1. ## Idempotent Linear Transformation

A:V->V is a linear transformation with A^2=A

Prove that im(A) intersect ker(A) = 0
I know from a previous exercise that for B, an arbitrary linear transformation, there exists an integer m such that im(B^m) intersect ker(B^m)= 0. I think I can use this and say that A^m=A.

Prove that V=im(A) direct sum ker(A)
I know that dim(V)=dim(im(A))+dim(ker(A))
But I don't know how to use idempotent-ness of A to get set equality. Can I use that the intersection is 0?

Prove that there is a basis of V such that A with respect to this basis is a diagonal matrix with only 0s and 1s.
This one I have no idea how to start.

2. Originally Posted by robeuler
A:V->V is a linear transformation with A^2=A

Prove that im(A) intersect ker(A) = 0
I know from a previous exercise that for B, an arbitrary linear transformation, there exists an integer m such that im(B^m) intersect ker(B^m)= 0. I think I can use this and say that A^m=A.
In previous section we showed that if $\displaystyle \text{im}(A^m) = \text{im}(A^{m+1})$ and $\displaystyle \ker (A^m) = \ker (A^{m+1})$ then $\displaystyle \text{im}(A^m) \cap \ker (A^m) = \{ 0 \}$. This is a special case with $\displaystyle m=1$ because $\displaystyle A=A^2$!

3. Originally Posted by robeuler
A:V->V is a linear transformation with A^2=A

Prove that im(A) intersect ker(A) = 0
I know from a previous exercise that for B, an arbitrary linear transformation, there exists an integer m such that im(B^m) intersect ker(B^m)= 0. I think I can use this and say that A^m=A.

Prove that V=im(A) direct sum ker(A)
I know that dim(V)=dim(im(A))+dim(ker(A))
But I don't know how to use idempotent-ness of A to get set equality. Can I use that the intersection is 0?

Prove that there is a basis of V such that A with respect to this basis is a diagonal matrix with only 0s and 1s.
This one I have no idea how to start.
Yes the first one I was pretty confident about.

4. Yes, similar to a group being written as a direct product, I think you only need to quote that dimV=dimim+dimker (G=HK) and im intersect ker = {0} (H intersect K = {0}). Anyone have ideas on the matrix part? All I know is that for that matrix MM=M, so M=I or M is not invertible (singular, detM=0), and then I-M also satisfies (I-M)(I-M)=I-M. Don't see how that gets me anywhere. Help?

5. Since A^2=A, you have that A restricted to im(A) is the identity. Since you showed that V=im(A)+ker(A), just pick a basis for im(A) and a basis for ker(A), and their union is a basis for V. What does the matrix of A look like under this basis?

6. Just a remark, since no one else has seemed to say this. The idempotent endomorphisms are precisely the projections.

$\displaystyle (i)\quad y\in\textrm{Im}A\Rightarrow \exists u \in V:y=Au \Rightarrow Ay=A(Au)=A^2u=Au=y$

$\displaystyle (ii)\quad Ay=y\Rightarrow y\in \textrm{Im}A$

So,

$\displaystyle \ker (A-I)=\textrm{Im}A$

Taking into account that

$\displaystyle V=\ker A\oplus \ker (A-I)$

then, $\displaystyle A$ is diagonalizable with diagonal matrix:

$\displaystyle D=\textrm{diag}(0,\ldots,0,1,\ldots,1)$

( eventually $\displaystyle D=0$ or $\displaystyle D=I$ ).

Fernando Revilla

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# examples of idempotent of linear transformation

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