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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$$$\left[\begin{array}{rr}{1} & {-5} \\ {-5} & {1}\end{array}\right]$$

$P=\left[\begin{array}{cc}{\mathbf{u}_{1}} & {\mathbf{u}_{2}}\end{array}\right]=\left[\begin{array}{cc}{1 / \sqrt{2}} & {-1 / \sqrt{2}} \\ {1 / \sqrt{2}} & {1 / \sqrt{2}}\end{array}\right]$$D=\left[\begin{array}{cc}{6} & {0} \\ {0} & {-4}\end{array}\right]$

03:48

Pagadala K.

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 1

Diagonalization of Symmetric Matrices

Introduction to Matrices

Missouri State University

McMaster University

Idaho State University

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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.

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