Find Dimension of the subspace

# Tag: subspace

## linear algebra – Intersection of a vector subspace with a cone

Given a set of vectors $S={v_1, v_2,…,v_d} subset mathbb{R}^{N}, , N>d$, is there any algorithm to decide if there exist a vector with all coordinates strictly positive in the generating subspace $langle S rangle$?

I am aware of results like Farka’s Lemma (or variants as Gordon’s Theorem, etc…).

Or some papers like: *Ben-Israel, Adi. “Notes on linear inequalities, I: The intersection of the nonnegative orthant with complementary orthogonal subspaces.” Journal of Mathematical Analysis and Applications 9.2 (1964): 303-314.*

But I am looking for an algorithm to decide yes or no.

## functional analysis – On a subspace that is isomorphic to a dense subspace

Let $X$ be an infinite dimensional Banach space and let $M$ be a dense subspace of $X$, i.e., $overline{M}=X$. Let $N$ be another subspace of $X$ such that $N$ is topologically isomorphic to $M$.

Then, is it true that $N$ is necessarily dense in $X$, i.e., $overline{N}=X$ ? I think (intuitively) this should be true. But, how to proceed to prove this fact? Could anybody provide a hint?

## linear algebra – Using Gram-Schmidt find an orthonormal base S0 of the subspace W = L (S) of the vector space V from the given base S of W in the case …

Using Gram-Schmidt find an orthonormal base S0 of the subspace W = L (S) of the vector space V from the given base S of W in the case V = R ^ 2, S = {f1 = ( 3,4)}

So “f1” = (3 ^ 2 4 ^ 4) ^ 1/2 = sqrt (265), can I choose another vector? Since f2 = (0,1), there would be ‖f2‖ = 1.

And the base would be f1 ‘= f1 / ‖f1‖ and f2’ = f2 / ‖f2‖

## Functional analysis of fa: subspace complemented by the direct sum of two Banach spaces

When I was reading a newspaper, I saw something like:

Yes $ F $ and $ E $ are Banach spaces with symmetrical bases (precisely, they are spaces with symmetric sequence), and $ F $ is isomorphic to a supplemented subspace of $ l_2 oplus E $, so $ F = l_2 $ or $ F $ is isomorphic to a supplemented subspace of $ E $.

The author claimed that the result is followed by standard elemental arguments. We omit the details. I don't know what the argument is. Any clue?

## linear algebra: how many dimensional subspaces $ k + 1 $ of $ Bbb {F} _q ^ n $ contain a dimensional subspace $ k $ $ S $

Given a $ k $three-dimensional subspace $ S $ from $ Bbb {F} _q ^ n $ ($ n $three-dimensional vector space over finite field of $ q $ elements), I want to know how many $ k + 1 $ dimensional subspaces of $ Bbb {F} _q ^ n $ Contains $ S $. My initial approach was that, given a basis of $ S $, any vector in $ Bbb {F} _q ^ n $ not contained in $ S $ will be linearly independent of that base, so you can add that vector to the base to produce a $ k + 1 $ dimensional subspace contained in $ S $. $ Bbb {F} _q ^ n $ have $ q ^ n $ elements and $ S $ have $ q ^ k $ elements so that means there is $ q ^ n-q ^ k $ vectors you can add to give you a $ k + 1 $ dimensional space For each of these vectors, all scalar multiples excluding zero will give you the same $ k + 1 $ dimensional subspace when added to the base, so we must divide by $ q-1 $. Therefore there $ (q ^ n-q ^ k) / (q-1) $ of these $ k + 1 $ dimensional subspaces in $ Bbb {F} _q ^ n $ containing $ S $. My problem is that I saw somewhere that the answer is indeed $ (q ^ {(n-k)} – 1) / (q-1) $ Which is clearly different from what I found, so I was wondering where I went wrong.

Any help is appreciated, regards.

## at.algebraic topology: inclusion of subspace with superior direct images that do not disappear

I'm looking for a concrete topological insight for the derived drive.

Leave $ f: X a Y $ be a continuous map The derived push $ mathbf Rf_ ast $ take a sheaf $ F $ to the sheafification of presheaf cohomology $ V mapsto mathrm H ^ bullet (f ^ {- 1} V, F) $. When $ f $ is identity, sheafification is zero for $ n geq 1 $.

Sheafification of a presheaf $ P $ can be built by taking mapping $ PU $ to the equivalence class of families of sections of $ P $ defined in an open deck of $ U $, where we identify families that match on sufficiently small open decks. A section $ s in PU $ it is assigned to the equivalence class it represents.

Hence the fact that the previous sheafification is zero when $ f = 1 $ expresses the fact that the sheaf's cohomology is global in nature (each cocycle supports one deck where it is zero).

Using the previous sheafification construct, a section on $ mathbf Rf_ ast F (V) $ is an equivalence class of a family of cociclos $ ( Gamma_i in mathrm H ^ bullet (f ^ {- 1} V_i, F)) $ where $ (V_i) twoheadrightarrow V $ it is an open cover, and we identify families if they match on a sufficiently small preimage of an open cover.

The sheafification map is not zero for $ n geq 1 $ if there is a cocycle of $ F $ in some preimage $ f -1 V $ which is not removed by any preimage from an open cover of $ V $.

When $ f: X subset Y $ is a subspace inclusion the above means that there is a cocycle of $ F $ in $ f ^ {- 1} V = X cap V $ such that it is not restricted to zero in any open neighborhood in $ f ^ {- 1} V_0 = X cap V_0 $ in $ X $ from some trouble spot $ x_0 in X $.

This cannot happen for locally closed inlays because their feed function is exact.

**Question 1.** What is an instructive example of a subspace inclusion whose top direct images are not zero?

**Question 2.** For "what kind of maps" $ f $ Are non-zero non-zero direct images expected? (Examples welcome)

Finally, I would appreciate references with explicit topological examples of the derived advance.

## reference request – Subgroup of $ mathrm {GL} _n $ symmetric stabilizing linear subspace stabilizing matrices

I am currently reading a document where it is stated that the subgroup of $ mathrm {GL} _n $ ($ n geq 4 $) which retains a generic subspace of $ bigwedge ^ 2 mathbb {C} ^ n $ larger than $ 3 $ must be finite There are no references in the document, where it is stated that this fact is easily verified. Unfortunately (for me), I can't prove it.

Here "preserve" means that $ g.w. {} ^ {t} g in W $, For any $ w in W $. Is this fact really known? I am looking for a reference of this fact, or a quick test.

## linear algebra – Basis for T-cyclic subspace

I have been going back to this test for a few days and cannot convince myself of the fact that $ T ^ (v) $ is in $ beta = {v, T (v), T ^ {2} (v), …, T ^ {j-1} (v) } $. I wonder if anyone could help me pinpoint this. The proof is as follows.

*Theorem $ 5.22. $ Let T be a linear operator in a vector space of finite dimensions. $ mathrm {V}, $ and let $ mathrm {W} $ denote the $ mathrm {T} $ cyclic subspace of $ mathrm {V} $ generated by a non-zero vector $ v in mathrm {V}. $ Leave $ k = operatorname {dim} ( mathrm {W}). $
Then $ left {v, mathrm {T} (v), mathrm {T} ^ {2} (v), ldots, mathrm {T} ^ {k-1} (v) right } $ is a basis for $ mathrm {W} $.
Test. (a) from $ v neq 0, $ set $ {v } $ It is linearly independent. Leave $ j $ be the largest positive integer for which
$$
beta = left {v, mathrm {T} (v), ldots, mathrm {T} ^ {j-1} (v) right }
$$
It is linearly independent. Such $ j $ must exist because $ V $ It is of finite dimension. Leave $ mathrm {Z} = operatorname {span} ( beta). $ Then $ beta $ is a basis for $ mathrm {Z}. $ Further, $ mathrm {T} ^ {j} (v) in mathrm {Z} $ by the linear independence theorem. We use this information to show that $ mathrm {Z} $ is a $ mathrm {T} $ invariant subspace of $ V. $ Leave $ w in Z. $ as $ w $ is a linear combination of the vectors of
$ beta, $ there are scalars $ b_ {0}, b_ {1}, ldots, b_ {j-1} $ such that
$$
w = b_ {0} v + b_ {1} mathrm {T} (v) + cdots + b_ {j-1} mathrm {T} ^ {j-1} (v)
$$
and therefore
$$
T (w) = b_ {0} T (v) + b_ {T} {2} (v) + cdots + b_ {j-1} T ^ {j} (v)
$$
So $ T (w) $ is a linear combination of vectors in $ Z $, and therefore belongs to $ Z $.
So $ mathrm {Z} $ it is $ mathrm {T} $ -invariant. Further, $ v in mathrm {Z} $. By exercise $ 11, mathrm {W} $ is the smallest T-invariant subspace of $ V $ that contains $ v, $ so that $ W subseteq $ Z. Clearly $ Z subseteq W $ and then we conclude that $ mathrm {Z} = mathrm {W} $. It turns out that $ beta $ is a basis for $ mathrm {W}, $ and therefore $ operatorname {dim} ( mathrm {W}) = j. $ So $ j = k. $ This proves (a).*

I definitely see it for that matter when $ j = dim (V) $ (i.e. the largest $ j $ for which $ beta $ is linearly independent *it is* the dimension of $ V $.)

Then $ T ^ (v) $ must be on $ beta $ why $ beta $ now encompasses all the $ V $ .

But in the case is that $ j <dim (V) $, which prevents exactly $ T ^ (v) $ of belonging to the span of those $ k-j $ vectors that span $ beta $ not enough? I think this is my problem. Is this implicitly resolved by assuming that $ j $ is the largest integer for which $ beta $ is it linearly independent?

Thank you very much in advance!

## real analysis: the dense subspace standard is the same

Yes $ X $ is the set of step functions in $ (0.2 pi) $ with the $ L ^ 2 $ rule. $ phi_n = int_ {0} ^ {2 pi} f (x) n cos (n ^ 2x) dx $ are linear functional in $ X $. Wts $ | phi_n | = | ncos (n ^ 2x) | $ 2

I know this has to do with Riesz's representation theorem. We know $ X $ it is a dense subspace of $ L ^ 2 ((0.2 pi)) $. Basically I think I just need to show $ | phi_n | _X = | phi_n | $ 2. I know since $ X $ it is dense that $ overline X = L ^ 2 $ so the rules are basically the same, except the limit points of $ X $. But I wasn't sure how to show this.