Let $X_1,\ldots,X_k$ be lineary independent local sections of a vector bunlde $V$, $rank(V)&gt;k$, defined on some open set $U$. Can I construct a local section $X_{k+1}$ on a (maybe smaller) open subset of $U$, such that $(X_1,\ldots,X_{k+1})$ are lineary independent on the smaller subset?  edit: I know that answer, however I would be interested in a more "elementary" argument.
120720
Let $\pi: E \to M$  be a smooth vector bundle of rank $n$, and suppose $s_1, \ldots, s_m$ are independent smooth local sections over an open subset $U \subset M$.   Can I prove the "basis extension theorem" for smooth sections? That is, for each $p \in U$, there are smooth sections $s_{m+1},\ldots,s_n$ defined on some neighborhood of $V$ of $p$ such that $(s_1, \ldots, s_n)$ is a smooth local frame for $E$ over $U \cap V$.
104883
It seems rather obvious to me that $\sin(x)$, $\sin(2x)$, $\sin(3x)$ are linearly independent in $\operatorname{Map}(\mathbb{R},\mathbb{R})$, but I'm not sure how to prove it (or disprove it if I'm wrong?). I know that for linear independence it must $$ a\sin(x)+b\sin(2x)+c\sin(3x)=0 $$ $$ \implies a=b=c=0$$ but how can I show this is true for any $x \in \mathbb{R}$?
84995
Let $X$ be a scheme over a field $k$, and let $x\in X$ be a rational point, that is, we have $k(x):=\mathcal{O}_x/\mathfrak{m}_x\cong k$.  Let $\alpha:\mathfrak{m}_x/\mathfrak{m}_x^2\rightarrow k$ be an element of $T_x=(\mathfrak{m}_x/\mathfrak{m}_x^2)^*$, the Zariski tangent space to $X$ at $x$.  Using $x$ and $\alpha$, I'd like to define a $k$-morphism of schemes Spec $k[\varepsilon]/\varepsilon^2\rightarrow X$.  In the course of the construction, I need to define a local map of local rings $\mathcal{O}_x\rightarrow k[\varepsilon]/\varepsilon^2$.  Using $\alpha$, what is a natural way to define such a map?  It might be helpful to note that I have an injection $k\hookrightarrow\mathcal{O}_x$ arising from the $k$-scheme structure on $X$.  I think this leads to a decomposition $\mathcal{O}_x/\mathfrak{m}_x^2=k\oplus(\mathfrak{m}_x/\mathfrak{m}_x^2)$ as $k$ vector spaces, from which the desired map would follow, but I can't seem to explicitly see what this decomposition is inside of $\mathcal{O}_x/\mathfrak{m}_x^2$.  This is part of exercise II.2.8 in Hartshorne.
103477
Let $F(\mathbb{R},\mathbb{R})$ be the vector space of all functions from $\mathbb{R}$ to $\mathbb{R}$. Prove that for all $n\in\mathbb{N}$, the set of vectors $$\left\{sin(x), sin(2x),sin(2^2x),\dots,sin(2^nx)\right\}$$   form a linearly independent subset of $F(\mathbb{R},\mathbb{R})$.  I know that to form a linearly independent subset the set must be closed under addition, closed under scalar multiplication, and contain the neutral element.  Showing that it contains the neutral element is relativley straight forward since when $x=0$ every element also equals $0$. Therefore this subset contains the neutral element. However, I am having a hard time coming up with a way to approach the parts showing that it is closed under scalar multiplication and addition for all $x$. Do you have a piece of advice or a hint on how to approach those part?
110644
Suppose we are given a $k$-dimensional complex vector space. Consider the second exterior power of $V$, that is $\Lambda^2V$. Denote $X=\{v_1\wedge v_2\colon v_1,v_2\in V\}$.  Now I have two questions:  1) Is it true that if $k=4$, then $x\in X$ if and only if $x\wedge x=0\in \Lambda^4 V$?  2) Let $k&gt;4$. Does there exist $x\in \Lambda^2 V\setminus X$ such that $x\wedge x\neq 0$?  Thank you, C.
102933
It's well known that, in $\mathbb{R}^n$:     (1) Open and Connected $\Rightarrow$ Path-connected   The proof essentially goes through the fact that (2) Every path-connected component will be open. Using this fact, we arrive at a contradiction if we suppose there are more than one path-connected component.  Well, trying to see where (1) would keep being valid, I arrived at the following:     If $X$ is a locally convex topological vector space, then (1) is valid.   But, to prove this, I proved (2). But for that, you use the (rather strong imo) convexity of a local base (to essentially repeat the argument for $\mathbb{R}^n$). But I'm not satisfied with this, as I think that we are using a lot of strong conditions (using the existence of a "line segment" to prove the existence of a "curve" seems very bazooka-like to me).  Anyway, given the previous considerations, my questions are:  Are there more general examples of spaces where (1) holds? Is there a characterization of spaces that satisfy this property?  And bonus question, since (2) implies (1) if the set is open.:  Are there more general examples of spaces where open sets satisfy (2)? Is there a characterization spaces that satisfy this property?
43037
This is a question from Finite-Dimensional Linear Algebra by Mark S. Gockenbach page 57 (exercise 2.5.15). I hope to check my proof.  Let $V$ be a vector space over a field $F$, and suppose $S$ and $T$ are subspaces of $V$ satisfying $S \cap T = \{0\}$. Suppose $\{s_1,s_2, ..., s_k\} \subset S$ and $\{t_1,t_2, ..., t_l\} \subset T$ are both linearly independent sets. Prove that $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$.  My proof) Suppose that $\alpha_1 s_1 +...+ \alpha_k s_k + \alpha_{k+1}t_1 +...+ \alpha_{k+l}t_l = 0$ for some $\{ \alpha_i\}_{i=1}^{k+l} \subset F$ Then $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = -(\alpha_1 s_1 + ... + \alpha_k s_k) \in S$ Since $\{ t_1,...t_l \} \subset T$ and $T$ is a subspace of $V$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_1 \in T $. Since $S \cap T = \{0 \}$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = 0$. Since $\{t_1, ... , t_l \}$ is a linearly independent set, $\alpha_{k+1} = ... = \alpha_{k+l} = 0$. By doing similarly, we can show that $\alpha_1 = ... = \alpha_k = 0$. Therefore, $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$. $\blacksquare$
85937
Prove:     If $v_1,...,v_r$ are eigenvectors that correspond to distinct eigenvalues $\lambda_1, ...,\lambda_r$ of an $n \times n$ matrix $A$, then the set $\{v_1,...,v_r\}$ is linearly independent.   Please give an example and tell me how this theorem works!
122542
Recall that a space $X$ is locally compact if for any point $x$ in $X$, and any neighborhood $U$ of $x$ in $X$,there is a neighborhood $V$ of $x$ in $X$,and a compact subspace $C$ of $X$ such that $x \in V \subset C \subset U.$ Let $X$ be a locally compact space, and let $f : X \rightarrow Y$ be a continuous, closed surjection. Suppose that $f^{-1}(y)$ is compact for each $y \in Y$ . Show that $Y$ is locally compact.
32156
I want to prove the fact that  If $W\subseteq V$ is a vector subspace of $V$ than   $dim(W)=dim(V)=n \implies W=V$   Without using the fact that if $dim(V)=n$ any set of $n$ indipendent vectors in $V$ is necessarily a basis of $V$, neither the fact that a basis of $\mathscr{L}=(\vec{v_1},...,\vec{v_n})$ is necessarily the set $\big\{ \vec{v_1},...,\vec{v_n} \big\}$. (Since these two facts are consequences of the proof).  By hypotesis I have that $W\subseteq V$ which means that all the vectors in $W$ are in $V$. I must prove the opposite inclusion $V\subseteq W$: all the vectors in $V$ are a linear combination of $n$ vector of a basis, but I don't know how to prove that all the vectors of that basis are contained in $W$.  Can I say that they must belong to $W$ since $W$ has $n$ vectors in its basis too? (But I think in this way I'm assuming the facts listed above).  Can you help me to finish the proof?   Thanks for your help
113065
I know that any subset of V with more than k vectors is always linearly dependent, but how do I include matrix A into the vectors given to prove the statement?   Also, for part (b), does the statement mean that Ax=0 only has the trivial solution since V intersects span{u1, u2, ..., uk}={0}?  Any help would be appreciated! Thank you!
90894
Suppose $Y=Z(f_1,\cdots ,f_r)\subseteq \mathbb{A}^n_k$ where each $f_i$'s are linear homogeneous polynomials which are $k$-linear independent. Then $Y$ is also a vector space over $k$. My question: Is the vector space dimension of $Y$ is same as the dimension of $Y$ as an affine variety? If yes how?    Thank you.
85180
The following problem is from Golan's linear algebra book. I have posted a proposed solution in the answers.   Problem: For $n\in \mathbb{N}$, consider the function $f_n(x)=\sin^n(x)$ as an element of the vector space $\mathbb{R}^\mathbb{R}$ over $\mathbb{R}$. Is the subset $\{f_n:\ n\in\mathbb{N}\}$ linearly independent?
7383
I'm having problems with the following problem:  Let $V$ be a vector space over the field $F$. Let $n \in \mathbb{N}$. Suppose that $\left( v_1, \ldots,v_n \right)$ is a linearly independent family of vectors in $V$. Show that the families $\left( {{v_1},{v_1} + {v_2}} \right),\left( {{v_1},{v_1} + {v_2},{v_1} + {v_2} + {v_3}} \right),$ $\ldots,$ $\left( {{v_1},{v_1} + {v_2}, \ldots ,{v_1} + {v_2} +  \ldots  + {v_n}} \right)$ are linearly independent.  This is my attempt:  I take a typical finite subset such as $\left( {{v_1},{v_1} + {v_2}, \ldots ,{v_1} + {v_2} +  \ldots  + {v_n}} \right)$,  where $n \in \mathbb{N}$. Then suppose that $\lambda_1{v_1}+\lambda_2\left({v_1} + {v_2}\right)+ \ldots +\lambda_n \left({v_1} + {v_2} +  \ldots  + {v_n}\right)=0$,  where $n \in \mathbb{N}$.  I know there is the trivial solution $\lambda_1=\lambda_2=\ldots=\lambda_n=0$, but I'm not sure how to check if there are others?
101150
I want to see if I am understanding the proof of the replacement theorem correctly.   Let $V$ be a vector space that is spanned by a set $G$ containing $n$ vectors.  Let $L \subseteq V$ be a linearly independent subset containing $m$ vectors.  Then $m\leq n$ and there exists a subset H of G containing exactly $n-m$ vectors s.t. $L \cup H$ generates $V$.   The proof in the book says to start with $m=0$. In that case $L=\varnothing$, the empty set.        Let's assume this theorem is true for some integer $m \geq 0$.   Let $L = \{v_j\}_{j=1}^{m+1}$ and define it as a linearly independent subset of $V$ consisting of $m+1$ vectors.  Since any subset of a linearly independent set is linearly independent as well ($S_1 \subseteq S_2 \subseteq V$), then $\{v_j\}_{j=1}^m$ is linearly independent also.   It then says to use the induction hypothesis to say that $m \geq n$. Well, ok, but that seems sort of pulled out of nowhere to me. The next step is to say that there is therefore another subset, $\{u_k\}_{k=1}^{n-m}$ of G such that $\{v_j\}_{j=1}^m \cup \{u_k\}_{k=1}^{n-m}$ spans $V$.   That being the case there are scalars $(a_j)_{j=1}^m$ and $(b_k)_{k=1}^{n-m}$ which you can multiply by the vectors $v_j$ and $u_k$. You can then add the two sets of vectors, yielding $$\sum_{j=1}^m a_j v_j + \sum_{k=1}^{n-m} b_k u_k = v_{m+1}$$  At this point I am not sure I understand things. Because we are assuming $n-m&gt;0$ -- otherwise $v_{m+1}$ is linearly dependent. But then it says not only is $n&gt;m $ but $n&gt;m+1$, and that is where I am losing the plot to get to the next step.
22818
The problem:  Consider the smooth quadric $Q=V(X_{0}X_{1}+X_{2}X_{3}+X_{4}^{2})\subset\mathbb{P}^{4}$ and the line $L=V(X_{0},X_{2},X_{4})$ contained in it. Prove that there exists a vector bundle $F$ on $Q$ with a section vanishing exactly at $L$.  The hardest thing is to get the transition matrices. Any help is appreciated.
96858
So I'm trying to prove the classic dimension theorem for linear transformations. Here is what I have so far:  Given a linear transformation $T:V\rightarrow W$  Let the basis of $\mathrm{Im}T$ be $\{u_1,...,u_m\}$ where $\forall u\in V:T(u)$.  Let the basis of $\ker T$ be $\{v_1,..,v_k\}$ where $\forall v:T(v)=0$.  Lets prove $B=\{u_1,...,u_m,v_1,..,v_k\}$ is a basis of $V$:  (1) $B$ is linearly independent: Assume (falsely) that $B$ isn't linearly independent. Since $\{u_1,...,u_m\}$ and $\{v_1,..,v_k\}$ are both the basis of $\ker T$ and $\mathrm{Im}T$ repectively, they must each be linearly independent.If this is the case, $\exists b\in V:b\in \{v_1,..,v_k\}\cap \{v_1,..,v_k\}$ however this contradicts $\mathrm{Im}T\cap \ker T=\emptyset$ so $B$ must be linearly independent.  (2)$\mathrm{span}(B)=V$: Assume (falsely) that $b_2\in V| b_2\notin \mathrm{span}(B)$. If this is the case, then $b_2\notin \{\forall u\in V:T(u)\}\cap\{\forall v:T(v)=0\}$. This is impossible because $b_2\in V$ so $\mathrm{span}(B)=V$  Since (1) and (2) are true, according to basis identities, $B$ is a basis of $V$. This means that $\dim V=|B|=\dim\ker T+\dim\mathrm{Im}T$ QED