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HW4, MAE 270A. Linear systems I. Fall 2005. UCI

Nasser M. Abbasi

June 21, 2014

Contents

1 Problem 3.12
2 Problem 3.13
 2.1 A1
 2.2 A2
 2.3 A3
 2.4 A4

1 Problem 3.12

Given

     (           )      (  )      (  )
     | 2  1  0  0|      | 0|      | 1|
     |           |      |  |      |  |
     || 0  2  1  0||      || 0||      || 2||
A  = |           |  b = |  |  ¯b = |  |
     ||           ||      ||  ||      ||  ||
     || 0  0  2  0||      || 1||      || 3||
     (           )      (  )      (  )
       0  0  0  1         1         1

What are the representation of A   w.r.t. the basis {b, Ab,A2b, A3b} , and the basis {¯b,A ¯b,A2¯b,A3¯b} respectively?

Answer

We are given the matrix A  and we want to find its repesentation under new basis, say Q  = {q1,q2,⋅⋅⋅,qn}

Let the new represntation Matrix be called A¯  .

The ith column of A¯  is the reprsentation of the base q
 i  under Q.

Now, the vector qi  itself is independent of its representation under different coordinates systems. Hence we can write

qi = A (qi representation under A ) = Q (qi representation under Q )

Hence let qi  representation under Q be the column x  , and let qi  representation underA  be column y  , then the above equation can be written as

qi = A (y ) = Q (x)
(1)

Now we solve for x  in the above equation since we know A  and y  and Q  .

This will give us one column of ¯
A  . We do the above again for each vector qi  , and at the end we have ¯A

Hence the algorithm is

input: Old basis A  , new basis matrix Q  , vector y  which is the representation of qi  under A

output: column x  which is the representation of qi  under Q  by solving equation (1).

Collect all these x′s  to build up the matrix ¯A

Hence we apply the algorithm for each of these vectors, one at a time.  

First we build the matrix Q  (the new basis), this is done by just calculating b,Ab,A2b, A3b  , this reults in

    (             )

    |  0  0  1  6 |
    ||             ||
    |  0  1  4  12|
Q = ||             ||
    ||             ||
    |  1  2  4  8 |
    (             )
       1  1  1  1

Hence to find the representation of y = b  under Q  , we solve for x  in equation (1). We do this for each y = b,y = Ab  , y = A2b  and y = A3b  as follows

pict

Solving for the last equation above is shown here.

(   )       (  )      (   )      (  )       (   )

| 24|       | 0|      | 0 |      | 1|       | 6 |
||   ||       ||  ||      ||   ||      ||  ||       ||   ||
| 32|       | 0|      | 1 |      | 4|       | 12|
||   ||  = x1 ||  ||  + x2||   || + x3 ||  ||  + x4 ||   ||
|   |       |  |      |   |      |  |       |   |
|| 16||       || 1||      || 2 ||      || 4||       || 8 ||
(   )       (  )      (   )      (  )       (   )
  1           1         1          1          1

Hence we get these 4 equations

pict

C-D= ⇒

15 = x2 + 3x3 + 7x4
(E)

B-E=⇒

17 = x3 + 5x4
(F)

A-F=⇒

7 = x4

Sub. into F we get x3 = 17 − 5 × 7 =  − 18  , sub into E we get x2 = 15 − 3 (− 18 ) − 7 (7) = 20  , sub into D we get x1 =  1 − 20 − (− 18) − 7 = − 8

Hence      (     )
       − 8
     ||     ||
     |     |
     ||  20 ||
x =  ||     ||
     | − 18 |
     |(     |)

        7

Hence A¯  , the form that the matrix A  takes w.r.t the basis Q  = {b,Ab, A2b,A3b } is

     (              )

     | 0  0  0  − 8 |
     ||              ||
     | 1  0  0   20 |
¯A =  ||              ||
     ||              ||
     | 0  1  0  − 18 |
     (              )
       0  0  1   7
(2)

Now that we have found the representation of A  under Q,  we are given a new basis Q ′ = {¯b,A¯b,A2 ¯b,A3¯b} and asked again find representation of A  under Q ′

Clearly the problem is not asking us to do this computation all over again for the new basis, becuase the algorithm would be the same and we can just repeat all the above steps again but using the new basis. So we need to try to find a short cut solution.

Start by finding the charactertic polynomial for A

pict

We know that a matrix companion form can be written directly if we know the matrix charaterstic equation.

If Δ (λ) = λ4 + α4λ3 + α3 λ2 + α2λ + α1   then the matrix A  companion form can be written as (see page 55 in text)

(              )

| 0  0  0  − α1|
||              ||
| 1  0  0  − α2|
||              ||
|              |
|| 0  1  0  − α3||
(              )
  0  0  1  − α4
(4)

Hence by comparing (4) and (3) we see that α1 =  8,α2 = − 20,α3 =  18,α4 = − 7  , hence the compaion form for A  is

(              )

|| 0  0  0   − 8||
|              |
|| 1  0  0   20 ||
|              |
|| 0  1  0  − 18||
|              |
(              )
  0  0  1   − 7
(5)

Compare matrix (5) and (2) we see they are the same. (I do not think I needed to show that (5) and (2) are the same here, this is just a confirmation).

Using Cayley-Hamilton, then

pict

Hence

A4 =  7A3 − 18A2 +  20A − 8
(6)

If we multiply the above by b  we get

pict

If we multiply (6) by any other vector, say ¯
b  we also get

 4      ( 3 )      (  2)      (   )    ( )
A ¯b = 7  A ¯b  − 18  A ¯b  + 20  A ¯b − 8  ¯b

But we found that the representation of A  under Q  =  {b,Ab,A2b, A3b } is

( (  )   (  )  (   )                                                                           )
||                                                                                              ||
||| | 0|   | 0|  |  0|                                                                           |||
|||| ||  ||   ||  ||  ||   ||                                                                           ||||
{ || 1||   || 0||  ||  0||                                                                           }
  |  |  ,|  | ,|   | ,(coeff. of polynomial expressing A4b in L.I. combination of A3b, A2b,Ab, b)
||| || 0||   || 1||  ||  0||                                                                           |||
||| |  |   |  |  |   |                                                                           |||
|||( (  )   (  )  (   )                                                                           |||)
    0      0      1

Hence the representation of A  under   ′   [¯  ¯   2¯   3¯]
Q  =   b,Ab,A  b,A  b is also

( (  )   (  )  (   )                                                                           )
||                                                                                              ||
||| | 0|   | 0|  |  0|                                                                           |||
||| ||  ||   ||  ||  ||   ||                                                                           |||
|{ | 1|   | 0|  |  0|                                                                           |}
  ||  ||  ,||  || ,||   || ,(coeff. of polynomial expressing A4¯b in L.I. combination of A3 ¯b,A2¯b,A ¯b,¯b)
||| ||  ||   ||  ||  ||   ||                                                                           |||
||| | 0|   | 1|  |  0|                                                                           |||
||| (  )   (  )  (   )                                                                           |||
(   0      0      1                                                                            )

But with the help of the cayley-hamilton theorm, we see that the coeff. of these polynomials (i.e. the values in the last column of  ¯
A  ) are the same if the basis were    3   2
{A b,A  b,Ab, b} or {  3   2       }
 A  ¯b,A ¯b,A ¯b,¯b .

Hence the representation of A  under       [             ]
Q ′ =  ¯b,A¯b,A2 ¯b,A3¯b is the same as the representation of A  under Q =  [b,Ab, A2b, A3b]

2 Problem 3.13

Find jordan form of the following matrices

     (          )        (             )        (          )        (              )
       1  4  10             0   1    0            1  0  − 1           0    4     3
     ||          ||        ||             ||        ||          ||        ||              ||
A  = ||          || ,A  =  ||             ||  ,A  = ||          ||  ,A  = ||              ||
 1   | 0  2   0 |    2   |  0   0    1 |    3   | 0  1   0 |    4   | 0   20    16 |
     (          )        (             )        (          )        (              )
       0  0   3            − 2  − 4 − 3           0  0   2            0  − 25  − 20

Answer

2.1 A1

For A1   ,

pict

Hence λ1 = 1, λ2 = 2,λ3 = 3 ⇒ 3 distinct roots. Hence Jordan form is the standard diagoonal form

                  |--------------|
(            )    | (        )   |
  λ1   0   0      |   1  0  0    |
||            ||    | ||        ||   |
|            |    | |        |   |
||  0   λ2  0 ||  = | || 0  2  0||   |
(            )    | (        )   |
   0   0   λ3     |   0  0  3    |
                  |              |
                  ----------------

2.2 A2

For A2,

pict

Try factor (λ + 1)  out, then we have λ3 + 3λ2 + 4λ + 2 =  (λ +  1)(f (λ ))  , hence         λ3+3λ2+4λ+2     2
f (λ) = ----(λ+1)--- = λ  + 2λ + 2  hence                          2
|A − λI | = 0 = (λ + 1)(λ  + 2λ + 2)  . But roots of the quadratic are    √ ------
−b±--b2−4ac   −2±√4−-4×2   −2±√-−4-  −-2±2j
     2a     =      2     =     2    =   2   = − 1 ± j

Hence λ1 =  − 1,λ2 = − 1 + j,λ3 = − 1 − j ⇒ 3 distinct roots, hence Jordan form is

                 |----------------------------|
(           )    | (                      )   |
                 |                            |
| λ1   0   0|    | |  − 1    0        0   |   |
||           ||    | ||                      ||   |
||  0  λ2   0||  = | ||  0   − 1 + j     0   ||   |
(           )    | (                      )   |
   0   0  λ      |    0      0     − 1 − j    |
           3     |                            |
                 -----------------------------

2.3 A3

For A
  3

pict

Hence λ1 = 1, λ2 = 1,λ3 = 2  . The eigenvalues are not all distinct.

For λ = λ1   , we have the matrix           (          )

          | 0  0  − 1|
          ||          ||
(A − I) = || 0  0   0 ||
          (          )
            0  0   1 which has rank of 1, hence nullity is 3 − 1 = 2  , hence we can find 2 independent vectors in the null space of the matrix (A − λ1I) , so we get 2 jordan blocks for λ =  1  of order 1. Hence the Jordan form is (        )
  1  0  0
||        ||
||        ||
| 0  1  0|
(        )
  0  0  2 or (        )
  2  0  0
||        ||
||        ||
| 0  1  0|
(        )
  0  0  1

One can also solve the above as follows:

The null space of A −  λ1I  is 2, hence we can find 2 L.I. vectors in this null space as follows. Let u1   be the first eigenvector associated with eigenvalue λ1   .

pict

Hence x3 = 0  , and so x1   and x2   can take any values. Hence      (  )

     | 1|
     ||  ||
u1 = | 0|
     |(  |)

       0

Now u2   is the second eigenvector associated with λ = 1  . It will have similar solution to the above, but since it is L.I. to u1   , we now pick x1 =  0  and x2 = 1  , hence      (  )
     | 0|
     |  |
u2 = ||  ||
     | 1|
     (  )
       0 or we could have found u2   as follows: write                     (          )  (   )       (   )    (  )
                      1  0  − 1     x           x        1
                    ||          ||  ||  1||       ||  1||    ||  ||
                    |          |  |   |       |   |    |  |
Au2  = λ1u2 + u1 ⇒  || 0  1   0 ||  || x2||  = λ1 || x2|| +  || 0||
                    (          )  (   )       (   )    (  )
                      0  0   2      x3          x3       0 hence (        )    (       )

| x1 − x3|    | x1 + 1|
||        ||    ||       ||
|   x2   |  = |   x2  |
|(        |)    |(       |)

    2x3           x3 hence x3 = 0  , then x1 = 0  and x2   any value, say 1. Hence      (   )

     |  0|
     ||   ||
u2 = ||  1||
     (   )
        0 as above.

Now to find u3   . This is the eigevector associated with λ =  2  .  Hence

pict

Hence x  = 0
  2  and x
 3   is any value, say 1 hence x  =  − 1
  1  hence      (    )
       − 1
     ||    ||
u  = ||    ||
 3   |  0 |
     (    )
        1

Note that this is L.I. to both u1   and u2

Hence                   (            )
                     1  0  − 1
                  ||            ||
                  |            |
Q  = (u1,u2,u3) = ||  0  1   0  ||
                  (            )
                     0  0   1 hence

pict

Which matches earlier solution.

2.4 A4

      (              )

      | 0   4     3  |
      ||              ||
A4 =  || 0   20    16 ||
      (              )
        0  − 25  − 20

pict

Hence λ =  0,0,0  hence we have multipicity of 3.

Matrix (A −  λI) = A  since λ =  0.   Rank of A is 2, hence null space is 1, hence we find one eigenvector

pict

But this gives (    )    (  )
  x1        0
||    ||    ||  ||
|    | =  |  |
|| x2 ||    || 0||
(    )    (  )
  x3        0 hence can not use.

Matrix                    (              ) (              )    (         )

                   | 0   4     3  | | 0    4     3 |    | 0  5  4 |
         2     2   ||              || ||              ||    ||         ||
(A −  λI)  = A  =  || 0   20    16 || || 0   20    16 ||  = || 0  0  0 ||
                   (              ) (              )    (         )
                     0  − 25  − 20    0  − 25  − 20       0  0  0

This has rank =1, hence nullity=2, so we still need one more eigevector. keep going.

Matrix              (              )  (        )    (        )

             | 0    4     3 |  | 0  5  4|    | 0  0  0|
             |              |  |        |    |        |
(A −  λI)3 = || 0   20    16 ||  || 0  0  0||  = || 0  0  0||
             |(              |)  |(        |)    |(        |)

               0  − 25  − 20     0  0  0       0  0  0 this has rank=0 and nullspace =3, hence we can find the last eigenvector from this null space.

pict

So pick vector u
 3   such that the above is true and also that (A  − λI)2 u ⁄=  0
           3

Try      (  )     (        )  (  )    (  )
       0        0  5  4     0       4
     ||  ||     ||        ||  ||  ||    ||  ||
     |  |     |        |  |  |    |  |
u3 = || 0||  ⇒  || 0  0  0||  || 0|| =  || 0||  ⁄= 0
     (  )     (        )  (  )    (  )
       1        0  0  0     1       0  hence we pick      (  )
       0
     ||  ||
     |  |
u3 = || 0||
     (  )
       1

So now we write                      (        ) (   )    (  )

                     | 0  5  4| | 0 |    | 4|
        2            ||        || ||   ||    ||  ||
(A − λI ) u3 = u2 ⇒  | 0  0  0| | 0 | =  | 0|  = u2
                     |(        |) |(   |)    |(  |)

                       0  0  0    1        0

And we finally write                      (              ) (  )    (     )
                     | 0   4     3  | | 0|    |   3 |
                     |              | |  |    |     |
(A − λI)1 u3 = u1 ⇒  || 0   20   16  || || 0||  = ||  16 ||
                     |              | |  |    |     |
                     (              ) (  )    (     )
                       0  − 25  − 20    1       − 20

Hence                   (             )

                  |    3   4  0 |
                  ||             ||
Q  = (u1,u2,u3) = ||   16   0  0 ||
                  (             )
                     − 20  0  1 hence

pict

Hence jordan form is (        )

| 0  0  1|
||        ||
| 1  0  0|
|(        |)

  0  0  0