matrix代写 | math – matrix

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matrix代写 | math – 该题目是一个常规的matrix的练习题目代写, 是比较有代表性的matrix等代写方向

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ECE 5253 – Applied Matrix Theory Problem Set 7 Professor Zhong-Ping Jiang

Problem 1.For the matrix

A=

1 0 1

0 2 0

0 0 1

identify the spacepg()and the principal verctors of grade 2. Solution. We have a given matrixAR^3 ^3. So,

det(IA) =det

1 0 1
0 2 0
0 0  1

= 0

which is 1 = 1 2 = 2 We know, (iIA)gp= 0 Now, for= 1,

 (IA)^2 = 0

0 0 1

0 1 0

0 0 0

0 0 1

0 1 0

0 0 0

p= 0


0 0 0

0 1 0

0 0 0

p 1
p 2
p 3

=

0

0

0

So, P 2 ( 1 = 1) ={p^1 |p^1 =col(, 0 ,)} Now, for 2 ,

 (2IA)^2 = 0

1 0 1

0 0 0

0 0 1

p 1
p 2
p 3

=

0

0

0

So, P 1 ( 2 = 2) ={p^2 |p^2 =col(0,,0)} Hence,

p^1 =


0

 p^2 =

0


0

Problem 2. express the following vectors as unique representations of principle vectors found in Problem 1:

x=

2

9

84

, x=

0

9. 3

0

Solution. i) Given:

x=

2

9

84

ExpressXas unique representations of principle vectors found in problem 1, from problem 1,

p^1


0

, p^2 =

0


0

Then,

x= 2

1

0

0

9

0

1

0

+ 84

0

0

1

ii) Given:

x=

0

9. 3

0

Use same method,

x= 9. 3

0

1

0

, where p^2 =

0


0

 and = 1

Problem 3.Can you transform the following matrix into a Jordan form:

A=

  
0  
0 0 

,  6 = 0
Solution.
A=

  
0  
0 0 

IA=

0  
0 0 
0 0 0

So, null(P 1 ) =< 1 , 0 , 0 >

(IA)^2 =

0  
0 0 
0 0 0

0  
0 0 
0 0 0

=

0 0 ^2
0 0 0
0 0 0

So, null(P 2 ) =< 1 , 0 , 0 >;< 0 , 1 , 0 >

(IA)^3 =

0 0 ^2
0 0 0
0 0 0

0  
0 0 
0 0 0

=

0 0 0

0 0 0

0 0 0

So, null(P 2 ) =< 1 , 0 , 0 >;< 0 , 1 , 0 >;< 0 , 0 , 1 > Hence,

v^1 =

0

0

1

v^2 = (AI)v^1 =

0  
0 0 
0 0 0

0

0

1

=



0

v^3 = (AI)v^2 =

0  
0 0 
0 0 0



0

=

^2
0
0

So,

P=

0  ^2
0  0
1 0 0

P^1 =

0 0 1

(^011) 0 ^2 ^120

Therefore,

P^1 AP=

0 0 1

(^01) ^10 ^2 ^120

  
0  
0 0 

0  ^2
0  0
1 0 0

=

 0 0
1  0
0 1