I don't know where to start on this proof. I know that a linearly independent set is a set where c1v1+c2v2+⋯+cnvn=0 has no non-trivial solutions other than c1=c2=...=cn=0.   How can I use that to figure out this proof?
45034
Is the above statement true? If I have already proven that the set is linearly independent, is it automatically given that the subset is also linearly independent? If so, how would I go about showing that as a proof.
39514
Suppose $V_1$ and $V_2$ are $k$-vector spaces with bases $(e_i)_{i \in I}$ and $(f_j)_{j \in J}$, respectively. I've seen the claim that the collection of elements of the form $e_i \otimes f_j$ (with $\left(i,j\right) \in I \times J$) forms a basis for $V_1 \otimes V_2$. But I seem to get stuck with the proof.  My question: What's the easiest way to see that the above set is indeed linearly independent?
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?
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 $x \in \mathbb{R}$?
38459
I have $n$ lambdas, which are all different real and positive numbers, where:  $\lambda_1 &lt; \lambda_2 &lt; \cdots &lt; \lambda_n$.  I then have to show that these functions are linearly independent:   $$e^{\lambda_1 t}, e^{\lambda_2 t}, \ldots, e^{\lambda_n t}$$  So I guess what I want to show is that the only solution to this equations:   $$c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}+ \ldots +c_n e^{\lambda_n t}=0, \quad \text{for all } t$$  is that all constants are zero. I am not entirely sure how to do this when it is for $n$ lambdas - I did it earlier for just 3 lambdas, where I differentiated the function, so I have tried to use the same approach.  I wanted to get $n$ equations with $n$ unknown, so I differentiated the function $n-1$ times, and then I chose to look at the case where $t = 0$, so I have something looking like this:   $$c_1+c_2+ \ldots +c_n = 0$$  $$c_1 \lambda_1+c_2 \lambda_2+ \ldots +c_n \lambda_n = 0$$  $$ \cdots $$  $$c_1 \lambda_1^{n-1}+c_2 \lambda_2^{n-1}+ \ldots +\lambda_n^{n-1} = 0$$  Then I put this into a matrix $Ac = 0$, and now I want to show that the $\det(A)$ doesn't equal zero, so that the only solution is c = 0, but I'm not quite sure how to do this, or whether there is an easier way to do it ?
113403
Let $V$ be an n-dimensional vector space over $\mathbb{R}$ and $T:V \rightarrow V$ be linear. Call a vector $v \in V$ cyclic if $V$ is spanned by $\{v, \ Tv, \ T^2v,..\}$.  Question:  Show that $v$ is cyclic for $T$ then there is some $k \geq 1$ such that   $$B=\{v, \ Tv,..T^{k-1}v \}$$ is linearly independent in $V$ and that $T^kv$ lies in  the span of $B$.  Thoughts:  If we suppose that no such $k$ exists, then every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent, or it is linearly dependent with $T^lv$ not lying in it span.  There are in effect then two cases to consider.  Suppose every subset of the form $\{v, \ Tv,..T^{l-1}v \}$ is linearly dependent. Then $T^{l-1}$ can be expressed as a linear combination of $v,...,T^{l-2}v$. Now, $\{v, \ Tv,..T^{l-2}v \}$ is also linearly dependent by assumption and so we can express $T^{l-2}$ as a linear combination of  $v,...,T^{l-3}v$ etc... Continuing in this way, we show that $T^{l-1}v= \lambda v$ for some $\lambda \in \mathbb{R}$. Thus $$B=\{ \mu v \ | \ \mu \ \text{is an eigenvalue of} \ T^mv \ \text{for some} \ m \}$$ This is 'at best' countably infinite, and therefore could not span $V$ even if it were 1-dimensional, wince our base field is $\mathbb{R}$, which is uncountable.  Suppose then that $B$ is linearly independent for some $k$ but that $T^kv$ does not lie in its span. I can't see how to progress this argument.  Note: This question is for second year Undergraduates, and I think there must be some more simple approach; even the first part surprises me if we need to argue using the uncountability of $\mathbb{R}$, so any alternative approaches are also appreciated.
121842
Friends,  Do you know of a neat proof of the linear independence (over $\mathbb{C}$) of the functions $f(t) = e^{at}$ and $g(t)=e^{bt}$ when $a$ and $b$ are linearly independent over $\mathbb{Q}$?
86878
$Q$: Find a subset of $ \mathbb R^3$ which is closed under $*$, but not $+$. $A$: The set $\mathbb R$$u$ $+$ $\mathbb R$$v$, where $u$ and $v$ are linearly independent vectors in $ \mathbb R^3$. Here $ \mathbb R$$w$ = $\{aw: a \in \mathbb R\}$ for a given vector $w$. This is closed under $*$, but not $+$.  By  $u$ and $v$ are linearly independent vectors in $ \mathbb R^3$  do they mean two sets of linearly independent vectors $u$ and $v$ in $ \mathbb R^3$ where $u$ could be  $u = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$? Generally, please help me understand how this set is not closed under $+$. Thanks.
25440
Prove that there exist uncountable family of subsets of $\mathbb{N}$ which is linearly (totally) ordered by realtion of inclusion.  I do not know how to prove it. Actually I cannot concieve that such family of set can be uncountable. I would be thankful for help
55527
If $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, and $(\beta_1,\beta_2,\beta_3) = (\alpha_1,\alpha_2,\alpha_3)C$. Prove that if $\mid C\mid \neq 0$, then $\beta_1,\beta_2,\beta_3$ are linearly independent.   This statement seems to be trivial, if $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, then $\det{(\alpha_1,\alpha_2,\alpha_3)} \neq0$, and $\det{(\alpha_1,\alpha_2,\alpha_3) C} = \det{(\alpha_1,\alpha_2,\alpha_3)}*\det{}C\neq 0$, which implies $\det(\beta_1,\beta_2,\beta_3) \neq0$, and hence the columns are linearly independent.  I'm not sure how to prove it rigorously, or if I'm missing any steps here. Would really appreciate any help! Thank you!
7401
Problem 4.3.32 in Linear Algebra, Lay:  Let $V$ and $W$ be vector spaces, let $T:V\to W$ be a linear transformation, and let $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ be a subset of $V$. Suppose that $T$ is a one-to-one transformation [...]. Show that if the set of images $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.  All I can think of doing: If $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then there exist weights $c_1, \dots, c_p$, not all zero, so that $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}$$ Since $T$ is linear: $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = T(c_1\mathbf{v}_1) + \cdots + T(c_p \mathbf{v}_p)=T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$$ So the set $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ must be linearly dependent. Am I missing something? And why must $T$ be on-to-one for this to be true? An example showing that it is not true if $T$ is not one-to-one would be great!
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
55784
I'd like to know if the following proof is valid.  The only thing I'm not sure about (though I can't see why it's invalid if it is) is if we can always use the Hausdorfness of $Y$ to separate an open set from $f(C)=g(C)$.  "Let $X$ be a space, $Y$ a $T_2$-space, and $f,g:X \to Y$ continuous functions.  Prove that $C:=\{x \in \,|\,f(x)=g(x)\}$ is a closed subset of $X$."  Let $x \in X$ be such that $f(x) \ne g(x)$.  Since $Y$ is $T_2$, there are open sets $U_\alpha \ni f(x)$ and $V_\alpha \ni g(x)$ such that $U_\alpha \cap O_\alpha=V_\alpha \cap O_\alpha=\varnothing$ for all $\alpha \in A$ where $A$ is the set indexing the points of $f(C)=g(C)$, and $O_\alpha$ is an open set containing $x_\alpha \in f(C)=g(C)$.  Let $U:=\displaystyle\bigcap_{\alpha \in A} U_\alpha$ and $V:=\displaystyle\bigcap_{\alpha \in A} V_\alpha$.  Then $U \cap f(C)=V \cap f(C)=\varnothing$.  Therefore, since $f,g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ disjoint from $C$.  Since $X\setminus A$ is the union of such open sets, it is open itself and therefore $C$ is closed.  Thanks.
118738
Let $V$ be a $K$-vector space,  $f: V \rightarrow V$ a linear map. Let $v \in V$. May a number $n ≥ 0$ exist, so that: $f^n(v) \not= 0$ and $ f^{n+1}(v) = 0$. Show that $v, f(v),...,f^n(v)$ are linearly independent.   What i know:   $f^0 = id, f^1 = f, f^2 = f \circ f, ...$ $v, f(v),...,f^n(v)$ are linearly independent if:  $a_1v + a_2f(v) + \cdots+a_nf^n(v)=0 \implies a_1 = a_2 =\cdots= a_n = 0$ $f$ is linear, hence $f(0) = 0$, $af(n) = f(an)$, $f(v_1 + v_2) = f(v_1)+f(v_2)$   My thoughts on it: the definition of linearly independent, in our case, can be rewritten as  $a_1v + f(a_2v) + f(f(a_3v)) + \cdots+f^n(a_nv)=0 \implies a_1 = a_2 =\cdots= a_n = 0$  Taking the function $f$ on both sides we can also obtain  $a_1f(v) + f(f(a_2v)) + f(f(f(a_3v))) + \cdots+f^{n+1}(a_nv)=0 \implies a_1 = a_2 =\cdots= a_n = 0$   ... I feel that i am on the right track, but i can't take it further from here.  So i have to show this sum only becomes $0$ when all $a_n$ are $0$. I thought about a proof by contradiction, but i wouldn't know where i can lead this to a contradiction.   Thanks for all help
108287
I wish to prove that if $f: \mathbb{C} \to \mathbb{C}$ is analytic and not of the form $az+b,$ then $z, f(z), f(f(z)), \dots$ are linearly independent functions over $\mathbb{C}.$  The cases $n=0, 1$ are trivial. To solve this problem, I'm starting on the first non-trivial scenario: assume $c_1 z + c_2 f(z) + c_3 f(f(z)) = 0$ for some $c_1, c_2, c_3.$ If I can solve this, I'll probably know how to solve the general case. But I'm getting nowhere.   The equations $z-f(z)-f(f(z))+f^3(z)=g(z)-g(g(z)) = 0$ for $f(z) = z+1, g(z) = |z|$ suggest we should impose analycity and non-linearity. Any hints or ideas?    Edit: I've obtained a major breakthrough, but one step is still missing. Define a function to be $n$-independent if $z, f(z), f(f(z)), \dots, f^{(n)}(z)$ are linearly independent over $\mathbb{C}$ on some non-empty open subset of $\mathbb{C}.$  Unproven Lemma: The set of $n$-dependent functions forms a vector space under addition.  $f$ being $n$-dependent easily implies $cf$ is $n$-dependent, so additivity is the only thing that stands in the way of proving this lemma. We need to figure out some way to handle terms like $(f+g) \circ (f+g) = f(f(z)+g(z))+g(f(z)+g(z))$ without messing up the rest of the terms.   First note that $h(z)=az+b$ is $2$-dependent since $z, h(z), h(h(z))$ are $3$ vectors in the $2$ dimensional vector space of polynomials with degree $\le 1.$  Suppose $f$ is analytic and non-linear. Let $n$ be minimal such that $f$ is $n$-dependent. Let $g(z) = f(z)-(f(1)-f(0))z-f(0).$ Since $f$ isn't linear, $n \ge 2,$ implying $g$ is $n$-dependepnt. Suppose $c_0 z + c_1 g(z) + \dots + c_n g^n(z) = 0.$ Setting $z=1,$ we get $c_0 = 0,$ so $c_1 z + \dots + c_n g^{n-1}(z) = 0$ on $\mathcal{O} = g(\mathbb{C}),$ which is open by the open mapping theorem since $g$ is analytic and non-constant.   Thus, $g$ is $n-1$ dependent. If $n \ge 3,$ this means $f$ is $n-1$ dependent, contradiction. If $n=2,$ then $g$ is $1$-dependent, so $g(z)=cz$ for some $c,$ which means $f$ is linear, contradiction.
13339
This is a response to the following exchange at    [Bill constructs a $\aleph_0$ linearly independent set in $\Bbb R$]  Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. –   I believe the answer to this question is no, for the following reason. Suppose we have a spanning set for $\Bbb R$ over $\Bbb Q$, of cardinality $\frak n$. Then by definition we have the cardinality equation ${\frak c}\le^*[{\frak n}]^{&lt;\omega}\cdot \aleph_0$ (where $[{\frak n}]^{&lt;\omega}$ is the cardinality of finite subsets, and $\frak m\le^*n$ means there is a surjection from $\frak n$ to $\frak m$). If $\frak n$ is finite, then we get ${\frak c}\le^*\aleph_0$ which is false, so $\frak n$ is infinite (but not necessarily Dedekind-infinite). If we assume the axiom of choice, then the equation reduces to $\frak c\le n$, hence every Hamel basis is size continuum. But without AC I don't know any way to simplify the original equation to get anything stronger than &quot;$\frak n$ is infinite&quot;. This is at best heuristic evidence for the original question, since it is considering a different problem (spanning sets instead of linearly independent sets). How could one show that there are no ZF-constructible $\Bbb Q$-linearly independent subsets of $\Bbb R$ of cardinality continuum?
77248
Given a set of vectors, we can compute the number of independent vectors by calculating the rank of the set, but my question is how to find a maximal linearly independent subset. Thanks!