Matrix, minimal polynomial

[Maybe_Memorie]

Homework Statement

A n × n-matrix A satisfies the equation A² = A.

(a) List all possible characteristic polynomials of A.

(b) Show that A is similar to a diagonal matrix

Homework Equations

The Attempt at a Solution

A² = A
so, A² - A = 0
A(A-I) = 0

Our minimal polynomial is x² - x = m(x)

Our eigenvalues are 0 and 1, and since in our minimal polynomial each one has a multiplicity of 1, A is similar to a diagonal matrix consisting of n jordan blocks with either 1's or 0's on the diagonal and 0 everywhere else since we have an nxn matrix and n jordan blocks.

So I've proved part (b) first.

For part (a) we don't know how many times each eigenvalue occurs in the characteristic polynomial, so p(t) = t^a(t-1)^b
where a,b=0,1,2,...,n.Is this correct?

 

[micromass]

All is correct, except

A(A-I) = 0

A = 0, or A = I

AB=0 does not imply A=0 or B=0. For example

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

is a matrix A that satisfied $A^2=A$, but which is not 0 or I.

 

[Maybe_Memorie]

Okay, thanks!

 

[micromass]

Hmm, couldn't the minimal polynomial also be a (polynomial) divider of x²-x?

It's not my day today. Thanks ILS...

 

[I like Serena]

It's not my day today. Thanks ILS...

No, I just deleted my post.
They're asking for all characteristic polynomials, not for all minimal characteristic polynomials.

 

[micromass]

No, I just deleted my post.
They're asking for all characteristic polynomials, not for all minimal characteristic polynomials.

No, you were correct. If a matrix satisfies $A^2=A$, then its minimal polynomial is not necessary $x^2-x$. I should have pointed that out.

That said, this doesn't change anything about the solutions of the OP. They remain correct.

 

[HallsofIvy]

If A^2- A= A(A- I)= 0, then, whatever the characteristic equation of A is, it must include factors of x and x- 1.