Solution Session3 Exercise2

Exercise Session 3 Exercise 2 (i) Done in class 2 2 1 2 (ii) A = 4 1 2 2

5

3 2 2 5:

Show that the vector equation Ax = x reduces to a single scalar equation. Find eigenvectors x; y corresponding to the eigenvalue 1 which have the property that xT y = 0: 2 In (i) we have shown that characteristic polynomial of matrix A is ( 1) ( 7): Eigenvalue = 1 is a root of characteristic polynomial of degree 2: It means that under linear transformation A; there is not only one eigenvector corresponding to = 1 that does not move, but there are two vectors that do not change direction under transformation A. As long as x and y are two eigenvectors of matrix A that correspond to eigenvalue = 1 (Ax = x; Ay = y); any linear combination of these vectors is also an eigenvalue of matrix A that corresponds to eigenvalue = 1: Let’s show this: take a and b arbitrary numbers s.t. (a; b) 6= (0; 0) : consider vector z = ax + by: Az = A(ax + by) = aAx + bAy = ax + by = z: Hence, Az = z and z is an eigenvector of A that corresponds to = 1: This means that there is in fact a hyperplane corresponding to = 1 that is not changed under transformation A. Any two nonparallel vectors de…ne a hyperplane, hence in our unchanged hyperplane we can choose any two nonparallel vectors that de…ne this hyperplane. In the question we are asked to choose two orthogonal vectors: xT y = 0. Let’s …nd to = 1 : 0eigenvectors, 1 0 corresponding 1 2 30 1 0 1 x1 0 2 1 1 2 x1 0 2 1 2 5 @ x2 A = @ 0 A (A I) @ x2 A = @ 0 A =) 4 1 x3 0 2 2 5 1 x3 0

2

1 4 1

1 1 2

2

3 2 2 1 2 5!4 1 4 1

1 1 1

3 2 2 5! 2

1

1

2

0

1 0 1 x1 0 @ x2 A = @ 0 A ! x3 0

x1 + x2 2x3 = 0: It means that the vector equation Ax = x reduces to a single scalar equation 0 1 x1 If x = @ x2 Ais an eigenvector corresponding to the eigenvalue 1; hence x3

1

x3 = x b=

x1 +x2 : 2

x kxk

Take x1 = 1; x2 = 1; x3 = 0 1 1 = p13 @ 1 A 1

1+1 2

= 1: Normalizing vector x :

0

1 y1 If y = @ y2 A is an eigenvector corresponding to the eigenvalue 1; hence y3

y3 =

y1 +y2 2 :

We are looking for 0 orthogonal bT y = 0 1vectors: x y1 p1 p1 p1 @ y2 A = p1 y1 + y2 + 3 3 3 3 y1 +y2 2

0 =) y1 =

y2 ; y 3 =

y1 +y2 2

=

y1

y1 2

y1 +y2 2

=

p3 (y1 3

+ y2 ) =

=0

1 y1 Hence, vector y of the form: y = @ y1 A is an eigenvector corresponding 0 to the eigenvalue 1 and y is orthogonal to x: Take y1 = 1 0

Normalizing vector y : yb =

y kyk

=

p1 1+1

0 @

1 0

1

0

1 A=@

p1 2 p1 2

0

1 A

So, we have found two orthonormal vectors corresponding to = 1 : x b= 1 0 p1 1 1 2 @ 1 A ; yb = @ p1 A : These two vectors de…ne a hyperplane that is not 2 1 0 changed under linear transformation A. 0

(iii) 0 Done in class 1 1 B zb = @

p

6 p1 6 p2 6

C A

(iv) Using spectral theorem for symmetric matrixes (matrix A is symmetric) we can write D = S 1 AS; where D is a diagonal matrix with principal diagonal composed of eigenvalues and matrix S is composed of corresponding eigenvectors. In our case: 2 1 3 2 3 p p1 p1 1 0 0 3 2 6 6 7 p1 p1 S = 4 p13 5; D = 4 0 1 0 5 2 6 2 p1 p 0 0 7 0 3 6 2

In class in was also shown that eigenvalues of symmetric matrix are orthonormal and S 1 = S 0 :

S

1

2

6 = S0 = 4

p1 3 p1 2 p1 6

Check that D = S 2 6 4

p1 3 p1 2 p7 6

p1 3 p1 2 p7 6

p1 3

0 14 p 6

32 76 54

p1 3 p1 2 1 p 6 1

p1 3

0 2

6 AS = 4 p1 3 p1 3 p1 3

p2 6 p1 3 p1 2 p1 6

p1 2 p1 2

0

3

7 5;

p1 3 p1 2 p1 6 3

p1 6 p1 6 p2 6

p1 3

0 p2 6

2

1 7 4 0 = 5 0

32

2 74 5 1 0 1 0

1 2 2 3

0 0 5 7

2

5

32 2 6 2 54

(v) It was shown in class that if A is symmetric then A 1 = SD 2 1 32 3 2 p1 p p1 p1 1 0 0 3 2 6 3 1 6 74 1 p1 p1 p1 56 0 0 A 1 = SD 1 S 1 = 4 p13 5 4 1 2 6 2 p1 p2 p1 0 0 17 0 3 6 6 2 1 32 1 3 2 3 1 p p1 p p p1 p1 1 2 6 2 7 6 3 3 6 p13 7 6 p13 7 4 7 1 6 7 72 5 1 p p1 p1 0 4 3 54 2 5= 7 7 7 2 7 6 2 2 2 3 1 2 1 2 p1 p p p p 0 7 7 7 3 7 6 6 6 6

3

1

p1 3 p1 3 p1 3

p1 2 p1 2

0

0

S :

p1 3 p1 2 p1 6

p1 3

0 p2 6

3

7 5=

p1 6 p1 6 p2 6

3

7 5=