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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$$$\left[\begin{array}{rrr}{3} & {-2} & {4} \\ {-2} & {6} & {2} \\ {4} & {2} & {3}\end{array}\right]$$

$D=\left[\begin{array}{ccc}{7} & {0} & {0} \\ {0} & {7} & {0} \\ {0} & {0} & {-2}\end{array}\right]$ and $P=\left[\begin{array}{ccc}{\mathbf{u}_{1}} & {\mathbf{u}_{2}} & {\mathbf{u}_{3}}\end{array}\right]=\left[\begin{array}{ccc}{1 / \sqrt{5}} & {4 / 3 \sqrt{5}} & {2 / 3} \\ {-2 \sqrt{5}} & {2 / 3 \sqrt{5}} & {1 / 3} \\ {0} & {5 / 3 \sqrt{5}} & {-2 / 3}\end{array}\right]$

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 1

Diagonalization of Symmetric Matrices

Introduction to Matrices

Campbell University

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University of Michigan - Ann Arbor

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Hello. In this case, we need to dive. Analyze the full of wind matrix. But we need to find an orthogonal matrix for the diagonal ization. So let's, uh let's start by getting the egg in value. So the egg in values for these military X corresponds to plumb the one that is seven. But in this case has multiplicity too on the second. Okay, Eigen value corresponds to minus two. Okay, so we got the Eigen values for these matrix. And the last thing that we need to do is to obtain the or the correspondent Eigen vectors on May make them or tonal to make the or tonal part. We're going to use the garnish meat, the banishment procedure. So basically, that's going to that's how we're going to obtain the orthogonal matrix. Okay, so let's start. So let's crypt. Let's get the Eigen vectors, Associate it to London one. So, in order to obtain that, we need to consider the following system that these a minus lambda I where I use the identity matrix. In this case, we're going to use Lambda one X is equals to zero, so we need to solve this system. Okay, so this system is equal to minus four minus two for minus two minus 1 to 4 to four. Multiply by x one X 23 equals 2000 Okay, so the next thing that way need to do is put these metrics here in the echelon form in the role echelon form. So we need to put some zeros and make some operations. Ah, that in this case, are the following. So if we take the second role and we put the second role minus one half off our one on the third role, we just some with the first role we obtain the following matrix the following Reduce it Rochon form of the metrics. So we have seen minus four minus 24 000000 times x one x 23 equals 20000 Okay, um, we can go further on divide this whole first roll by four. So at the end, we obtain one on half minus 1000000 We divide by minus four. So here is Are one is equal to 14 off our one minus one for X 123 I hear these equals 2000 So the solution for the system for this yes, for the system is that X one is equal to minus one half next to minus. Sorry. Plus, here is X two plus three unwell X to the next three or independent. So the solution in this case is so in this case, the solution is minus one half X two plus three x two x three and we can separate this. I Okay, so this yeah, can be separated into two. So those that are X to this one half 10 last x three 101 Okay, so this here corresponds to the Eigen vectors. Associate it to to number one. So I'm going to call them be One is minus one half, 10 on V to these equals toe 101 Okay, so now we need to make them or to normal. Okay, so how to do that? I'm going to start with this V two because it is easier to normalize. So v two right? The normalized because I'm going to call it you too. Okay, So you two is going to be V two divided the norm off to in this case, that the norm off to corresponds to the square root off to. So these YouTube is equals to one divided the square root off to 01 divided by the square root off to. Okay, so this is the first vector that is normalized the normal victim. Now, we need to make this V one. Actually, I'm going to call it to make it simple. This I'm going to go You want? Okay? Because it's our first normal vector. Now, we need to make this V one or Ciano on further or to normal. So we need to use the Grand Schmidt procedure. So basically, that means that YouTube will be V one, minus the dot product. Okay, off B one times two divided B one time to on Dhere V two. Okay, so with that, like, uh huh here. Yeah. Is B one story here is, uh, he got a mistake Here is V two times we too. Okay. Yes. Perfect. So just buy a comment, and these geometrically mean that both that here we got to here is V one. We want to make this V one or two or told not to be, too. So basically, we're subtracting the projection to be to so here we're subtracting this part, this part in the direction off. Me too. So here, this length we're obtaining by doing this part here. Okay. On multiplying by b two. We're thinking the direction off, too. So by subtracting that part, we're going to obtain on or tonal vector in this case, YouTube. Okay, so that's what we're going to do. So this result into you to minus is opposed to Well, the minus 1 to 0 minus one half or blast one half off 101 on this vector is equal to minus one half. Mhm. Mhm Two on one half. Okay. What? On the last thing? Actually, I'm going to call this in the in another way, because this is previously previous making it normal. I'm going to go to Alfa two. Okay, so it's just a naught owner Victor, to be one to be to. Okay, on. You too is going to be Alfa two divided. Norm off. Alfa two. This is equal to minus one. Divided three x square root off to two thirds square off to um, one divided three square root off to let me. Great. This better minus one three divided. The square root off to here is two thirds the square off to and here one divided three school off. So this is the next, um, or to normal vector to you one. So until now, we got two vectors you won. That is one divided. The square root off to zero. One divided, squared off to a new to that corresponds to minus one. Divided three years square root off to two thirds the square root off to on Dhere 13 squared off. Okay, so until now, we've got these two vectors are or tonal and even more. There they are. Normal. Uh uh. Normal vectors. So the last thing that we need to do is compute the Eigen vector associated to the Lambda too. Okay, so Linda, too, is minus two. So again, we need to solve the system a minus minus lambda to the identity. X equals to zero. So that is equivalent to say to fault the following system five minus 24 minus 28 242 on five. Here here x one x two x three and thes is equals to 000 Okay, so we got the system. We need to reduce these matrix here to the role Asian form on board were after performing some raw operations, we obtain the following matrix That is one 01 011 half 000 6123 and these is equals to 0000 Okay, so we got this system now. So after solving this system, Okay, so here we got that. Yeah, X one is equals to minus three on the X two is equal to minus one half x three. So the solution for the system is equals two x three minus one, minus one half on one. So, as you may see, this corresponds to only one vector on the Actually, this this vector is does this. So we three corresponds to minus one minus one half and one. Okay, So the last thing that we need to do is to normalize this vector. So, after normalization off this vector, do you three will be be three divided normal B three, this is equal. Tu minus the third, minus one third on two thirds. And this is the remaining vector the thing. The last thing that we need to do is well, the diagonal ization will be given by the Matrix de that corresponds to the Eigen vectors on the diagonals. So 70007000 and minus. So here we got the again the Eigen values for the metrics A on B that is the metrics that corresponds toe in each column, we're going to put each off this again vectors the normalized Aiken factors. So here we're going to put u one u two unused. Three. If we spend this expression, the matrix B looks like this one divided this. Cool it off to zero. One of the major squared off to minus one here, the third squared off to one divided three squared off to minus two thirds minus one third. Finally, the third's and that is a matrix B for the diagonal ization off a so thes. Here is the solution

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