# Jordan Canonical form proof

1. May 10, 2008

### 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.

2. May 10, 2008

### 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.

3. May 10, 2008

### 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

4. May 10, 2008

### krcmd1

I think I am confusing myself. Please forgive.

5. May 10, 2008

### 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?

6. May 10, 2008

### tiny-tim

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
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

7. May 10, 2008

### 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.

8. May 10, 2008

### tiny-tim

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.
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.

9. May 10, 2008

### 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?

10. May 10, 2008

### tiny-tim

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.

11. May 10, 2008

### krcmd1

thank you all!

12. May 10, 2008

### 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)?

13. May 10, 2008

### 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.

14. May 10, 2008

### 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" ?

15. May 11, 2008

### 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. Lets 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 lets 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 alot of what is already have been said, just trying to say it different, hope it helps.

16. May 11, 2008

### tiny-tim

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.

17. May 11, 2008

### 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.

18. May 11, 2008

### tiny-tim

Never fear … come back here!

19. May 11, 2008

### 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.

Last edited: May 11, 2008
20. May 11, 2008

### 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.

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