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

[gysush]

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).

 

[Mark44]

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.

 

[Mark44]

Try again:
$$A^2 = A \Leftrightarrow A^2 - A = 0$$

 

[gysush]

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}.

 

[Dick]

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.

 

[Dick]

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.

 

[gysush]

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?

 

[Dick]

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.