# Thread: All matrix such that A^2 = A?

1. ## All matrix such that A^2 = A?

The matrix A = [[a, i][i, b]], where i^2 = -1, a = (1/2)(1 + sqrt(5)) and
b = (1/2)(1 - sqrt(5)) has the property that A^2 = A. Describe all 2 X 2 matrices A with complex entries such that A^2 = A.

I don't know why they give us the matrix A, as in, I don't know how to use it to solve the problem. would a method involving something like this work?

[[a, b][c, d]] [[a, b][c, d]] = [[a, b][c, d]]

and try to solve for a, b, c, d?

[[a, b][c, 1-a]], where b and c are arbitrary and a is any solution of the quadratic equation a^2 - a + bc = 0.

2. You're told $\mathbf{A}^2 = \mathbf{A}$, so

$\left[\begin{matrix}a&i\\i&b\end{matrix}\right]\left[\begin{matrix}a&i\\i&b\end{matrix}\right]=\left[\begin{matrix}a&i\\i&b\end{matrix}\right]$

$\left[\begin{matrix}a^2+i^2&ai+bi\\ai+bi&i^2+b^2\end{mat rix}\right]=\left[\begin{matrix}a&i\\i&b\end{matrix}\right]$

$\left[\begin{matrix}a^2-1&(a+b)i\\(a+b)i&b^2-1\end{matrix}\right] = \left[\begin{matrix}a&i\\i&b\end{matrix}\right]$

So equating the like elements gives

$a^2 - 1 = a$
$(a+b)i=i$
$(a+b)i=i$
$b^2-1=b$.

Now try to equate $a$ and $b$.

3. OK, so what was the point in giving i^2 = -1, a = (1/2)(1 + sqrt(5)) and b = (1/2)(1 - sqrt(5))?

4. Originally Posted by hashshashin715
OK, so what was the point in giving i^2 = -1, a = (1/2)(1 + sqrt(5)) and b = (1/2)(1 - sqrt(5))?

Are you kidding? Solve the equations $a^2-1=a\Longleftrightarrow a^2-a-1=0\,,\,\,b^2-1=b\Longleftrightarrow b^2-b-1=0$ and you'll see why!

Tonio

5. OK, so when I solve those equation, i'll get a = (1/2)(1 + sqrt(5)) and b = (1/2)(1 - sqrt(5))

That's only one matrix that satisfies the equation A ^ 2 = A, which they give you in the first place. Am I missing something here?

6. Well, it obviously isn't the only matrix that satisfies $A^2 = A$ since the identity matrix also does.

I don't think there is a reason they give you the matrix $A$; it's just used as motivation so that you get a chance to see that there even exist matrices other than the identity. Another example is the matrix $\begin{pmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}$.

And yes, your original proposed method is exactly what you should do, although it is more efficient to note $A^2 = A \Rightarrow A(A - I) = 0$ and solve for the equation

$
\begin{pmatrix}a & b \\ c & d \end{pmatrix} \begin{pmatrix}a - 1 & b \\ c & d - 1\end{pmatrix} = 0
$

7. Originally Posted by hashshashin715
Describe all 2 X 2 matrices A with complex entries such that A^2 = A.

...

$\begin{bmatrix}a&b\\c&1-a\end{bmatrix}$, where b and c are arbitrary and a is any solution of the quadratic equation a^2 - a + bc = 0.
That clearly is not the complete answer because, as Guy has pointed out, the identity matrix satisfies the condition that it is equal to its square, although it is not of the form indicated. The zero matrix is another example not included in that solution.

8. Originally Posted by hashshashin715
OK, so when I solve those equation, i'll get a = (1/2)(1 + sqrt(5)) and b = (1/2)(1 - sqrt(5))

That's only one matrix that satisfies the equation A ^ 2 = A, which they give you in the first place. Am I missing something here?

No, it's not "only one", since Prove it's answer showed you that ALL the possible matrices are of this form...

Tonio

9. Originally Posted by tonio
No, it's not "only one", since Prove it's answer showed you that ALL the possible matrices are of this form...

Tonio
The question is asking for the form all 2x2 idempotent matricies, not all idempotent matricies of the form $\begin{pmatrix} a & i \\ i & b\end{pmatrix}$.

10. Originally Posted by Guy
The question is asking for the form all 2x2 idempotent matricies, not all idempotent matricies of the form $\begin{pmatrix} a & i \\ i & b\end{pmatrix}$.

So I see now....well, he/she needs now to solve similar equations as the ones from Prove it's answer.

Tonio