Show that Z cannot be turned into a vector space over any field.   So, we have 2 cases here.   Case 1:lets suppose the charF=P,  n does not equal 0, then (1+1+...+1)n=1n+1n+...+1n=n+n+...+n=pn=wchich does not equal zero  =0n  =0               ................................................           a contradiction  Case 2: CharF=0 ->Q  ?????
122369
Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field.  Proof.  Surpose that: $\phi : (A, +) \rightarrow  \mathbb{Z} $ is an isomorphism. Then there is some $r∈A $ such that $ϕ(r)=1_\mathbb{Z}$. We have $$ 1_\mathbb{Z}=ϕ(r)=ϕ(2(r/2))=2ϕ(r/2)$$  or equivalently that $ϕ(r/2)=\frac{1}{2}$. But this is not in $ \mathbb{Z}$, so there can be no such morphism.  Is my proof above correct? Thanks
61897
Let $V$ be a finite-dimensional vector space, $V_i$ is a proper subspace of $V$ for every $1\leq i\leq m$ for some integer $m$. In my linear algebra text, I've seen a result that $V$ can never be covered by $\{V_i\}$, but I don't know how to prove it correctly. I've written down my false proof below:  First we may prove the result when $V_i$ is a codimension-1 subspace. Since $codim(V_i)=1$, we can pick a vector $e_i\in V$ s.t. $V_i\oplus\mathcal{L}(e_i)=V$, where $\mathcal{L}(v)$ is the linear subspace span by $v$. Then we choose $e=e_1+\cdots+e_m$, I want to show that none of $V_i$ contains $e$ but I failed.  Could you tell me a simple and corrected proof to this result? Ideas of proof are also welcome~  Remark: As @Jim Conant mentioned that this is possible for finite field, I assume the base field of $V$ to be a number field.
56074
Given vector space $V$ over $\mathbb R$ such that the elements of $V$ are infinite-tuples. How to show that any basis of it is uncountable?
89298
Are $\mathbb{Z_p},\mathbb{Q},\mathbb{R},\mathbb{C}$ above themselves vector space? Is a field above anoother field a vector space?   As for 1. we know that $\Bbb R^n$ is a vector space so in particular it is true for $n=1$  For 2. by definition a vector space is $V$ with addition and multiplication over a field, therefore 2 is true.
88732
Is there an example of a vector space $V/F$ such that $0 \ne c \in F$, $\vec{0} \ne v \in V$, but $c \cdot v = 0 $?  I couldn't think of an example.
22569
Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:     Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field  and let $c \in  F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$   has no root  in $F$. [Hint: consider two cases: char$(F) = p$ and   char$(F)$ different of $p$.]   I have attempted this a lot. Anyone has an answer?
37473
Let $\mathbf{V}(\mathbb{K}_1,V),$ and $\mathbf{W}(\mathbb{K}_2,W)$ two vector spaces over different fields ( as an example: $\mathbb{K}_1=\mathbb{C}$ and $\mathbb{K}_2=\mathbb{R}$).  We can generalize the notion of  a linear transformation $T:\mathbf{V}\to\mathbf{W}$ for such two space?  My idea is that we have no problems for additivity: $$ T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+ T (\mathbf{y}) $$ but we have some truble with homogeneity since for $T(\alpha \mathbf{x})$ we cannot define $\alpha T(\mathbf{x})$.  It seems that we must have some function $\lambda :\mathbb{K}_1 \to \mathbb{K}_2 $ so that we can write something as $$ T(\alpha \mathbf{x})=\lambda(\alpha)T(\mathbf{x}) $$   but there is some ''natural'' definition of such function $\lambda$ that preserve the intuitive meaning of linearity?
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.
75296
Note that (1) all norms on finite dimensional vector space are equivalent.  And here (2) norms on vector space are equivalent iff they have same topology.   Question : How can we prove (2) ? Please recommend reference  Thank you in anticipation
114244
I have two small questions that I can't solve. So, there are 8 axioms for vector spaces. One of them is that $1 \cdot a = a$ for all $a \in V$ and of course $1$ is multiplicative identity in $\mathbb{F}.$ How can I be sure this axiom cannot be derived from other 7 axioms? Also, why is there no definition of $\alpha a= a\alpha, \forall \alpha \in \mathbb{F}, \forall a \in V$? Why is multiplication of scalars and vectors not defined in commutative way?
78166
I was reading up on the difference between countable and uncountable sets, and was wondering if there was a basis of uncountable size.  I now know there are, however they all seem to be covering rather high level areas in math.  So I was wondering; what is the simplest example of a basis of uncountable size for a vector space over a field?
115553
I would appreciate a hint on the following problem:     Let $V$ be a finite dimensional vector space over $F$. There are two scalar products such that: $$ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=0 \implies \langle v,w\rangle_2=0\Big) $$ Show that $$ \exists \ c \in F \ \ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=c \langle v,w\rangle_2\Big) $$   I have tried to define an orthonormal basis with respect to $\langle \cdot,\cdot\rangle_1$ hoping that the transformation matrices would be different by a constant, yet it seems to lead nowhere.
29716
If $U_1$, $U_2,\ldots,U_n$ are proper subspaces of a vector space $V$ over a field $F$, and $|F|\gt n-1$, why is $V$ not equal to the union of the subspaces $U_1$, $U_2,\ldots,U_n$?
19974
I am having a difficulty setting up the proof of the fact that two basis of a vector space have the same cardinality for the infinite-dimentional case. In particular, let $V$ be a vector space over a field $K$ and let $\left\{v_i\right\}_{i \in I}$ be a basis where $I$ is infinite countable. Let $\left\{u_j\right\}_{j \in J}$ be another basis. Then $J$ must be infinite countable as well. Any ideas on how to approach the proof?
109064
I'm kicking myself over this one, but I just can't seem to make the argument rigorous. From Axler's Linear Algebra Done Right:  for a vector space $V$ with an underlying field $F$:  Take an element $a$ from $F$ and $\vec{v}$ from $V$. $a\vec{v}=\vec{0}\implies a=0 $ or $ \vec{v}=\vec{0}$  After only being able to come up with half of a direct proof, I tried doing this by proving the contrapositive $a\neq 0 \wedge \vec{v} \neq \vec{0} \implies a\vec{v}\neq \vec{0}$  Say $a\vec{v}=\vec{u}$. Since $a$ is non-zero, we can divide both sides by $a$.  $$\vec{v}=\frac 1 a \vec{u}$$  If $\vec{u}$ were $\vec{0}$ then by  $$\frac 1 a \vec{0}=\frac 1 a (\vec{0}+\vec{0})\implies\frac 1 a \vec{0}=\vec{0}$$ $v$ would be $0$ as well. Since it isn't by assumption, $\frac 1 a \vec{u}$ cannot be zero and so $\vec{u}$ cannot be as well.   Is this fully rigorous? It seems like a very simple question, but I'm not sure about it. Namely, the last step of $\frac 1 a \vec{u}\neq 0 \implies \vec{u}\neq 0$ doesn't seem obvious. I think I need to use the $1\vec{v}=\vec{v}$ axiom, but I'm not sure how.  Is there a more direct proof? This whole contrapositive business seems a bit clunky for something so simple.
19864
In my opinion both are almost same. However there should be some differenes like any two elements can be multiplied in a field but it is not allowed in vector space as only scalar multiplication is allowed where scalars are from the field.  Could anyone give me atleast one counter- example where field and vector space are both same. Every field is a vector space but not every vectorspace is a field. I need an example for which a vector space is also a field.  Thanks in advance. (I'm not from mathematical background.)
7683
Let $V$ and $W$ be vector spaces over a field $F$. Let $L(V,W) = \{T:V\to W : T \text{ is linear} \}$, that is, $L(V,W)$ is the collection of all linear functions from $V$ to $W$. For $S,T \in L(V,W)$ and $a \in F$, define new functions $(S+T):V \to W$ and $ aT :V \to W$ by $(S+T)(x) = S(x)+T(x)$ ; $(aT)(x) = a(T(x))$.   Prove that $S+T$ belongs to $L(V,W)$ and $aT$ belongs to $L(V,W)$.