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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$$$\left[\begin{array}{llll}{4} & {3} & {1} & {1} \\ {3} & {4} & {1} & {1} \\ {1} & {1} & {4} & {3} \\ {1} & {1} & {3} & {4}\end{array}\right]$$

$$D=\left[\begin{array}{llll}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 5 & 0 \\0 & 0 & 0 & 9\end{array}\right]$$

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 1

Diagonalization of Symmetric Matrices

Introduction to Matrices

Missouri State University

Campbell University

Baylor University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

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Orthogonally diagonalize t…

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Diagonalize the matrices i…

11:18

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were given a matrix and were asked towards orginally diagonal eyes The matrix giving north ogle matrix P in a diagonal matrix t You'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'll keep that in mind. I noticed that a transposed is equal to Dagnall stays the same 44 for four And then we'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'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's like in Victor'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'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'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'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'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'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'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'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'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'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'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'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'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'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'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'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'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'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

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