Question

Suppose a matrix A, such that A^2=A and A does not equal 0. Does this imply that A is invertible?

Answer 1

no not necessary

Answer 2

What matrix would disprove this? Wouldn't A have to be the identity?

Answer 3

basic rule of invertiblity of matrice says that it should be singular

Answer 4

Sorry, I'm confused. What would be a matrix that doesn't have an inverse where A^2=A? I can't find one :$

Answer 5

Sorry... didn't read the question thoroughly

Answer 6

I know, I thought 0 too, but that's not allowed :$

Answer 7

let's say A has an inverse
A A = A
multiply by Ainv
Ainv*A*A= Ainv*A
A= I
It seems A must be I

Answer 8

but we can have matrices A*A= A
which are not I, so it seems you can not assume this means A is invertible.

Answer 9

On second thought... what about a square matrix with just a 1 in the upper left corner? That's not invertible

Answer 10

it says in this page

http://en.wikipedia.org/wiki/Idempotent_matrix

that "With the exception of the identity matrix, an idempotent matrix is singular"

Answer 11

\[\huge A=\left[\begin{matrix}1 & 0 \\ 0 & 0\end{matrix}\right]\]
And I think you'll find that
\[\huge AA=A\]

Answer 12

so if A is not the identity matrix, then A is not invertible

Answer 13

The A=[1,0
0,1] does work. So it isn't necessarily the identity.

Answer 14

you can form A using v * v^T / (v^T v)
where v is a vector

Answer 15

10
00 sorry

Answer 16

10
00

isn't invertible though because it has a determinant of 0

Answer 17

So... that answers the question? :)

Answer 18

A A = A does not imply A is invertible

Answer 19

It implies if A is invertible, A is I, otherwise A is not invertible

Answer 20

So, what matrix would disprove this? What matrix has A^2=A but A isn't invertible?

Answer 21

v * v^T / (v^T v)

Answer 22

10
00

is a good counter-example

Answer 23

??
Doesn't A already do that?
\[\huge A=\left[\begin{matrix}1 & 0 \\ 0 & 0\end{matrix}\right]\]