1. ## rank

let A be an nxn matrix st A^2 = A and rank(A) = n.

then the dimension of A should be nxn= n^2 right?

and i was wondering, if the rank(A) =n, then the det A= 1 ( since there will be n leading ones by row operations). then A is invertible.

but should A be invertible, then shouldnt the dimA= n^2 be equal to the dim of the image which in this case is n? there seem to be a contradict

2. I'm sorry, I don't understand what you mean by "dim(A)". A matrix is NOT a vector space and so does not have a dimension. It also does NOT follow that if A is invertible, then det(A)= 1. If A is invertible then dim(A) is not 0 but can be any non-zero number.

If A is invertible, then its rank is equal to the dimension of its domain space and so is the rank of $A^2$, $A^3$, etc.

3. hmm the question that i was given was " if A is an nxn matrix such that A^2 = A and rank(A) =n, prove that A is the identity matrix".

and the answer to that:

" the rank can be defined as the size of the largest non zero determinant of a matrix. if rank(A)=n, then det(A) is not zero, thus A is invertible. multiplying the given relation with A inverse on the left, gets A=I"

which i dont really get..

4. What part don't you get? If rank(A)= n, then A maps $R^n$ onto all of $R^n$. That is, A is both one-to-one and "onto" and so is invertible. $A^{-1}(A^2)= A^{-1}A$ so $A= I$. By trying to talk about "dim(A)", which makes no sense, you made the problem much harder!

5. sorry.

isint the rank= dim of the image?
i dont get how the rank(A)= n means that A maps R^n onto R^n.