(1) Consider the matrix A=[-253 -232 -96 1088 280; 213 204 93 -879 -225; -90 -90

Question

(1) Consider the matrix

A=[-253 -232 -96 1088 280; 213 204 93 -879 -225; -90 -90 -47 360 90; -38 -36 -18 162 40; 62 64 42 -251 -57]

(a) Find all eigenvalues of A.

(b) Determine a basis for each eigenspace.

(c) If the matrix is diagonalizable, construct a matrix that will diagonalize it. You may not explicitly use the diagonalize command in MATLAB.

(2) Consider the matrix

B=[ 0 1 1 0; 1 -2 2 1; 1 2 -2 1; 0 1 1 0]

Since this matrix is symmetric, it is orthogonally diagonalizable. Again, without using the diagonalize command in MATLAB, find an orthogonal matrix that P diagonalizes B. Make sure you show that P is orthogonal.

Explanation / Answer

code:

a)

A=[-253 -232 -96 1088 280; 213 204 93 -879 -225; -90 -90 -47 360 90; -38 -36 -18 162 40; 62 64 42 -251 -57];

[P,D]=eig(A);

which gives:

P =

0.7201 0.7246 0.5533 0.7128 0.5842
-0.6172 -0.6012 0.0376 -0.6040 -0.0837
0.2572 0.2581 -0.4651 0.2622 -0.2207
0.1029 0.1098 -0.0817 0.1165 -0.0981
-0.1543 -0.1867 0.6851 -0.2116 0.7703

D =

7 0 0 0 0
0 -2 0 0 0
0 0 -2 0 0
0 0 0 3 0
0 0 0 0 3

where eigen values are diagonal elements of D:

7,-2,-2,3,3

b)

Basis in coulmn vectors of P corresponding to eigen values of 7,-2,-2,3 and3 repectively:

P =

0.7201 0.7246 0.5533 0.7128 0.5842
-0.6172 -0.6012 0.0376 -0.6040 -0.0837
0.2572 0.2581 -0.4651 0.2622 -0.2207
0.1029 0.1098 -0.0817 0.1165 -0.0981
-0.1543 -0.1867 0.6851 -0.2116 0.7703

c)

let D be diagonal vector:

D=inverse of(P)*A*P

code:

diagonal=inv(P)*A*P;

which gives:

diagonal =

7.0000 0.0000 0.0000 0.0000 0.0000

0.0000 -2.0000 0.0000 0.0000 0.0000

0.0000 0.0000 -2.0000 0.0000 0.0000
0.0000 0.0000 0.0000 3.0000 0.0000
0.0000 0.0000 0.0000 0.0000 3.0000

2)

PTP=I

PT=P-1

code:

B=[ 0 1 1 0; 1 -2 2 1; 1 2 -2 1; 0 1 1 0];

[P,D]=eig(B)

which give:

P =

0.0000 -0.5000 -0.7071 -0.5000
0.7071 0.5000 0.0000 -0.5000
-0.7071 0.5000 -0.0000 -0.5000
0 -0.5000 0.7071 -0.5000

D =

-4 0 0 0
0 -2 0 0
0 0 0 0
0 0 0 2

which gives:

check if P is orthoganal or not

code:

transpose(P)*P

which gives:

ans =

1.0000 0.0000 -0.0000 -0.0000
0.0000 1.0000 -0.0000 0.0000
-0.0000 -0.0000 1.0000 -0.0000
-0.0000 0.0000 -0.0000 1.0000

which implies P is orthogonal matrix

diagonalizing B:

PTBP=digonal matrix

code:

transpose(P)*B*P

which give diagonal matrix:

ans =

-4.0000 -0.0000 -0.0000 0.0000
-0.0000 -2.0000 -0.0000 -0.0000
-0.0000 -0.0000 -0.0000 0.0000
-0.0000 -0.0000 0.0000 2.0000

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