Jordan Canonical form proof

[krcmd1]

I'm trying to teach myself math for physics (a middle aged physicist wannabee). Wikipedia's proof for the exisitence of a JC form for matrix A in Cn,n states:

"The range of A − λ I, denoted by , is an invariant subspace of A"

I'm having trouble seeing why any element of Ran(A − λ I) is in the range also of A.

Also, where can I find instructions on putting equations in this forum?


Thankyou!

Ken C.

 

[mathwonk]

range (A-tI) is A-invariant if applying A to any vector in range A-tI, yields another such vector.

but if we have a vector w = (A-tI)v, then Aw = AAv - tAv = (A-tI)(Av).

thus whenever w is in range A-tI, then Aw is also.

the point is that all polynomials in A commute, and both A and A-tI are such polynomials.

 

[krcmd1]

Thank you, but it's not the invariant part that I don't get. How should I know that any w = (A-tI)v is also = Ay

 

[krcmd1]

I think I am confusing myself. Please forgive.

 

[krcmd1]

Maybe I am misunderstanding the statement "is a subspace of A." Does "Ran (A-tI) is a subspace of A" imply that any y = (A-tI)x is a linear combination of the columns of A?

 

[tiny-tim]

The range of A − λ I, denoted by Ran(A − λ I), is an invariant subspace of A"

I'm having trouble seeing why any element of Ran(A − λ I) is in the range also of A.

Hi krcmd1!

Because any element of Ran(A − λ I) is of the form (A − λ I)V, for some vector V.

So A((A − λ I)V) = AAV - λAV = (A − λ I)AV, which is in the range of A.

For the definition of "invariant subspace", see http://en.wikipedia.org/wiki/Invariant_subspace

Also, where can I find instructions on putting equations in this forum?

There's an introduction somewhere … but I can't find it …

If you look harder than I have, you'll find it!

And bookmark http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000
and maybe
http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000

 

[krcmd1]

does this show that (A-λI)V must be in the range of A? I can see that you've demonstrated that if ran(A-λI) is a subspace it is invariant, but I still don't see that every (A-λI)x = Ay for all x.

And it seems to me that this isn't even true, so maybe I just don't understand what is meant by Ran(A-λI) is a subspace of A.

 

[tiny-tim]

So A((A − λ I)V) = AAV - λAV = (A − λ I)AV, which is in the range of A.

oops … I left out - λ I … so it should read:

So A((A − λ I)V) = AAV - λAV = (A − λ I)AV, which is in the range of A - λ I.

does this show that (A-λI)V must be in the range of A?.

Sorry … my mistake has misled you …

(A-λI)V isn't in the range of A …

A of (A-λI)V is in the range of A-λI.

In other words, A sends A-λI of anything into A-λI of something else.

 

[krcmd1]

Thanks. I really appreciate your patience.

Is there a type in the wiki article?

"A proof
We give a proof by induction. The 1 × 1 case is trivial. Let A be an n × n matrix. Take any eigenvalue λ of A. The range of A − λ I, denoted by Ran(A − λ I), is an invariant subspace of A."


Doesn't this imply that for every x, there is a y s.t. (A-λI )x = Ay?

 

[tiny-tim]

The range of A − λ I, denoted by Ran(A − λ I), is an invariant subspace of A."

Doesn't this imply that for every x, there is a y s.t. (A-λI )x = Ay?

No, it implies that for every x, there is a y s.t. A(A-λI)x = (A-λI)y.

(If x is V, then y is AV)

btw, mathwonk said the same thing, only he got it right first time!

Have another look at the wiki article on invariant subspaces.

 

[krcmd1]

thank you all!

 

[krcmd1]

It bothers me that I'm missing basic stuff. Why does Wiki say:


"... is an invariant subspace of A." instead of simply "...is an invariant subspace." (i.e. of Cn,n)?

 

[mathwonk]

i have completely and correctly proved the assertion made in your post.

but your assertion, that range(A-tI) is in the range of A, is different, and is also false.

(take A = 0.)

in spite of your claim otherwise, you seem not to know what the phrase "is an invariant subspace of A" means.

 

[krcmd1]

Please let me know when I am starting to impose.

I fully accept what you are saying, and I'm trying to correct my misconception of invariant subspace. I have been assuming that a subspace is a subset of vectors with certain closure properties. So is it correct that a subspace of A does not necessarily belong to the image of A? Is it closure under the multiplication by A that makes ran (A-lambdaI) a subspace "of A" ?

 

[mrandersdk]

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.

 

[tiny-tim]

Why does Wiki say:

"... is an invariant subspace of A." instead of simply "...is an invariant subspace." (i.e. of Cn,n)?

Hi krcmd1!

Because there's no such thing as "an invariant subspace, period".

It has to be "an invariant subspace of a function or operation" (in this case, the matrix A).

It is a subspace of C(n,n), and it is invariant under A.

 

[krcmd1]

Thank you, very much.
This I understood, as you explained it before. I was trying to prove an implication that evidently the statement doesn't even have, specifically that A defines a space, i.e. the vectors formed by linear combinations of its columns, and that similarly so does (A-lambdaI), and that the second space is a subspace of the first space that is invariant.

Now that you've helped me sort that I out I fear I am in for similar misconceptions on every page as I read on.

 

[tiny-tim]

Now that you've helped me sort that I out I fear I am in for similar misconceptions on every page as I read on.

Never fear … come back here!

 

[mathwonk]

have you finally understood that "invariant subspace of A" means "subspace which is invariant under action by A",

i.e. a subspace S such that for all vectors v in S, Av is also in S?

notice this definition is explicitly given in the first sentence of the link in post 6 to wikipedia, as no doubt it also is in your book. in understanding math it is crucial to read and digest the definitions of the terms.

 

[trambolin]

well, I think a term like "this subspace is A-invariant" is more helpful in the beginning. Because, it does not relate to any subspace. Just tells that A does not send these elements to the outside of the subspace where you initially took them. Hence the term "invariant"

Maybe it is your cup of tea.

 

[mathwonk]

good point. things we take for granted are often the stumbling block for beginners. like when my third grade teacher tried to teach me about "bringing down the remainder" in long division.

she thought that "down the page" which was lying flat on my desktop, meant inwards toward my stomach, while i thought "downwards" meant toward the floor, so i thought she wanted me to put the remainders on the floor.