Linear Algebra - polynomial functions of matrices

  • Thread starter Snoogx
  • Start date
  • #1
22
0

Homework Statement


Given p(x) = x4+2x2+1 and
A =
[[1 1 -2 0]​
[0 1 0 2]​
[1 1 -1 1]​
[0 0 -2 -1]]​
p(A) = 0
Find a polynomial q(x) so that q(A) = A-1
a) What is q(x)?
b) Compute q(A) = A-1

Homework Equations


I found the Cayley-Hamilton theorem, which states: p(x) = det(A-xIn).


The Attempt at a Solution


I found the inverse of A, which is:
[[-1 -1 2 0]​
[4 1 -4 -2]​
[1 0 -1 -1]​
[-2 0 2 1]]​
From here I thought I could set q(x) = A-1.
[[-1-x -1 2 0]​
[4 1-x -4 -2]​
[1 0 -1-x -1]​
[-2 0 2 1-x]]​
Then I should just solve out to find the det(). I used the 2nd column as a reference point.
Doing this I end up with the equation, x4-6x2+1.
so q(x) = x4-6x2+1.

This seems to be the wrong answer. I'm not at all sure if I did this right, or if I made an error in my calculations. Would really appreciate if someone could look over and point out my error.

Thanks
 

Answers and Replies

  • #2
35,294
7,152

Homework Statement


Given p(x) = x4+2x2+1 and
A =
[[1 1 -2 0]​
[0 1 0 2]​
[1 1 -1 1]​
[0 0 -2 -1]]​
p(A) = 0
Find a polynomial q(x) so that q(A) = A-1
a) What is q(x)?
b) Compute q(A) = A-1

Homework Equations


I found the Cayley-Hamilton theorem, which states: p(x) = det(A-xIn).


The Attempt at a Solution


I found the inverse of A, which is:
[[-1 -1 2 0]​
[4 1 -4 -2]​
[1 0 -1 -1]​
[-2 0 2 1]]​
From here I thought I could set q(x) = A-1.
[[-1-x -1 2 0]​
[4 1-x -4 -2]​
[1 0 -1-x -1]​
[-2 0 2 1-x]]​
Then I should just solve out to find the det(). I used the 2nd column as a reference point.
Doing this I end up with the equation, x4-6x2+1.
so q(x) = x4-6x2+1.

This seems to be the wrong answer. I'm not at all sure if I did this right, or if I made an error in my calculations. Would really appreciate if someone could look over and point out my error.

Thanks

You went at it the hard way.

Did you check that p(A) = 0? I did, and it turns out that this is so.

p(A) = 0 ==> A4 + 2A2 + I = 0
==> -A4 - 2A2 = I
If you can factor an A out of the left side, you'll be almost where you need to go.
 
  • #3
2
0
I have a similar problem, and was wondering what part b was asking for.
 
  • #5
2
0
That's what I thought, but it wants a polynomial. Thank you.
 
  • #6
35,294
7,152
And A-1 will be that polynomial in A; i.e., q(A).
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,847
966
As Mark44 said before, [itex]-A^4 - 2A^2 = A(-A^3- 2A)= I[/itex]
Now what is [itex]A^{-1}[/itex]?
 

Related Threads on Linear Algebra - polynomial functions of matrices

Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
6K
Replies
2
Views
9K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
625
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
4
Views
2K
Top