## Find dimension of the subspace

Find Dimension of the 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 $$j , which prevents exactly $$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.