite these rows as 00 to Nick. Prisoner 01 Negative. One 01 zero one. And this last row is going to be zero one negative one to this. Could be written in reduced. So says one one negative. 13 zero 101 01 negative. 12 and zero 01 Negative one. Simplify. I will subtract one of road to from row one and what I&#x27;ve wrote to from row three. So we have one one minus 10 Inedible to minus zero is negative. 13 minus one is to 0101 You&#x27;re a minus zero is zero one minus 10 Negative one minus heroes. Negative one to minus one is one and last row stays the same. 001 Negative one. And we see that we can eliminate the fourth row and obtained the matrix 10 Negative. 12 01 01 00 one, Negative. 10000 In fact, all right, this little bit differently. So multiply this by a negative one and then I&#x27;m going toe Also, subtract or not to track, but add row three to row one. So we have one plus zero is 1000 negative. One plus one is zero and to post negative one is one. And so we obtained the system of equations. The 21 plus V 24 equals zero. The 22 plus the 24 equals zero and the to three minus feet to four equals zero. This tells us that the 21 is equal to negative V 24 v 22 is equal to negative V 24 and that&#x27;s V 23 is also equal to B 24 And so, if we take V 24 to be parameter negative t and we get the general form for the Eigen vector V two is the vector t to negative T and negative T can be written as t times the vector 11 negative one negative one. In setting t one, we obtain that the Eigen vector B 21 is one one negative one negative one. So v 21 is 11 negative One negative one we have it. The norm of the 21 is equal to the square root of one plus one plus one plus one which is the square root of four, which is to And so we have the unit Vector you three is equal to this is V 21 over the norm of V 21 which is going to be 1/2 1 half negative, 1/2 negative. 1/2. We have that since you three is a unit vector and is an eye conductor and as a basis we have This is in or so normal basis for the Eigen space associated with Lambda too. Finally. But Lambda Three be the last remaining Eigen Vector, which is nine. Then we have that any for Eigen value than the eye conductor. The three will satisfy the equation. A minus nine i the three equals zero vector the matrix a minus nine. I is the matrix A Except the diagonals are now going being negative five. So Negus five native, five native, five negative size and then 311113 And use symmetry to fill in the rest. Using row reduction. All right, this first as 11 Three negative. Five 11 negative. 53 and three. Negative. 511 Negative. 53 11 Then I was attract Row one from road to it&#x27;s attract three of row one from row three and add five for one to row for so I get matrix 113 Negative five one minus 101 minus 10 Negative. Five minus three is negative. Eight and three minus negative. Five is positive free minus 30 Negative five minus three is negative. One minus nine is native eight and one plus 15. 16. Negative. Five plus 50 three plus five eight one plus 15 is 16 and one minus 25 is negative. 24 and this can be simplified to the matrix with entry 00 Then this is one negative one. This changes to 11 negative too. And this will be one to negative three. We can rearrange this to obtain the matrix 113 native five 00 one negative one. But instead that will have zero 11 negative too. 01 to negative three and then 001 negative one With further reduction, I&#x27;ll subtract row two from row one so that I get when minus. Here was one one minus one is zero three minus one is to and negative five minus negative too. Is negative. Three 011 negative too. And I&#x27;ll subtract Row two from row three to get zero minus 001 minus 10 Tu minus one is one and negative. Three minus negative. Two is negative one and finally 001 negative one. And this reduces to the matrix one. This time I&#x27;ll subtract two of row three from row one. So we get one minus zero is here 10 minus 00 Tu minus two a zero and negative three minus two times negative. One is negative. Three plus two, which is negative. One zero. That&#x27;s attracts one of row three from road to So I have zero minus 001 minus zero is one one minus one is zero and negative. Two minus negative. One is negative. One 001 negative one. And finally, it&#x27;s attracting row three from row 40000 from which I obtained the system of equations. The 31 minus the 34 equals zero. The three to minus B 34 equals zero and V 33 minus B 34 equals zero. So we have that. The 31 equals V 32 equals B 33 equals V 34 And so if we take B 31 to be the parameter t, we have the general form for the I conjecture me three is T t T t or tee times 1111 And if we take TV one, we obtain the I in vector V three, which is 111 one. Um, we have that The 11 We&#x27;re not the 11 Sorry. The three is a basis for this Eigen space. The norm of the three is equal to the square root of one plus one plus one plus one just the square to four to so that the unit vector you four is equal to vector. The three divided by form of the three is equal to 1/2. 1/2 1/2 1/2 And so we have that you four is going to be in Ortho normal basis for this Agon space Since a is symmetric. It follows that the set of unit vectors You won. You too, You three and you four, our orthe ogle and he&#x27;s form a basis. So this is going to be a complete worth. The normal set of Eigen vectors for a And so if we take P to be the Matrix Who&#x27;s calling vectors? Are you one? You too? You three. And you four. This is the same as the Matrix. One of a route to negative one of a route to 00 00 One of her too negative. One of two. Then You three is 1/2 1 half negative. 1/2 negative on behalf New four is 1/2 1/2 1/2 1/2. So this is the Matrix P in the Matrix D is going to be diagnosed matrix whose entries are the Eigen values. So we have that Lambda one was an Eigen value multiplicity to since the wagon space mention of to So we have to ones five and a nine. The rest are zeros. So this is our matrix D

Bobby Barnes · Answer

you in there? It&#x27;s my night. No, I have to see it. Money. Lambert, You go nine minus lambda. My for what do you by solving the question you get then and remember like No. When Lambda wanted, I convicted. We want equality minus one by two. One. And when Lamba Week was by then, I need to Who won? No, they&#x27;re not. You want me minus one bagel bite? Nobody look like And you equal Who? One day who? No, let&#x27;s be calls minus one little bite. Nobody. A little light a second. Coleman off one, buddy. And that I gonna be Ben Eagle people. Right? Thank you for watching.

Problem

Orthogonally diagonalize the matrices in Exercise…

Problem 16 Medium Difficulty

Answer

$P=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}{2} & {-1} \\ {1} & {2}\end{array}\right), D=\left(\begin{array}{cc}{5} & {0} \\ {0} & {10}\end{array}\right)$

More Answers

Related Courses

Linear Algebra and Its Applications

Related Topics

Discussion

Top Algebra Educators

Anna Marie V.

Kayleah T.

Kristen K.

Michael J.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 7

Video Transcript

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

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$P=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}{2} & {-1} \\ {1} & {2}\end{array}\right), D=\left(\begin{array}{cc}{5} & {0} \\ {0} & {10}\end{array}\right)$

Linear Algebra and Its Applications

Anna Marie V.

Kayleah T.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Anna Marie V.

Kayleah T.

Kristen K.

Michael J.

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

Linear Algebra and Its Applications