Let $V_{n}(F)$ be a vector space over the field $F=\mathbb Z_{p}$ with $\dim V_{n} = n$, i.e., the cardinality of $V_{n}(\mathbb Z_{p}) = p^{n}$. What is a general criterion to find the number of bases in such a vector space? For example, find the number of bases in $ V_{2}(\mathbb Z_{3})$. Further, how can we find the number of subspaces of dimension, say, $r$?   I need a justification with proof. I have a formula, but I am unable to understand the basic idea behind that formula.
75055
I am asked to find how many there are $k$-dimensional subspaces in vector space $V$ over $\mathbb F_p$, $\dim V = n$.  My attempt: 1) Let's find a total number of elements in $V$: assume that $\{v_1, v_2, \cdots, v_n\}$ is a basis in $V$. Then, for every $v \in V$ we can write down $$ v = a_1 v_1 + a_2 v_2 + \cdots + a_n v_n $$ and since the coordinates ($a_1, \cdots, a_n$) are from $\mathbb F_p$ there are $p^n$ vectors in $V$;  $p-1$ without the zero vector.  2) Let's look at a situation where $k=1$. Let's call this 1-dimensional space $V&#39;$. $$\forall v&#39; \in V&#39;. v&#39; = a_1 v_1$$ where $v_1$ is a basis in $V&#39;$. We know that if there are two non-zero vectors $u \in V&#39;_1$ and $v \in V&#39;_2$,  they are not linear dependant. So, every 1-dimensional subspace has $(p-1)$ basises. Therefore, there are $\frac{p^n - 1}{p-1}$ possible 1-dimensional subspaces in $V$  3) k-dimensional subspace is defined by the set of it's basises. Since basis can not contain zero vectors we can write down the formula for selecting $k$ linear independent vectors: $C^k_m (p-1)^k$, where $m = \frac{p^n - 1}{p-1}$. Here we first choose $k$ 1-dimentioanl subspaces and then we choose one of $(p-1)$ non-zero vectors from each of the subspaces.  4) .. unfortunately, this is where I am stuck. My intuition says that the answer may be  $\frac{p^n - 1}{(p-1)^k}$, but this might be completely wrong and I don't know how to go about finishing the problem.  Thanks in advance.
88273
Can we define a vector space structure on $\mathbb {R}^n$ other than usual scalar multiplication and usual addition such that the dimension of $\mathbb {R}^n$ over $\mathbb {R}$ is not $n$ but some $m$ not equal to $n$?
123394
Let $V$ be an $n$-dimensional vector space over $\mathbb R$. Prove that $V$ has a subspace of dimension $r$ for each $0 \le r \le n$.  Is this as simple as saying that $V$ has a basis of $n$ elements, and then take $r$ elements from this basis and this will generate an $r$-dimensional subspace? Or am I missing something? It seems too simple.  Thanks
83713
I was looking at the definition on wikipedia of a vector space (similar/equivalent definitions are everywhere, but I thought I'd list it here for completion):  A vector space over a field $F$ is a set $V$ together with two binary operations that satisfy the eight axioms listed below. Elements of $V$ are called vectors. Elements of $F$ are called scalars. In the two examples above, our set consists of the planar arrows with fixed starting point and of pairs of real numbers, respectively, while our field is the real numbers. The first operation, vector addition, takes any two vectors $v$ and $w$ and assigns to them a third vector which is commonly written as $v + w$, and called the sum of these two vectors. The second operation takes any scalar a and any vector $v$ and gives another vector $av$. In view of the first example, where the multiplication is done by rescaling the vector $v$ by a scalar $a$, the multiplication is called scalar multiplication of $v$ by $a$.  To qualify as a vector space, the set $V$ and the operations of addition and multiplication must adhere to a number of requirements called axioms. In the list below, let $u$, $v$ and $w$ be arbitrary vectors in $V$, and $a$ and $b$ scalars in $F$.   Associativity of addition: $u + (v + w) = (u + v) + w$ Commutativity of addition: $u + v = v + u$ Identity element of addition: There exists an element $0 ∈ V$, called the zero vector, such that $v + 0 = v$ for all $v ∈ V$. Inverse elements of addition: For every $v ∈ V$, there exists an element $−v ∈ V$, called the additive inverse of $v$, such that $v + (−v) = 0$ Compatibility of scalar multiplication with field multiplication: $a(bv) = (ab)v$ Identity element of scalar multiplication: $1v = v$, where $1$ denotes the multiplicative identity in $F$. Distributivity of scalar multiplication with respect to vector addition: $a(u + v) = au + av$ Distributivity of scalar multiplication with respect to field addition: $(a + b)v = av + bv$   I was wondering if there were any examples of vector spaces where only the law for the identity element of scalar multiplication fails? The only one I could think of would be to redefine a true vector space's scalar multiplication to multiply the result by a constant factor. Are there others?
100325
In Dummit and Foote problem 12 of section 10.4, I need to show that if $V$ is some vector space over a field $F$ and $v,v'$ are nonzero elements of $V$, supposing we have that $v \otimes v' = v' \otimes v$ then this implies that $v = av'$ for some $a \in F$. I was trying to consider some basis $\mathcal{B} = \{e_i\}_{i\in I}$ for the vector space $V$, show that we can write $v$ and $v'$ as linear combinations of this basis elements and then argue that since we have $v \otimes v' = \sum_{i \in I}r_i e_i \otimes \sum_{i\in I} s_i e_i = \sum_{i\in I} s_i e_i \otimes \sum_{i \in I}r_i e_i = v' \otimes v$ then if we look at a particular element of the sum on LHS and RHS, namely $r_ie_i \otimes s_je_j = s_je_j\otimes r_i e_i$ then we can send the $r_i$ to the other side or do some manipulation of that sort and obtain the desired result.   I know this argument is not rigorous at all, so I am having a hard time doing it more rigorous (if this is a good approach) or actually attack this problem efficiently. I was wondering if this would be the right approach, like working with basis elements and so on or if there is some better way to look at this problem. Perhaps some use of universal property for modules would be helpful? Any help or suggestion with this problem is highly appreciated! Thanks so much!
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?
108374
Consider the complex vector space $\mathbb{C}^3$ and the subset $$S = \{(1, 0, i), (i, 2, -3), (2-i, 1+i, i)\}$$  Let $\textsf V$ denote $\operatorname{span}(S)$. Find a subset of $S$ that forms a basis for $\textsf V$.  How do I find the subset for something? I can put the vectors into matrix form and make them equal to $x, y, z$ respectfully but from there, I don't know how to proceed.  Am I missing a definition and and a series of steps? I know that a basis in simple terms is :     The minimal amount of vectors that are linearly independent and spans    $\textsf V$.
84189
There is a conjecture: "The only subspace of a normed vector space $V$ that has a non-empty interior, is $V$ itself." (here, the topology is the obvious set of all open sets generated by the metric $||\cdot||$).  I have a proof for the case $V$ is finite dimensional. Because, let $V$ have dimension $n$ and a subspace $S$ of $V$ have dimension $m &lt; n$. Let $\{v_1,v_2,\ldots,v_m\}$ be a basis for $S$, extended to the basis $\{v_1,\ldots,v_m,\ldots,v_n\}$ of $V$. Now, suppose that $p = b_1v_1+\cdots+b_mv_m$ is an interior point of $S$. Now, consider the norm $N(a_1v_1 +\cdots+a_nv_n)=\max(|a_1|,\ldots,|a_n|)$. Then, there is an $r&gt;0$, such that $||x-p||&lt;r$ and $x$ is in $V$ $\implies$ $x$ is in $S$, since on a finite dim. space, all norms are equivalent. Let $m&lt;k\leq n$,and chose $v$ in $V$ as:  $$v=\left(b_1+\frac{r}{2}\right)v_1+\left(b_2+\frac{r}{2}\right)v_2+\cdots+\left(b_m+\frac{r}{m}\right)v_m+\frac{r}{2}v_k$$  Then, $N(v-p)\leq\frac{r}{2}&lt;r$, so $v$ is in $S$ and by the subspace property of $S$, $v_k$ is in $S$ too, a contradiction to $m&lt;n$.    I have primarily 2 questions:  (1) Is there a simpler method to proof the conjecture for the finite dimensional case?  (2) Is the conjecture true for the infinite dimensional case?  Sorry, if the question admits a very trivial answer. The motivation behind my question , is the fact that an open interval in $\mathbb{R}$ is not open in $\mathbb{R}\times\mathbb{R}$, etc.
29482
Suppose that we have a vector space $V$ and the subspaces $U=\{(x,x,y,y)\in F^4:x,y\in F\}$ and $W=\{(x,x,x,y)\in F^4:x,y\in F\}$. Then how comes $$U+W=\{(x,x,y,z)\in F^4:x,y,z\in F\}?$$
56351
I'm having a hard time counting. I need to count the number of elements for the multiplicative group of invertible $2\times 2$ matrices $GL_{2}\left( \mathbb{F}_{p}\right) $ with elements from the field $\mathbb{F}_{p}$ where $p$ is any prime number. David S. Dummit and Richar M. Foote's Abstract Algebra (3rd edition) has a solution on page 413 but that approach is by counting the possible number of basis of vector spaces but I wanted to see if there was a direct way. The book says that $\left\vert GL_{n}\left( \mathbb{F}_{p}\right) \right\vert =\left( p^{n}-1\right) \left( p^{n}-p\right) ...\left( p^{n}-p^{n-1}\right) $ so my case reduces to $p^{4}-p^{3}-p^{2}+p$. For any matrix $\left(  \begin{array}{cc} a &amp; b \\  c &amp; d% \end{array}% \right) $ with $a,b,c,d\in \mathbb{F}_{p}$, I understand that $p^{4}$ is the total number of 2x2 matrices and that we need to subtract the possible combinations for $ad=bc$ (non-invertible elements) but I can't account for the rest. In particular, that plus sign is absolutely baffling. Any hint would be appreciated. I'd then take this up for the general $n$x$n$ case but first I need to dispose the 2x2 case.  P.S. I tried looking for a pre-existing answer and a related answer is present  but it's by the same approach as the book and I'd like to have an approach without resorting to linear independence. This , too, was a case for $p=3$ but I couldn't see the pattern.
30893
Let be $ V $ vector space :   a) If we have an infinite chain of subspaces $ {U}_{1}\mathrm{\subseteq}{U}_{2}\mathrm{\subseteq} $  Show that $ {U}\mathrm{{=}}\mathop{\mathrm{\cup}}\limits_{{k}\mathrm{{=}}{1}}\limits^{\mathrm{\infty}}{U}_{K} $  is a subspace .   b) if  $ V $ is finite dimensional show that there is a some subspace ,call it  $ {U}_{P} $ ,such that $ {U}\mathrm{{=}}{U}_{p} $   and hence $ {U}_{P}\mathrm{{=}}{U}_{{P}\mathrm{{+}}{1}}\mathrm{{=}}{\mathrm{....}} $
48722
Definition. A subspace  of a vector space V is a subset H of V that has three properties  (a) The zero vector of V is in H.  (b) $\forall$ u, v $\in H$, we have  u + v $\in H$.   (c) $\forall$ u $\in H$, c scalar, we have c  u $\in H$.  For $n \geq$ 0, denote $\mathbb{P}_n$ to be the set of all polynomials of degree at most $n$.  Which of the following are subspaces of $\mathbb{P}_n$ for an appropriate value of $n$? Support your answer with reason.  (a) All polynomials of the form $p(t) = a + t^2$, where $a \in \mathbb{R}$.  (b) All polynomials of degree at most 3, with integers as coefficients.  Solution:  Both the statements (a) and (b) above are not subspaces of $\mathbb{P}_n$.  Reason:  For part (a), All polynomials of the form $p(t) = a + t^2$, where $a \in \mathbb{R}$, is not a subspace because it failed to meet the standards of the first property stated above, that is, the coefficient of the $t^2$ term is always a 1, therefore the zero vector is not in this subspace of polynomials.    For part (b), All polynomials of degree at most 3, with integers as coefficents is also not a subspace becuase it as well failed to meet the standards of the third property, that is, integer coefficents, when multiplied by a real scalar, will not necessarily be of integer type leading to integer coefficents. Thusly, the resulting polynomial will not be in the subset of polynomials $\mathbb{P}_n$. $\Box$  Can someone please look at this and see if I have the right idea and approach to this question. Thank You
43478
Given the following problem     Let $U$ and $V$ be finite dimensional vector spaces, such that $\dim(U) = \dim(V)$. Prove that $U$ and $V$ are isomorphic.   I was just wondering if somebody could critique and validate my proof.  Proof  Two vector spaces $U$ and $V$ are isomorphic if and only if there exists a bijection $\phi: U \to V$. That is, an invertible linear map between the two. Let $\dim(U) = \dim(V) = n$ and let $\left\{u_1, \dots, u_n\right\}$ and $\left\{v_1, \dots, v_n\right\}$ be bases for $U$ and $V$, respectively. It follows that any vector $u \in U$ can be written as  $$ u = c_1u_1 + \cdots + c_nu_n $$  We define $\phi: U \to V$ as  $$ \phi(c_1u_1 + \cdots + c_nu_n) = c_1v_1 + \cdots + c_nv_n $$  Observe that $\phi$ is invertible. That is, given a vector  $$ v = d_1v_1 + \cdots + d_nv_n $$  in $V$, we define $\phi^{-1}: V \to U$ as  $$ \phi^{-1}(d_1v_1 + \cdots + d_nv_n) = d_1u_1 + \cdots + d_nu_n $$  Since there exists a bijection between $U$ and $V$, they are isomorphic.
108421
Let $V$ be the set of all symmetric $3 \times 3$ matrices. (Recall that $V$ is a subspace of $M_{3\times 3}.$) Find a basis of $V$ , and show that it is a basis.  I'm not sure what the question means by "basis of $V$." Any help would be appreciated.
8520
The question is from Axler's Linear Algebra text. The $\mathcal{L}(V)$ stands for the space of linear operators on the vector space $V$.          Suppose that V is a complex vector space with dim $V=n$ and $T \in \mathcal{L}(V)$ is such that     $$\text{null} \ T^{n-2} \neq \text{null} \ T^{n-1}  $$     Prove that $T$ has at most two distinct eigenvalues.      I fist thought of solving this by contradiction. That is, I thought, suppose there were three distinct eigenvalues. Then, there would be an equation like  $$(x-\lambda_1I)^{d_1}(x-\lambda_2I)^{d_2}(x-\lambda_3I)^{d_3} $$ where the $d_i$'s are positive integers that sum to dim $V$. Call this polynomial $q(x)$ the characteristic poly. Thus, by Cayley's theorem, $$q(T)=(T-\lambda_1I)^{d_1}(T-\lambda_2I)^{d_2}(T-\lambda_3I)^{d_3}=0 $$ Then multiplying out and setting dim $V = d_1+d_2+d_3 = n$, I could then get a poly with the various powers of $T$ (this is a little tricky to write down). In particular, I wanted to see the powers $n-2$ and $n-1$. I thought, ok, so now, rewrite the poly in terms of each of these and using the fact that $\text{null} \ T^{n-2} \neq \text{null} \ T^{n-1}$ you get some vector $v \in V$ such that, $$T^{n-1}v = (\text{poly}_1)v = 0 $$  but  $$T^{n-2}v = (\text{poly}_2)v = k \neq 0 $$ I can think of some interesting things about $(\text{poly})_1$ and $(\text{poly})_2$, in particular, each has the monic term $T^n$. At this point, I'm not sure any of this helped.  Well, anyways, I'm sure someone has a much better approach! Thanks in advance to anyone who read this.
91321
I'm trying to prove the following statement:     In a vector space $V$ over a field $\mathbb{F}$, if $\{u,v,w\}$ is a basis for $V$, then $\{u+v, u+v+w, v+w\}$ is also a basis.   $$\underline{\text{My proof}}$$  Suppose that $\{u,v,w\}$ is a basis for $V$. We need to show that $\{u+v, u+v+w, v+w\}$ is linearly independent and spans $V$.  Now since $\{u,v,w\}$ is a basis, it is linearly independent, so the only solution to $\alpha u+\beta v+\gamma w=0$  is the trivial solution $\alpha=\beta=\gamma=0.$ In addition, $\{u,v,w\}$ spans $V$, so, for all $v_i \in V,$ we have $$v_i=au+bv+cw$$ for some $a,b,c \in \mathbb{F}.$  Proof that $\{u+v, u+v+w, v+w\}$ is linearly independent: Suppose that $\delta(u+v)+\varepsilon(u+v+w)+\zeta(v+w)=0.$ Then we have $$(\delta+\varepsilon)u+(\delta+\varepsilon+\zeta)v+(\varepsilon+\zeta)w=0.$$  But we've just proved that $\alpha u+\beta v+\gamma w=0 \implies \alpha=\beta=\gamma=0$, therefore, if we take $$\alpha=\delta+\varepsilon, \beta=\delta+\varepsilon+\zeta, \gamma=\varepsilon+\zeta,$$ then we have $$\begin{cases} \delta+\varepsilon=0 \\ \delta+\varepsilon+\zeta=0 \\ \varepsilon+\zeta=0\end{cases} \quad ,$$ the only solution to which is $\delta=\varepsilon=\zeta=0,$ hence $\{u+v, u+v+w, v+w\}$ is linearly independent.   Now we need to show that any $v_i$ can be written as some linear combination of $u+w, u+v+w, v+w.$  Let $a,b,c \in \mathbb{F}.$ If we can show that $$x_1(u+v)+x_2(u+v+w)+x_3(v+w)=au+bv+cw$$ has a solution (for $x_1, x_2, x_3 \in \mathbb{F}$, then it follows immediately that $\{u+v, u+v+w, v+w\}$ spans $V$.  Proof that $\{u+v, u+v+w, v+w\}$ spans $V$: Now, $$x_1(u+v)+x_2(u+v+w)+x_3(v+w)=au+bv+cw \\ \ \\\iff (x_1+x_2)u+(x_1+x_2+x_3)v+(x_2+x_3)w=au+bv+cw \\ \ \\ \iff \begin{cases} x_1+x_2=a \\ x_1+x_2+x_3=b \\ x_2+x_3=c\end{cases}$$  By substitution (a trivial exercise, whose steps I have left out), we get $$\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}=\begin{bmatrix} b-c \\ a-b+c \\ b-a\end{bmatrix} \quad ,$$ hence every $v_i \in V$ can be written in the form $x_1(u+v)+x_2(u+v+w)+ x_3(v+w)$, hence $\{u+v, u+v+w, v+w\}$ spans $V$.  Since $\{u+v, u+v+w, v+w\}$ is both linearly independent and spans $V$, it is a basis for $V$. $\square$ Note: you may be wondering: why are $b-c, \quad a-b+c, \quad b-a \in \mathbb{F}$? Answer: Since $\mathbb{F}$ is a field, it is closed under addition and subtraction, and $a,b,c$ were chosen to be in $\mathbb{F}.$  Question:     Is this proof correct? Can it be shortened?   Thanks!
36814
I need to prove that if $V$ is a finite dimensional vector space over a field K with a non-degenerate inner-product and $W\subset V$ is a subspace of V, then: $$ (W^\bot)^\bot=W $$ Here is my approach:  If $\langle\cdot,\cdot\rangle$ is the non-degenerate inner product of $V$ and $B={w_1, ... , w_n}$ is a base of $V$ where ${w_1, ... , w_r}$ is a base of $W$ then I showed that  $$ \langle u,v\rangle=[u]^T_BA[v]_B $$ for a symmetric, invertible matrix $A\in\mathbb{R}^{n\times n}$. Then $W^\bot$ is the solution space of $A_rx=0$ where $A_r\in\mathbb{R}^{r\times n}$ is the matrix of the first $r$ lines of $A$. Is all this true?  I tried to exploit this but wasn't able to do so. How to proceed further?