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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$$$\left[\begin{array}{cccc}{4} & {0} & {1} & {0} \\ {0} & {4} & {0} & {1} \\ {1} & {0} & {4} & {0} \\ {0} & {1} & {0} & {4}\end{array}\right]$$

$$D=\left[\begin{array}{llll}3 & 0 & 0 & 0 \\0 & 3 & 0 & 0 \\0 & 0 & 5 & 0 \\0 & 0 & 0 & 5\end{array}\right]$$

Algebra

Chapter 7

Symmetric Matrices and Quadratic Forms

Section 1

Diagonalization of Symmetric Matrices

Introduction to Matrices

Campbell University

Oregon State University

University of Michigan - Ann Arbor

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were given a matrix A and were asked toward fog only diagonal eyes. This matrix giving an orthogonal matrix diagram matrix were also given that the Eigen values for this Matrix or Lambda equals three and five. The Matrix A is 401004101104001 04 Noticed that a transposed says the Matrix has the same diagonal for four. For four. We exchanged zero with zero one with one zero with zero zero with zero one with one and zero with zero and recognize that this is the same as the matrix. A. So it follows. That's since a transfers equals a is a symmetric matrix. Therefore, it follows that a is north Ogden Aly Diagonal Izabal Your thought only diagonal eyes A when he defined complete set of orthe normal vectors. Eigen vectors for a So take lambda One to be the first Eigen value three. Then we have that. If the one is an Eigen vector then satisfies the equation. A minus three i v one equals zero vector The matrix a minus three I is the same as matrix except for before has become ones So we obtained the matrix. 10100101 10100101 in this simply reduces to the matrix. 10100101 0000 0000 from which we obtained a system of equations V 11 plus V one of three equals zero and V 12 plus view and four equals zero. This tells us that the 13 equals negative view on one and the 14 equals negative b 12 And so if we take V 11 to be the parameter s and he one to be the parameter t, we get that The general form for the I conductor V one is the vector s T negative s negative T which would be written as a linear combination s times 010 negative one or zero 01 My mistake s times 10 negative 10 plus t times +010 Negative one. So we see that this wagon spaces two dimensional. And if we take s to be one and t v zero, we obtained one Aiken Victor called V one is the Agon vector 10 negative one zero it would take Has to be zero. On TV one, we obtain a second Eigen Vector V two, which is linearly independent from V one, which is the connector 01 zero negative one. And so we have that set V one be to is a basis for this Eigen space. They also have that V one dotted with feet too is equal to zero plus zero plus zero plus zero, which is zero. And so we have that V one is or thought Donald to V two. And therefore we have this set view and be two of the north Ogle set norm of V one is the square root of one plus one, which is Route two in the normal. V two is the square root of one plus one, which is Route two. And so we have the unit vectors. You one, which is V one over. Norm of the one is equal to 1/2 zero negative. One of a route to zero. And the unit after you too, is the vector V two over the normal vector he to which is 0 1/2 zero negative 1/2. And we have that both. You want you to our Eigen vectors and are linearly independent in North Ogle. So that set you on you too. Isn't Ortho normal basis for this Eigen space? Next, Let claimed to be the second and last Eigen value five. Then if the three is an Eigen vector associated with this Eigen value, we have that A minus five I times feet three is equal to the zero vector matrix. A minus five. I is the sings matrix A Except for the diagonals now negative one. Instead of positive one, we obtain the matrix. Negative. 1010 ones. Zero negative. One 01 10 Negative. 10 and 010 Negative one. This could be reduced to the matrix with reduction 10 Negative. 10 010 Negative. One and 00000000 from which we can obtain these system of equations. The 31 minus V 33 equals zero and be three to minus fee. 34 equals zero. So we have that. These 31 is equal to be 33 and V 32 is equal 33 4 So if we take B 31 to be the parameter s and V 32 to be the parameter t yet the general form for the I conductor V three is S T yes, T is equal to the linear combination s times 1010 plus T times +01 zero one And if we take us to be won t v zero weekends the Eigen Vector V three, which is 10 10 And if we take s to be zero and t to be one, get second Eigen Victor, before which is 0101 and by construction we have that these two wagon vectors are linearly independent and thus form a basis for this wagon space. We also have that V three died with before is equal to zero, which implies that V three is orthogonal to before. So it's actually in orthogonal basis for this vector space. The norm of V three is equal to the square root of one plus one which is to in the normal before is the square root of one plus one which is equal to. And so we have the unit Vector you three, which is equal to V three over the norm of you three, Our normal V three. My mistake is a vector one over route to zero 1/2 zero. And the vector You four is a unit vector, which is V four over the norm of the four. This is the vector zero. Whatever too zero one of the route to and we have that both u three and U four our Eigen vectors for this wagon space and are valuable and are unit vectors. So it follows that set you three. You four is an Ortho Normal aces for this Eigen space. Since our Matrix A is symmetric, it follows that you won and you two are Ortho or thought Donald U three and U four. So it follows that to set u one u two you three you for is a complete Ortho normal set of Eigen vectors for this matrix. And we have that. If we take P to be the Matrix who's calling Victor's Air U one U two, you three and you four. This is the Matrix one of a right to zero negative one of a route to 01 of a route to zero negative 1/2 and whatever to zero Whatever. Route to zero and zero one of a route to 0 1/2 when we take matrix D to be a matrix whose entries our gagan values on the diagonal. So you have each Agnelli was multiplicity too. And so this is the matrix 3355 and then zeros at the worlds. And using these two matrices p and G, we can diet.

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