Can anybody please help me to solve?  Let $F$ be a field. We define $n$ linear functionals on $F^n$, for $n \geq 2 $, by: $f_k(x_1,\ldots,x_n) = \sum \limits_{j = 1}^n (k-j)x_j $. What is the dimension of the subspace annihilated by $f_1,f_2,\ldots,f_n$?  I have tried to prove that the $n$ functionals are linear independent.
122123
Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals on $F^n$ ($n \geq 2$) by $f_k(x_1, \dotsc, x_n) = \sum_{j=1}^n (k-j) x_j$, $1 \leq k \leq n$. What is the dimension of the subspace annihilated by $f_1, \dotsc, f_n$?   Approaches I've tried so far:   Construct a matrix $A$ whose $k$th row's entries are the coefficients of $f_k$, i.e., $A_{ij} = i - j$, and compute the rank of the matrix. Empirically, the resulting matrix has rank 2 for $n = 2$ to $n = 6$, but I don't see a convenient pattern to follow for a row reduction type proof for the general case. Observe that $f_k$, $k \geq 2$ annihilates $x = (x_1,\dotsc, x_n)$ iff $$\sum_{j=1}^{k-1} (k-j)x_j + \sum_{j = k+1}^n (j-k)x_j = 0$$ iff $$k\left(\sum_{j=1}^{k-1} x_j - \sum_{j=k+1}^n x_j\right) = \sum_{j=1}^{k-1} jx_j - \sum_{j=k+1}^n jx_j,$$ and go from there, but I do not see how to proceed.     Note: This is Exercise 10 in Section 3.5 ("Linear Functionals") in Linear Algebra by Hoffman and Kunze; eigenvalues and determinants have not yet been introduced and so I am looking for direction towards an elementary proof.
17159
Suppose $E$ is a normed vector space. Let $f$ be a continuous linear functional on $E$ and denote by $M$ the Kernel of $f$. Let $x\in E$. How to show that $$\operatorname{dist}(x,M)=\displaystyle\inf_{y\in M}\|y-x\|=\frac{|f(x)|}{\|f\|}\, ? $$
22490
I was just wondering whether anyone knows an answer to the following:  Suppose that ${\mathbb F}$ is a subfield of a field ${\mathbb G}$ and that $v_1,\ldots ,v_k$ are linearly independent vectors in ${\mathbb F}^n$ (over $\mathbb F$). Is it necessarily true that $v_1,\ldots ,v_k$ are also linearly independent when considered as vectors in ${\mathbb G}^n$ (over $\mathbb G$})?  Any help would be much appreciated. Thanks!
12185
Let $f : U \to V$ and $g : V \to W$ be linear transformations on the vector spaces $U$, $V$, and $W$. Supposedly,  $$ \dim(\ker(g \circ f)) = \dim(\ker(f)) + \dim(\ker(g) \cap \operatorname{im}(f)). $$  How might I go about proving that?    (Attempt:) The $\dim(\ker(g)\cap\operatorname{im}(f))$ term suggests to me that I should define a vector space $V' = \ker(g) + \operatorname{im}(f)$ to invoke the theorem that  $$ \dim(V') = \dim(\ker(g)) + \dim(\operatorname{im}(f)) - \dim(\ker(g) \cap \operatorname{im}(f)), $$  but I don't really see where to go from there.
80906
Can anyone give an explicit basis of the $k$-vector space $k(X) = \operatorname{Quot}(k[X])$ of rational functions over $k$?   The dimension is given by $$\dim_k k(X) = \max(|k|, |\mathbb N|).$$  If $k$ is infinite, this follows from $|k| \leqslant |k(X)| \leqslant |k[X] \times k[X]| = |k \times k| = |k|$ and the linear independence of $\lbrace \frac{1}{X - \alpha} \mid \alpha \in k\rbrace.$ If $k$ is finite, the result can be obtained similarly from $|k(X)|=|\mathbb N|$ and the linear independence of the monomials $X^n$, $n \geqslant 0$.
111704
Suppose $X$ is a possibly non-separable Banach space and let $X^*$ be its dual. Also, let $(f_n)_{n=1}^\infty$ be a sequence of [EDIT: linearly independent] norm-one functionals in $X^*$. Does there exist an element $x\in X$ such that $f_n(x) \in (0,\infty)$ for all $n$?
75041
I have a problem as follows: $W_1$ and $W_2$ are subspaces of a finite-dimensional vector space $V$. $W^0$ is the annihilator of $W$.  (a) Prove $(W_1 + W_2)^0 = W_1^0 \cap W_2^0$.  (b) Prove $(W_1 \cap W_2)^0 = W_1^0 + W_2^0$.  Thoughts so far:  By definition of Annihilator,  $W^0$ is the set of linear functionals that vanish on $W$. I feel that I also should use dual, but not sure how to put things together.
121665
Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$?
123661
Prove that the set    $$ W= \{f \in C([0, 1]):f(0)=0\}$$   is a closed linear subspace of $C([0,1]).$   Here $C([0, 1])$ is the space of continuous functions equipped with the uniform norm  $||f||_\infty = \sup_{x\in[0,1]}|f(x)|$.   I need some help in how to proceed  with this problem.   First, I want to prove that $W$ is, indeed, a subset of $C([0,1])$. For this, we have that for any two functions $f,g\in W$, $f+g\in W$, since $f(0) + g(0) = 0$. Then, also for any $\lambda \in \mathbb{R}$, we have that $\lambda f \in W$, since $\lambda f(0) = 0$. Finally, we also have $0 \in W$. So $W$ is a linear subspace of $C([0, 1])$.  My problem is how to prove that $W$ is closed. Based on the definition of a closed set, I wanted to show that a sequence of continuous functions $f_n$ converges uniformly to $f$ and that this $f$ is also in $W$. I started with something like this:  Let $f_n$ be a sequence of continuous functions such that $f_n(0) = 0$ for any $n\in \mathbb{N}$. We need to prove that $f_n \to f$, uniformly. i.e $$ \|f_n - f\|_\infty = \sup_{x\in[0,1]} |f_n(x) - f(x)|\to 0$$ as $n\to \infty$.   How can I proceed?
13799
I'll go ahead and give you the problem first, and then explain my trouble with it.  Which of the following subsets are subspaces of the vector space C(-$\infty$,$\infty$) defined as follows: Let V be the set of all real-valued continuous functions defnined on $\mathbb{R}^1$. If $f$ and $g$ are in $V$, we define $(f + g)(t) = f(t) + g(t)$. If $f$ is in $V$ and $c$ is a scalar, we define $c$ · $f$ by $(c · f)(t) = cf(t)$. Then $V$ is a vector space, which is donatedby C(-$\infty$,$\infty$).  (a) All nonnegative functions (b) All constant functions  There are more, but I have a feeling that after the (a) I'll be able to get the rest myself. Anyay, my problem is that the problem says it required calculus, and I don't understand how calculus even comes into it.  As far as I can tell, (b) would be a subspace (because it's are closed under operations of V), but (a) would not because V could make a negative value positive.  I may be completely wrong though; any suggestions?
108687
Let $V$ be a finite dimensional vectorspace over a field $\mathbb{ F}$. It's easy to show that if $U$ and $V$ are subspaces of $V$ then $U \cap V$ is a subspace. But what if there are an infinite number of them? I have seen sometimes in pure maths that you take an arbitrary finite collection of sets and show the statement you are trying to prove is true for them to deduce it is true for the whole infinite collection. Why is this technique allowed? Can I do it for this case? Why?
90554
Let $K$ be a field and $n \in \mathbb N$. Show the following: (i) Let $V \subset K^n$ be a subspace and $I_V$ the subset of $M_n(K)$ consisting of all the matrices whose rows belong to $V$. Prove that $I_V$ is a left ideal of $M_n(K)$. (ii) Show that every left ideal of $M_n(K)$ is of the form defined in (i).  I think I could show (i). I am having problems with (ii) For (i), take $M \in M_n(K)$ and $N \in I_V$. Let $P=MN$, I want to show that $P \in I_V$. Let $P_i=(P_{i1} ... P_{in})$ be the i-th row of $P$. Then, by definition of matrix multiplication, $$P_i=(\sum_{k=1}^n M_{ik}N_{k1} ... \sum_{k=1}^n M_{ik}N_{kn})$$ $$=\sum_{k=1}^n M_{ik} (N_{k1} ... N_{kn})$$ $$=M_{i1}(N_{11} ... N_{1n})+...+M_{in}(N_{n1} ... N_{nn})$$ This means that the $i-th$ row  of $P$ is a linear combination of the rows of $N$. From here it follows $P=MN \in I_V$. This proves $I_V$ is a left ideal of $M_n(K)$. I don't know how to show (ii), I would appreciate some help with that part.
108200
Let $V$ be a vector space. Given $U\le V$, define $U^\perp=\{f\in V^\ast\mid f(u)=0\,,\;\forall u\in U\}$ and given $W\le V^\ast$, define $W^\perp=\{u\in V\mid f(u)=0\;,\;\forall f\in W\}$. This defines a Galois connection between subspaces of $V$ and subspaces of $V^\ast$. If $V$ is finite dimensional, why is $\dim U+\dim U^\perp=\dim V$? Unfortunately, I am not very comfortable with linear algebra. Any hint will be appreciated.
63689
Let $U,W$ be subspaces of a vector space $V$. Show that   $$\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)$$   Hint: Show that the map given by $L:U×W\to V$ given by $L(u,w)=u-w$ is linear.   I can show that $L:U×W\to V$ given by $L(u,w)=u-w$ is a linear map. I also know that the dimension of $U×W$ is $\dim(U)+\dim(W)$. What do I do next? Any hints?
35477
Let  $f_1,\ldots,f_n$ be linearly independent linear functionals on a vector space $X$. Show that there exist $x_1,\ldots,x_n\in X,$ such that the $n\times n$ matrix $\big(f_i(x_j) \big)_{i,j=1}^n$ is non-singular.
123984
Let $E$ be a Banach space and $F$ a closed subspace. It is well known that the quotient space $E/F$ is also a Banach space with respect to the norm  $$ \left\Vert x+F\right\Vert_{E/F}=\inf\{\left\Vert y\right\Vert_E\mid y\in x+F\}. $$ Unfortunately in a set of lecture notes on (Lie) group representations (material for our study group) the author accidentally used here $\min$ instead of $\inf$. Probably a mostly harmless booboo, because at that point it was only needed to get a Banach space structure on the quotient, and we will probably be concentrating on Hilbert spaces anyway, where the problem does not arise.   Namely from Rudin's Functional Analysis I could not find a proof that the minimum should always be attained. Except in the case of a Hilbert space, where an application of parallelogram law (the sum of the squared norms of the two diagonals of a parallelogram equals that of the four sides) allows us to find a Cauchy sequence among a sequence of vectors $(y_n)\subset x+F$ such that  $$\lim_{n\to\infty}\left\Vert y_n\right\Vert_E=\left\Vert x+F\right\Vert_{E/F}.$$  But anyway, the suspicion was left that the infimum is there for a reason (other than conveniently allowing us to sweep this detail under the rug at that point of the development of theory), so in the interest of serving our study group I had to come up with a specific example, where the minimum is not achieved. It's been 25 years since I really had to exercise the Banach space gland in my brain, so it has shrunk to size of a raisin. Searching this site did help, because I found . There we have $E=C([0,1])$, the space of continuous real functions on $[0,1]$ equipped with the sup-norm. If we denote by $\Lambda$ the continuous functional $$ \Lambda: E\to\mathbb{R},f\mapsto\int_0^{1/2}f-\int_{1/2}^1f $$ and let $F=\ker\Lambda$, then the answer to the linked question proves that there is no minimum sup-norm function in the coset $\Lambda^{-1}(1)$.   So I have a (counter)example, and the main question has evolved to:     When can we use minimum in place of infimum in the definition of the quotient space norm?   My thinking:   It seems to me that the answer is affirmative, if $F$ has a complement, i.e. we can write $E=F\oplus F'$ as a direct sum of two closed subspaces such that the norm on $E$ is equivalent to the sum of the norms on $F$ and $F'$-components. But the first point also raises the suspicion that the question may be a bit ill-defined (and uninteresting) in the sense that the answer might depend on the choice of the norm $\left\Vert\cdot\right\Vert_E$ among the set of equivalent norms. However, if we, for example, perturb the sup-norm of $C([0,1])$ in the above example by multiplying the functions with a fixed positive definite function before taking the sup-norm, the argument seems to survive, so may be replacing the norm with an equivalent one is irrelevant?   So to satisfy my curiosity I also welcome "your favorite example" (one with a finite-dimensional $F$ would be nice to see), where we absolutely need the infimum here. Bits about any sufficient or necessary conditions for the minimum to be sufficient or (as a last resort :-) pointers to relevant literature are, of course, also appreciated.
56555
I am trying to prove that given a subspace $\mathbf{W}$ in $\mathbb{R^{n}}$, the subspace and its orthogonal complement 'cover' whole of $\mathbb{R^{n}}$ through '+' where we define $\mathbf{W}$+$\mathbf{W^{\perp}}$ as linear combinations of vectors both in the subspace and in its orthogonal complement. It seems intuitively right, and I can prove that the sum of their dimensions adds up to n, but I am not sure how to prove the question I am looking at. Thanks!