Find Jordan Form of A/Complex F which satisfies A^2=A

  • Thread starter gysush
  • Start date
  • #1
gysush
26
0

Homework Statement



I want to find the Jordan form of a matrix over the complex field such that A^2=A

The Attempt at a Solution



So I've tried writing the entries of A as a 2x2 matrix such as {{a+bi,c+di},{e+fi,g+hi}} and multiplying it by A and then setting it equal to A. Needless to say, it seems like to me this NOT the way to approach it. Also, I've fondled around with Mathematica trying to guess the solution ,but I can not find it. Can I just use the normal identity matrix? After all, 1 is still a complex number (with a=1 and b=0).
 

Answers and Replies

  • #2
36,334
8,293

Homework Statement



i want to find the jordan form of a matrix over the complex field such that a^2=a

The Attempt at a Solution



so i've tried writing the entries of a as a 2x2 matrix such as {{a+bi,c+di},{e+fi,g+hi}} and multiplying it by a and then setting it equal to a. Needless to say, it seems like to me this not the way to approach it. Also, i've fondled around with mathematica trying to guess the solution ,but i can not find it. Can i just use the normal identity matrix? After all, 1 is still a complex number (with a=1 and b=0).
a2 = a <==> a2 - a = 0

Edit: The equation above is supposed to be in terms of A, but for some reason it won't take cap letters.
 
  • #3
36,334
8,293
Try again:
[tex]A^2 = A \Leftrightarrow A^2 - A = 0[/tex]
 
  • #4
gysush
26
0
Ah, so the Cayley-Hamilton theorem tells us this is equivalent to the characteristic polynomial t^2 + t = 0...then the information about it being over F is just to let us know that it splits into t(t-1)...thus the roots are 0 and 1 with multiplicity 1. Thus, a Jordan matrix (up to permutations) is {{1,0},{0,0}} or {{0,0},{0,1}.
 
  • #5
Dick
Science Advisor
Homework Helper
26,263
620
Think about what the Jordan form looks like. It's block diagonal for one thing. So each block B also has to satisfy B^2=B.
 
  • #6
Dick
Science Advisor
Homework Helper
26,263
620
Ah, so the Cayley-Hamilton theorem tells us this is equivalent to the characteristic polynomial t^2 + t = 0...then the information about it being over F is just to let us know that it splits into t(t-1)...thus the roots are 0 and 1 with multiplicity 1. Thus, a Jordan matrix (up to permutations) is {{1,0},{0,0}} or {{0,0},{0,1}.

That won't do. If A=0, then A^2=A=0, but it's characteristic polynomial isn't t^2-t=0.
 
  • #7
gysush
26
0
If I write the general form of matrix A satisfying A^2 - A = 0...then, for the entires of A being {{a,b},{c,d}}

A^2 - (a+d)A + (ad-bc) = 0

Then for our problem => ad=bc
a+d=1

Solving for d and a and leaving b and c free then,

A(+/-) = {{1-1/2(1 +/- sqrt(1-4bc)),b},{c,{1/2(1 +/- sqrt(1-4bc)}}

Am I thinking about this correctly now? I'm not exactly sure how to put this in Jordan form.

What are the eigenvalues? If I think about A^2=A...then I get if x is egn vector with egn value c...Ax=cx...(A^2)x=(c^2)x...=> Ax=(c^2)x => Ax=(c^n)x ...How does this tell me anything about the egn values?
 
  • #8
Dick
Science Advisor
Homework Helper
26,263
620
If I write the general form of matrix A satisfying A^2 - A = 0...then, for the entires of A being {{a,b},{c,d}}

A^2 - (a+d)A + (ad-bc) = 0

Then for our problem => ad=bc
a+d=1

Solving for d and a and leaving b and c free then,

A(+/-) = {{1-1/2(1 +/- sqrt(1-4bc)),b},{c,{1/2(1 +/- sqrt(1-4bc)}}

Am I thinking about this correctly now? I'm not exactly sure how to put this in Jordan form.

What are the eigenvalues? If I think about A^2=A...then I get if x is egn vector with egn value c...Ax=cx...(A^2)x=(c^2)x...=> Ax=(c^2)x => Ax=(c^n)x ...How does this tell me anything about the egn values?

You don't have to put explicitly put a general matrix into Jordan form. You just want to know what the Jordan form of A looks like. If J is the Jordan form of A, then J^2=J. They only want to know what J looks like (in general terms). What are the possible eigenvalues of J? Can J have any off diagonal elements? Stuff like that.
 

Suggested for: Find Jordan Form of A/Complex F which satisfies A^2=A

Replies
6
Views
320
Replies
2
Views
329
Replies
14
Views
468
Replies
4
Views
210
  • Last Post
Replies
3
Views
376
Replies
3
Views
283
Replies
2
Views
377
Replies
4
Views
540
Replies
12
Views
707
Top