Proving Equality of Image and Eigenspace for Eigenvalue 1

[chrisb93]

Homework Statement

It's given or I've already shown in previous parts of the question:
$A \in M_{nxn}(F)\\<br /> A^{2}=I_{n}\\<br /> F = \mathbb{Q}, \mathbb{R} or \mathbb{C}\\<br /> ker(L_{I_{n}+A})=E_{-1}(A)$
Eigenvalues of A must be $\pm1$

Show $im(L_{I_{n}+A})=E_{1}(A)$ where E is the eigenspace for the eigenvalue 1

(I also need to show that $im(L_{I_{n}-A})=E_{-1}(A)$ but I think that should be simple once I've done one of them)

Homework Equations

The Attempt at a Solution

I know that I need to show both sets are contained within the other set so,

Show $im(L_{I_{n}+A}) \subseteq E_{1}(A)$
$y=L_{I_{n}+A}(x)$ Let y be a general element of the image
$=x+Ax$ By definition of the transformation
$\Rightarrow A y = A x + A^{2} x$ Multiply through by A
$= A x + x$ As A² is the identity element
$\Rightarrow A y = y \in E_{1}(A)$ As $E_{1}(A) := \{ x | A x = x \}$

I've no idea how to show $E_{1}(A) \subseteq im(L_{I_{n}+A})$

 

[HallsofIvy]

You don't seem to have defined "$L_{I_n+ A}$" in all that.

 

[chrisb93]

You don't seem to have defined "$L_{I_n+ A}$" in all that.

It's never specifically defined in the question but I believe the subscript is the matrix representing the map so $L(x) = (I_{n} + A)x$

 

[micromass]

Take an eigenvector with eigenvalue 1. So Ay=y.

You need to find x such that (A+I)x=y.

What if you take x=y??

 

[chrisb93]

Take an eigenvector with eigenvalue 1. So Ay=y.

You need to find x such that (A+I)x=y.

What if you take x=y??

I don't see how that works, if x=y then Ax=x therefore (A+I)x=Ax+x=2x=y which contradicts itself.