Find all Complex 2x2 matrices A

[beetle2]

Homework Statement

Find all Complex 2x2 matrices A such that A² = A

Homework Equations

$\[ \left( \begin{array}{cc}<br /> a & b \\<br /> c & d \\ \end{array} \right)^2\] = A<br />$

The Attempt at a Solution

A²=A

I need to solve the equations

$a^2+bc=a$
$<br /> (a+d)b =b$
$<br /> (a+d)c = c\\$
$<br /> d^2 + bc = d\\$
What is the best way to do this?

Should I use some kind of rearrangement like

using equation 1 and 4 we get

$a^2-d^2=a-d$

any help greatly appreciated

 

[LCKurtz]

I would think about two cases, that where A^-1 exists and where it doesn't exist.

 

[beetle2]

Are you talking about when A is an orthogonal matrix so A^-1 = A^T?

 

[Office_Shredder]

The question is: Is A invertible? If so, then just take A^-1 of both sides and what do you get?

 

[beetle2]

A^-1 exist when $ad-cb \neq 0$

so
case 1 $ad-cb \neq 0$
case 2 $ad-cb = 0$

 

[Office_Shredder]

Stop looking at the series of equations you have

If
A²=A

And A^-1 exists

In one step you can find A

 

[beetle2]

If A is invertable and A²=A
AA=A
A^-1AA=A^-1A
IA=I

So A = I

 

[Office_Shredder]

Now if A is not invertible, what can you say about how A acts on im(A), its image?

 

[HallsofIvy]

Think about the fact that if $A^2= A$ then $A^2- A= A(A- I)= 0$

 

[beetle2]

Would that mean that the only way that
$A^2- A= A(A- I)= 0$
is that either A = the trivial solution.

$\[ \left( \begin{array}{cc}0 & 0 \\0 & 0 \\ \end{array} \right)\] = A$
or

$\[ \left( \begin{array}{cc}-1 & 0 \\0 & -1 \\ \end{array} \right)\] = A$

regards

 

[Dick]

Would that mean that the only way that
$A^2- A= A(A- I)= 0$
is that either A = the trivial solution.

$\[ \left( \begin{array}{cc}0 & 0 \\0 & 0 \\ \end{array} \right)\] = A$
or

$\[ \left( \begin{array}{cc}-1 & 0 \\0 & -1 \\ \end{array} \right)\] = A$

regards

That's completely wrong thinking. If A and B are matrices AB=0 does NOT mean A=0 or B=0. And [[-1,0],[0,-1]] doesn't work at all. Here's some other ones that do [[1,0],[0,0]], [[1,1],[0,0]], [[1/2,1/2],[1/2,1/2]]. There's LOTS of them. You already followed up on the excellent hint that if A is invertible, then A=I is the only solution. Now suppose A is not invertible. Then det(A)=0, so you can add the condition ad-bc=0 to your original list of equations. That makes it easier. Now go to those equations and start working on cases. You should be able to write down a two parameter family of solutions that satisfy A^2=A besides the zero matrix.

 

[Hurkyl]

For the record, responders have been suggesting two different methods of approach. Office_Shredder's is a geometric one, and offers more insight into why idempotent matrices are interesting. The other was more algebraic.


I confess I would have suggested yet a different approach, one that starts by looking at the equation (a+d)b=b.

 

[Dick]

For the record, responders have been suggesting two different methods of approach. Office_Shredder's is a geometric one, and offers more insight into why idempotent matrices are interesting. The other was more algebraic.


I confess I would have suggested yet a different approach, one that starts by looking at the equation (a+d)b=b.

Why is that different from the algebraic approach? I'm just saying that the equations the OP originally posted are a lot easier to solve if you restrict to the case of noninvertible matrices, in which case you can put ad=bc. LCKurtz already suggested this, and it's a very good suggestion.

 

[beetle2]

As idempotent matrices must eigenvalues of 1 and 0 would it be all the Matrices that can produce the diagonal matrix

$\[ \left( \begin{array}{cc}1 & 0 \\0 & \\ \end{array} \right)\] = A$

 

[Dick]

No, I already pointed out that [[1/2,1/2],[1/2,1/2]] works.

 

[Hurkyl]

As idempotent matrices must eigenvalues of 1 and 0...

Figures you'd go for yet another solution! There are (at least) three things you need to do to turn this into a complete solution and a proof.

(1) What about non-diagonalizable matrices?

(2) For the diagonalizable solutions, you haven't covered all possible diagonal forms.

(3) You still need to simplify your characterization of the solutions into a form that would be considered "found".

 

[beetle2]

So I have the first case where the matreix is invertable


If A is invertable and A^2=A
AA=A
A-1AA=A-1A
IA=I

So A = I

The other case is when A is not invertable so then the det(A) = 0

A^2 = A
AA = A
AA - A = A - A
AA - A = 0
A(I-A) = 0


SO what complex matrix will do this?

 

[Dick]

So I have the first case where the matreix is invertable


If A is invertable and A^2=A
AA=A
A-1AA=A-1A
IA=I

So A = I

The other case is when A is not invertable so then the det(A) = 0

A^2 = A
AA = A
AA - A = A - A
AA - A = 0
A(I-A) = 0


SO what complex matrix will do this?

I already told you in post 10 that that factorization isn't particularly useful. Go back to your original set of equation and add the condition det(A)=0=ad-bc. You'll find they are pretty easy to solve.

 

[beetle2]

For det(A)= 1 = invertable

$\[ \left( \begin{array}{cc}1 & 0 \\0 & 1 \\ \end{array} \right)\]$


for det(A) = 0 = non invertable

b or c can be any value

$\[ \left( \begin{array}{cc}1 & b \\0 & 0 \\ \end{array} \right)\]$

$\[ \left( \begin{array}{cc}1 & 0 \\c & 0 \\ \end{array} \right)\]$

$\[ \left( \begin{array}{cc}0 & b \\0 & 1 \\ \end{array} \right)\]$

$\[ \left( \begin{array}{cc}0 & 0 \\c & 0 \\ \end{array} \right)\]$

$\[ \left( \begin{array}{cc}0 & 0 \\0 & 0 \\ \end{array} \right)\]$

also you have
for all square matrices nXn

$\[ \left( \begin{array}{cc}\frac{1}{n} & \frac{1}{n} \\\frac{1}{n} & \frac{1}{n} \\ \end{array} \right)\]$


I'm not sure how to get that last one. I just noticed that DIck posted that one and by trying different options noticed that it worked for all nXn matrices

 

[Dick]

Those are all one parameter special cases. At least the ones that actually work are. The really general solution has two free parameters b and c, both in the same matrix. Can you find it? It includes the [[1/2,1/2],[1/2,1/2]] case. I keep telling you to go back to your original equations and add ad=bc.

 

[beetle2]

I start out with.

$a^2+bc=a$
$(a+d)b=b$
$(a+d)c=c$
$d^2+bc=d$

I set them all to equal zero.

$a^{2}-a+bc=0$$(1)$
$(a+d-1)b=0$$(2)$
$(a+d-1)c=c$$(3)$
$d^{2}-d+bc=0$$(4)$]
and
$ad-bc=0$$(5)$

From equation (2) either a+d=1 or b=0

So take $b\neq 0 \rightarrow a+d=1$ than (3) and (2) are both satisfied.
from (1) $c=\frac{a-a^2}{b}$ since $b\neq 0[$

for d we can use from a+d=1 we have d = 1-a

So using (4) = $(1-a)^2-(1-a)+((a+1-a)\times(\frac{a-a^2}{b}))$
= $(1-a)^2-(1-a)+((a-a^2)$
=$-a+a^2+a-a^2=0$


So for b $neq 0$ we have the general matrix


$\[ \left( \begin{array}{cc}a & b \\\frac{a-a^2}{b} & 1-a \\ \end{array} \right)\]$

I'm sure there is an easier way to do this

 

[Dick]

I start out with.

$a^2+bc=a$
$(a+d)b=b$
$(a+d)c=c$
$d^2+bc=d$

I set them all to equal zero.

$a^{2}-a+bc=0$$(1)$
$(a+d-1)b=0$$(2)$
$(a+d-1)c=c$$(3)$
$d^{2}-d+bc=0$$(4)$]
and
$ad-bc=0$$(5)$

From equation (2) either a+d=1 or b=0

So take $b\neq 0 \rightarrow a+d=1$ than (3) and (2) are both satisfied.
from (1) $c=\frac{a-a^2}{b}$ since $b\neq 0[$

for d we can use from a+d=1 we have d = 1-a

So using (4) = $(1-a)^2-(1-a)+((a+1-a)\times(\frac{a-a^2}{b}))$
= $(1-a)^2-(1-a)+((a-a^2)$
=$-a+a^2+a-a^2=0$


So for b $neq 0$ we have the general matrix


$\[ \left( \begin{array}{cc}a & b \\\frac{a-a^2}{b} & 1-a \\ \end{array} \right)\]$

I'm sure there is an easier way to do this

You've got it. Yes. Was that really so hard? a=1/2 and b=1/2 gives you the [[1/2,1/2],[1/2,1/2]] solution. And now you can find a lot more solutions where none of the matrix entries are zero. Like M=[[2,3],[-2/3,-1]]. M.M=M. That's a=2, b=3. That's exactly what I was looking for.