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:
quot;The range of A #8722; #955; I, denoted by , is an invariant subspace of Aquot;
I'm having trouble seeing why any element of Ran(A #8722; #955; I) is in the range also of A.
Also, where can I find instructions on putting equations in this forum?Thankyou!
Ken C.
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.
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
I think I am confusing myself. Please forgive.
Maybe I am misunderstanding the statement quot;is a subspace of A.quot; Does quot;Ran (A-tI) is a subspace of Aquot; imply that any y = (A-tI)x is a linear combination of the columns of A?
Originally Posted by krcmd1 The range of A #8722; #955; I, denoted by Ran(A #8722; #955; I), is an invariant subspace of Aquot;
I'm having trouble seeing why any element of Ran(A #8722; #955; I) is in the range also of A.
Hi krcmd1!
Because any element of Ran(A #8722; #955; I) is of the form (A #8722; #955; I)V, for some vector V.
So A((A #8722; #955; I)V) = AAV - #955;AV = (A #8722; #955; I)AV, which is in the range of A.
For the definition of quot;invariant subspacequot;, see 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 ~dubois/...00000000000000
and maybe
~dubois/...00000000000000
does this show that (A-#955;I)V must be in the range of A? I can see that you've demonstrated that if ran(A-#955;I) is a subspace it is invariant, but I still don't see that every (A-#955;I)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-#955;I) is a subspace of A.
Originally Posted by tiny-tim So A((A #8722; #955; I)V) = AAV - #955;AV = (A #8722; #955; I)AV, which is in the range of A.
oops … I left out - #955; I … so it should read:
So A((A #8722; #955; I)V) = AAV - #955;AV = (A #8722; #955; I)AV, which is in the range of A - #955; I. Originally Posted by krcmd1 does this show that (A-#955;I)V must be in the range of A?.
Sorry … my mistake has misled you …
(A-#955;I)V isn't in the range of A …
A of (A-#955;I)V is in the range of A-#955;I.
In other words, A sends A-#955;I of anything into A-#955;I of something else.
Thanks. I really appreciate your patience.
Is there a type in the wiki article?
quot;A proof
We give a proof by induction. The 1 × 1 case is trivial. Let A be an n × n matrix. Take any eigenvalue #955; of A. The range of A #8722; #955; I, denoted by Ran(A #8722; #955; I), is an invariant subspace of A.quot;Doesn't this imply that for every x, there is a y s.t. (A-#955;I )x = Ay?
Originally Posted by krcmd1 The range of A #8722; #955; I, denoted by Ran(A #8722; #955; I), is an invariant subspace of A.quot;
Doesn't this imply that for every x, there is a y s.t. (A-#955;I )x = Ay?
No, it implies that for every x, there is a y s.t. A(A-#955;I)x = (A-#955;I)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.
thank you all!
It bothers me that I'm missing basic stuff. Why does Wiki say:quot;... is an invariant subspace of A.quot; instead of simply quot;...is an invariant subspace.quot; (i.e. of Cn,n)?
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 quot;is an invariant subspace of Aquot; means.
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 quot;of Aquot; ?
maybe I read you all wrong, but let me try too formulate:
quot;The range of A #8722; #955; I, denoted by , is an invariant subspace of Aquot;
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 #8722; #955; I, is a subspace of C^n.
Lets proof this:
let then there exist such that
so let a,b be complex numbers, then
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
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
claerly Ax is in C^n lets call Ax = w, then you have
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.
Originally Posted by krcmd1 Why does Wiki say:
quot;... is an invariant subspace of A.quot; instead of simply quot;...is an invariant subspace.quot; (i.e. of Cn,n)?
Hi krcmd1!
Because there's no such thing as quot;an invariant subspace, periodquot;.
It has to be quot;an invariant subspace of a function or operationquot; (in this case, the matrix A).
It is a subspace of C(n,n), and it is invariant under A. |
Post at 2011-5-25 14:55
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