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[M] Orthogonally diagonalize the matrices in Exercises $37-40$ . To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and, for each eigenvalue $\lambda,$ find an orthonormal basis for $\mathrm{Nul}(A-\lambda I),$ as in Examples 2 and $3 .$ $$\left[\begin{array}{rrrrr}{8} & {2} & {2} & {-6} & {9} \\ {2} & {8} & {2} & {-6} & {9} \\ {2} & {2} & {8} & {-6} & {9} \\ {-6} & {-6} & {-6} & {24} & {9} \\ {9} & {9} & {9} & {9} & {-21}\end{array}\right]$$

The eigen values are $-30,6,15,30 .$ So the diagonal matrix will be $5^{*} 5$ with diagonal elements as the eigen values and test $0 .$ The p matrix will be matrix of all normalized eigen values. For normaliaing. Just divide the elements by square of sum of elements.

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 1

Diagonalization of Symmetric Matrices

Introduction to Matrices

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were given a matrix, you were asked two orthogonal E di act allies. This matrix Matrix's 8 to 2 negative. 69 282 Negative 69228 Negative 69 Negative six. Negative six. Negative six 24 nine 99 nine nine Native 21 and I could read out the transpose and show that it's equal to a But I think, just been looking at this matrix. It's pretty clear that the entries opposite each other on the diagonal across the diagonal are equal. So it's clear that a transpose is equal to a which means that A is a symmetric matrix and therefore follows that a is fourth organ elite diagonal Izabal. So this problem makes sense to our Thorgan only diagonal eyes. A first we'll find the icon dies of a characteristic polynomial of a is the determined of a minus. I am toe I, and this is a five by five metrics, so calculating the determinant could take a very long time. Instead, I'll use a computer program to do this, and that went to do the same and pause here. So, using a computer algebra system, I calculated that the characteristic polynomial is in factored form Lambda plus 30 times Lambda minus 30 times Linda minus 15 times Linda minus six squared. You want to find the solutions to the characteristic equation? Just once equals zero. So every to the Paul, no meal. And so we have their Eigen values are Lambda one equals negative. 30 I'm 22 equals positive 30 and the three equals 15. And Linda four, is equal toe. Linda five is equal to six. So it has a multiplicity of to If he is an Eigen vector associated with Lambda one, it satisfies the equation. A plus three i times v equals zero vector to find B. Let's reduce the matrix a plus 30. I pause so you can do this yourself. After reduction, we obtain the matrix 10004 or when fourth I mean so you're a 100 1/4 zero zero's 10 1/4 000114 This red last year was actually just zero room and so we obtained the equations. The one is equal to negative 1/4 before, which is also equal to the two, which is equal to be three And so if we take the four to be the parameter negative 40 and we obtain general form of the Eigen vector, the is negative. 14 attempts in 84 has won t t t native 40 core tee times the vector 111 negative 14 Therefore I mean to take TV one. We obtained the Eigen vector one, which is 111 negative four the normal V one. Well, we have that The unit vector you one is the one over the norm of the one which is not everyone is one plus one plus one plus 16 square rooted, which is Route 19. So that's going to be one of a Route 19 one of the Route 19 one of her 19 and negatives 4/19. I actually didn't make the steak. This should be 1 1/4 here as well and then another row of zeros. And so they should actually be that this is all equal to negative. 1/4 the five. And so if we take the five to be the parameter negative 40 he obtained T T t t native 40 which is 111 one negative four now the norm is 16 plus one. 17 18 19 squared 20 Which is going to be to rid five. This is one over to route five. One of her to rid five one over to room five and negative 4/2 5 which is negative to over Route five. Likewise, if he isn't Aiken Victor associated with Linda to it satisfies a minus 30 I the equals zero vector. Find a V will have to reproduce a minus 30. I might wanna pause here this time we obtain Matrix 100 1/3 0 010 1/3 0 in 001 When 3rd 0 0000 one in a row of zeros. So we obtain the system of equations. V one is equal to negative 1/3 before V two is also equal to negative on third before, as is the three we have that fee five is equal to zero. So if we take the four to be the parameter negative three t, then we have the General I conductor V. It's going to be t t t negative three t zero or tee times a vector 111 negative 30 If we take tea to be one we obtained the Eigen vector be to which is 111 negative three zero unit Vector U two is V two over the normal V two and the normal V two is square root of 12 which is 233 And so this is going to be one over to three. 1/2 or three. 1/2 or three. Negative 3/2 3 and zero. Next. If V is an Eigen vector associated with Lambda three, it satisfies a plus 15. I, our a minus 15. I mean, I times v equals a zero vector. So we want to reduce matrix a minus 15 i And when a pause and do this yourself after row reduction, we obtain the matrix. 1000 negative. 10100 neck of 10010 Negative warning 0001 negative one And the rest of zeros. From this we obtain the system of equations. The one equals B five, as does B two b three and before. So if we take the five to be the parameter t and we have General form V is simply t t t t t or tee times, all ones. So if you take TV one we obtained you I connector V three, which is all ones, and we have that you threw. The unit vector is B three over the normal V three, which is normal. V three is going to be the square root of five. So this is one of unverified. Whatever Route five. Whatever. Route five. One of every five. Whatever. Five. And next we have that. If he isn't Eigen vector associated with Lambda Four or lamed of five that falls that it satisfies the equation. A minus six i times V equals zero. So we want to reproduce the matrix a minus six. I I want to pause while you do this on your own. After a reduction, we obtain the matrix. 11100 0001000001 And the rest of the entries are zero. So we obtained the system of equations. Excuse me. V one equals negative two minus 83 and V four and the five are equal to zero. And so if we take V two to be parameter negative s in the three to be the parameter negative t we obtain That V is the vector s plus t negative s native T 00 is equal to the linear combination s times one negative. One 000 plus two times one zero negative one 00 And if we take s to be one and t v zero, we obtained the Eigen vector before, which is one negative 1000 And if we take s to be zero in t to be one, we obtain the second Eigen vector the five She's 10 negative warm 00 by construction. We have that before and the fiber linearly independent. However, we have that since their dot product is non zero even though they're linearly independent and said they're a basis of I didn vectors you have that before out of it be five is equal to one which is not zero and therefore we have it before is not orthogonal to be five Tina North Ogle basis of Eigen vectors will use the Gram Schmidt or thought generalization procedure. So let's w four equal to be four ticked up You five to be be five minus the projection of the five under the four his before That would be five over the normal before times before this is equal to the five. This one's your negative. 1000 Here is your I guess, minus. We calculated this to be one over the normal before, which is route two times before, which is one negative. 1000 This is equal to one minus one of the route to zero plus one of route to negative one minus zero and 00 They should actually be the norm of the four squared here. So instead of being divided by route to this is divided by two. So this is actually not one minus one of her two, but one minus 1/2 which is 1/2 of illnesses. Also, 1/2 this is negative ornament. 00 And so we have that. You four is the unit vector, which is W four over the normal w four just going to be one of the route to negative one of two 000 And do you five this unit vector W five over the norm of w fine, which is equal to instead doing this. I'll do to W five over normal to W five sense or thought banality, and I can tell you stole preserved and this is going to be. While the norm of two w five is one plus one plus four square rooted just squared. Six. This is going to be one of a Route six 1/6 negative two of the route six 00 and we have that u four and u five former North Normal basis where this wagon space. And so we have that the set of vectors u one you too. U three u four and you five are a complete set of work, the normal Eigen vectors of a and so we can diagnose eyes A by taking p to be matrix. Whose column vectors are you went there, You five War one of her to rid 51 of her to rid 51 of her two word five negative two of every five. You, too, which is one of her two or 3 1/2 or 31 of her two or three negative 3/2 or 30 one of a Route 51 of the routes life one of a roof life in every fight whenever. Five one of her to negative one of 2000 And whatever Route 61 of a Route six Negative two of her Route six since year zero. This is one option for P and for B take you to be the Matrix just diagonal into his entries of the Eigen values in order. So we start with negative 30 then 30 then 15 and six and six. The rest of the entries are zero. This is our answer.

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