Find all Complex 2x2 matrices A

[beetle2]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

A²=A

I need to solve the equations

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

Should I use some kind of rearrangement like

using equation 1 and 4 we get

<br /> 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.

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

\[ \left( \begin{array}{cc}-1 &amp; 0 \\0 &amp; -1 \\ \end{array} \right)\] = 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.

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

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

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

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


for det(A) = 0 = non invertable

b or c can be any value

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

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

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

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

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

also you have
for all square matrices nXn

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


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


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

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


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

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.