maybe I read you all wrong, but let me try too formulate:
"The range of A − λ I, denoted by , is an invariant subspace of A"
more clearly, I think you read it wrong. Let A go from C^n to C^n
the forst thing it says is:
The range of A − λ I, is a subspace of C^n.
Lets proof this:
let w,v \in Ran(A-\lambda I) then there exist x,y \in C^n such that
(A-\lambda I)x = w
(A-\lambda I)y = v
so let a,b be complex numbers, then av+bw = a (A-\lambda I)y+ b (A-\lambda I)x = (A-\lambda I)(ay+bx)
because C^n is a vector space ay+bx is also in C^n, that is (A-\lambda I)(ay+bx) is in Ran(A-\lambda I), so so by equality above av+bw is in ran(A-\lambda I), so ran(A-\lambda I) is closed under scalar multiplication and vector addition, and is thus a subspace.
the next thing it says is that the subspace W=ran(A-\lambda I) is an invariant of A, maybe more clearly, the subspace W is invariant under A, which means that
A(W) \subset W
so when you take some element of W and use A on it then it is again in W. Let's se this:
Let v be in W, then again there is x in C^n such that (A-\lambda I)x = v, and then you get
Av = A(A-\lambda I)x = AAx-\lambda IAx = A(Ax)-\lambda I (Ax) = (A-\lambda I) (Ax)
claerly Ax is in C^n let's call Ax = w, then you have
Av = (A-\lambda I) w
that is the element v from ran(A-\lambda I) is again in the range of (A-\lambda I), and we have shown that the subspace ran(A-\lambda I), is invariant under A.
I know I said a lot of what is already have been said, just trying to say it different, hope it helps.
