Query 28

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course Mth 279

5/8 2

Query 28 Differential Equations*********************************************

Question: Diagonalize the matrix [2, 3; 2, 3].

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Your solution:

For A = [2,3;2,3]

Has eigenvalues

`lambda_1 =5, `lambda_2 =0

Giving us he eigenvector

x_1 = [1; 1], x_2 = [-(3/2),1]

So that T matrix equals,

[1, -(3/2);1,1]

Finding T^-1

[(2/5),(3/5);-(2/5),(2/5)]

So that

T^-1*(AT) = [(2/5),(3/5);-(2/5),(2/5)]*( [2,3;2,3]* [1, (-3/2); 1,1]) =[(2/5),(3/5);-(2/5),(2/5)]* [5, 0; 5,0]

=[5,0;0,0]

Where [5,0;0,0] is our Diagonalized matrix.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix

[7, -2, 2; 8, -1, 4; 8, 4, -1 ].

The characteristic equation of this matrix is (lambda - 3)^2 ( lambda + 1).

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Your solution:

For lambda = 3, algebraic mult. = 2 and geometric mult. = 2

For lambda = 1, algebraic mult. = 1 and geometric mult. = 1

T=[1,1,0;2,0,0;0,-2,0]

?????I can’t figure out what I’m doing incorrectly,I get the vectors,x1 =[1;2;0] and x2=[1;0;-2] for `lambda=3. For lambda=1 I get x3=[0;0;0]?????

I think I see my problem, we can’t find a diagonalizable matrix because x3 isn’t linearly independent right?????

I’m going to say that A = [7, -2, 2; 8, -1, 4; 8, 4, -1 ], cannot be diagonalize.

@& The problem is that I apparently had a typo in the matrix. It doesn't yield the equation (lambda-3)^2 ( lambda + 1) = 0*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: Find the algebraic and geometric multiplicity of each eigenvalue and, if possible, diagonalize the matrix

[ 5, -1, 1; 14, -3, 6; 5, -2, 5 ].

The characteristic equation of this matrix is (lambda - 2)^2 ( lambda -3).

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Your solution:

For lambda = 2, algebraic mult. = 2 and geometric mult. = 1

For lambda = 1, algebraic mult. = 1 and geometric mult. = 1

A is not diagonalizable due to lambda=2 having geometric mult of 1.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: Solve the system

y ' = [ -4, -6; 3, 5 ] y + [e^(2 t) - 2 e^t; -e^(2 t) + e^t]

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Your solution:

First we diagonalize the matrix where,

A= [ -4, -6; 3, 5 ]

Has eigenvalues

`lambda_1 =-1, `lambda_2 =2

Giving us he eigenvector

x_1 = [2; -1], x_2 = [-1,1]

So that T matrix equals,

[2, -1;-1,1]

Finding T^-1

[1,1;1,2]

So that

T^-1*(AT) = [1,1;1,2]*( [-4, -6; 3, 5]* [2, -1;-1,1]) =[1,1;1,2]* [-2, -2; 1,2]

=[-1,0;0,2]

Where [-1,0;0,2] is our Diagonalized matrix.

Letting y(t) = T*z(t) we get

d/dt[z1;z2] = [-1,0;0,2]*[z1;z2] + [e^(2 t) - 2 e^t; -e^(2 t) + e^t]

Which we can transform into 2 uncoupled scalar equations

z1’ = -z1 + e^(2 t) - 2 e^t

z2’ = 2(z2) - e^(2 t) + e^t

Giving us solutions

z1(t) = -1-e^(t) + c1e^(-t)

z2(t) = -(1/2)e^(4t)+e^(3t) + c2e^(2t)

Since y(t) = T*z(t)

y(t) = [2, -1;-1,1]*[ -1-e^(t) + c1e^(-t); -(1/2)e^(4t)+e^(3t) + c2e^(2t)]

@& The particular solution to the first equation would be of the form

z_1_P = A e^(2 t) + B e^(-t).

The equation would give us

(2 A + A) = 1 so that A = 1/3

B + B = -2 so that B = -1

z_1_P would therefore equal 1/3 e^(2 t) - e^(t)

Similar steps would give us z_2_P.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: Solve

x '' = [ 6, 7; -15, -16] x

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Your solution:

Rewritten as

x’’ +(-Ax)= 0

Where -A = [-6,-7;15,16]

First we diagonalize the matrix where,

A= [-6,-7;15,16]

Has eigenvalues

`lambda_1 =1, `lambda_2 =9

Giving us he eigenvector

x_1 = [1; -1], x_2 = [-1,(15/7)]

So that T matrix equals,

[1,- 1;-1,(15/7)]

Finding T^-1

[(15/8),( 7/8);(7/8),(7/8)]

So that

T^-1*(AT) = [(15/8),( 7/8);(7/8),(7/8)]*( [-6,-7;15,16]* [1, 1;-1,(15/7)]) =[1,1;1,2]* [-2, -2; 1,2]

=[1,0;0,9]

Where [1,0;0,9] is our Diagonalized matrix.

We now rearrange into the form

w’’ + Dw = 0

So when uncoupled we have

w_1’’ + w_1 = 0

w_2’’ + 9w_2 = 0

Solving characteristic equation for the equations we get r = -i,i for w_1 and r= -3i and 3i for w_2. So we have

w_1(t) = Acos(t) + Bsin(t)

w_2(t) = Acos(3t) + Bsin(3t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Would you believe I have worked through this whole assignment before I just realized that the diagolzination is related directed to the lambda values. Is that a coincidence that that the lambda values for almost every problem I’ve worked out was the values that were on the diagonal or is that how it always works out????????

I feel SOOO stupid sometimes!!!!

@& If you were stupid you wouldn't be thinking about these things and never would have noticed.

The property of the diagonal matrix is that when you multiply it by an eigenvector you get a multiple of the eigenvector, and that multiple is equal to the eigenvalue times the eigenvector.

Since in the eigenvector basis the vectors [1, 0, ..., 0], [0, 1, 0, ..., 0] etc. represent the eigenvectors, which when transformed become [lambda_1, 0, ..., 0], [0, lambda_2, 0, ..., 0] , the transformation matrix must therefoe have lambda_1, lambda_2, ... on the diagonals and zeros everywhere else.

In a 2-credit linear algebra class you simply can't cover change of basis in depth, but this is what's going on with a similarily transform.*@

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Self-critique rating:

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&#Good responses. See my notes and let me know if you have questions. &#