Let $V$ be a finite dimensional vector space.  I'm trying to see if any subspace $A$ of $V^*$ must be an annihilator of some subspace in $U$ of $V$.  I.e. I'm trying to see if $A=U^0$ for some subspace $U\subseteq V$.  So first I let $A$ be a subspace of $V^*$.  Then I denote the set of all $v\in V$ such that $\phi(v)=0$ for every $\phi\in A$ by $U$.  Lemma: $U$ is a subspace. Proof: Clearly $0\in U$.  Let $u,v \in U$ and $\lambda \in \Bbb F$.  Then for every $\phi \in A$ $$\phi(u+\lambda v) = \phi(u) + \lambda\phi(v) = 0 \implies u+\lambda v\in U$$  So $U$ is indeed a subspace of $V$. $\square$  Because $V$ is finite dimensional, so is $V^*$ and thus $A$.  Let $\{\phi_1, \dots, \phi_k\}$ be a basis for $A$.  Let $u\in U$.  Then $\phi_i(u)=0$ by definition.  Thus $\phi_i\in U^0$ for all $i=1,\dots k$.  So $A\subseteq U^0$.  So the only part that I still need is to show the reverse inclusion or that $\dim A = \dim U^0$.  But I can't figure out how to do it.    Here are my problems with the duplicate question (or more specifically its answers):   user218931 uses the double dual which is introduced in a later exercise. Rafael Deiga uses $\dim(\operatorname{null}\phi_1 \cap \cdots \cap \operatorname{null}\phi_m) = \dim V-m$ which is proven in a later exercise. Aweygan uses some functional analysis theorem called the Hahn-Banach theorem.   So all of the answers use tools that I don't currently have in my mathematical toolkit.  And yes, I realize that there's nothing theoretically wrong with using a result from later in the problem set as long as the proof of that result doesn't rely on this one, but I feel like Axler wouldn't have written out the exercises so that you needed to do that.  So to me it feels sorta like cheating to do them out of order.  Does anyone see a way to finish this problem without using the above three results?
121915
Exercise 26 page 115 of Linear Algebra Done Right by Sheldon Axler is the following:     Suppose $V$ is finite-dimensional and $\Gamma$ is a subspace of $V'$. Show that $$\Gamma=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}^0$$   where $V'$ is the dual space of $V$ and, for any $S\subset V$, $S^0$ is the annihilator of $S$.  Attempt:  Let $S=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}$. Clearly, $\Gamma\subset S^0$.I tried to show that $S^0\subset\Gamma$ using some bases of $V$ and $V'$, but I failed.  I also tried to show that $\dim{\Gamma}=\dim{V}-\dim{S}=\dim{S^°}$. If $s_1,\dots,s_n$ is a basis of $S$ and $s_1',\dots,s_n'$ its dual, and if $\psi_1,\dots,\psi_p$ is a basis of $\Gamma$, it's easy to see that $s_1',\dots,s_n',\psi_1,\dots,\psi_p$ is a linearly independent list of $V'$. Also, if you extend $s_1,\dots,s_n$ to a basis $s_1,\dots,s_n,v_1,\dots,v_m$ of $V$, then its dual $s_1',\dots,s_n',v_1',\dots,v_m'$ is a basis of $V'$; in fact $S^0=\text{span}\{v_1',\dots,v_m'\}$. Two remarkable facts:$$\forall i\in[1,m],\,\exists j\in[1,p],\,\psi_j(v_i)\neq0$$ because $v_j\notin S$, and $$\forall i\in[1,p],\,\exists j\in[1,m],\,\psi_i(v_j)\neq 0$$ because $\psi_i\neq 0$; in other words the matrix of the inclusion map from $\Gamma$ to $V'$ with respect to the basis of $\Gamma$ and the dual base of the chosen basis of $V$ has no $0$ row nor $0$ column. But this doesn't seem to provide a way to prove that any linear comination of the $(v_i')_{1\le i\le m}$ is a linear combination of the elements of the basis of $\Gamma$. I believe that $v_1,\dots, v_m$ should be chosen more carefuly but I fail to.  Could you please help me? Thank you in advance!
120065
It is not difficult to prove that the union of a chain (or, more generally, a directed family) of subspaces of a vector space $V$ is a subspace of $V$.   Given a family $\mathcal{F}$ of subspaces of a vector space $V$ such that the union of $\mathcal{F}$ is a subspace of $V$, is it true that $\mathcal{F}$ is a directed family?  If not, is there a "nice" characterization of families of subspaces whose union is a subspace?
61362
This problem is from Kunze Hoffman book. I think I go in the right direction to solve this but I miss some point to finish it. Can anyone help me?          Suppose $U$ is a self-adjoint unitary linear operator on a finite-dimensional inner product space $V$. Prove that we can find a subspace $W$ such that, if we express every $\alpha$ in $V\,$ in the form $\alpha = \beta + \gamma$, here $\beta$ in $W$ and $\gamma$ in $W^{\bot}$, then we have $U(\alpha) = \beta - \gamma$      If we denote $W = range(U+I)$, $W' = range(U - I)$,then must have $W \subset null(U-I)$ and $W \subset W^{'\bot}$. We can finish the proof if we can prove that $W' = null(U+I)$ or $W' = W^{\bot}$. But I can't prove this...  Thanks
111914
Suppose that $V$ is a vector space, and $W$ is a subspace of $V$. If   $V$ is finite dimensional, then prove $W$ too must be finite dimensional.   It seems intuitively obvious that the dimension of $W$ must be less than or equal to that of $V$, but I am uncertain of how specifically to show this to be true analytically.  This question has been asked before (). However, there is not a great degree of explanation in the given answers. The answer which labels itself as a hint is most similar to my thoughts towards an approach to proving this particular theorem.  I was thinking that if $\exists n\in\mathbb{R}:n=\text{dim}(V) \implies \exists B = \{\vec{v}_1,...,\vec{v}_n\}$, a basis for $V$, with $n$ vectors $\vec{v}_1, ..., \vec{v}_1 \in V$. Since $W$ is a subspace of $V \implies \beta$, a basis for $W$, which has $m$ vectors in it can be constructed by removing $k\in\mathbb{R}$ vectors from $B$ such that $B\setminus\{\vec{v}_i,...,\vec{v}_j\}=\beta$ where $\vec{v}_i,...,\vec{v}_j$ are those $k$ vectors. This means a basis for $W$ has $m=n-k$ vectors in it, and by the definition of bases $\implies \text{dim}(W)=m$, and since $n-k \in \mathbb{R}$ and $n-k=m \implies m\in\mathbb{R}$. Thus $W$ is finite. $\square$  It seems like there may be some fault with this argument, though.
2477
Here is the statement :  Let $E$ a finite $\mathbb{K}$-vectorial space and $\phi_1, \ldots, \phi_p, \psi$ a collection of linear forms on $E$ such that : $\bigcap \limits_{i=1}^p\ker(\phi_i)\subset \ker(\psi)$. Then $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_p\}$.  The suggestion is to make an induction on $p\ge 1$ (I already know that this method is not the fastest way) : For $p=1$ then $\ker(\phi_1)\subset \ker(\psi)$. If $\phi_1 \equiv 0$ then $\ker(\phi_1)=E$ so $\psi \equiv 0$ and the statement is true. If not, we have the fact that $\phi_1$ and $\psi$ are linearly dependent so $\psi \in \operatorname{span} \{\phi_1\}$ and the statement is also true. Now we want to check the statement for $p+1$. We suppose $\bigcap \limits_{i=1}^{p+1}\ker(\phi_i)\subset \ker(\psi)$ and we want to prove that  $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_{p+1}\}$. If all the linear forms are null then $\psi \equiv 0$ and the statement is true. If not, we have $\bigcap \limits_{i=1}^{p+1}\ker(\phi_i)\subset \bigcap \limits_{i=1}^p \ker(\phi_i) \subset \ker(\psi)$ so by the induction's hypothesis, $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_p\}$. Moreover (not sure) $\ker(\phi_{p+1})\subset \ker(\psi)$ so $\psi \in \operatorname{span} \{\phi_{p+1}\}$. Then I'm stuck... Thanks in advance
62659
I have a conjecture. I have a problem proving or disproving it.      Let $w \in \Lambda^k(V)$ be a $k$-vector. Then $W_w=\{v\in V: v\wedge w = 0 \}$ is a $k$-dimensional vector space if and only if $w$ is decomposable.   For example, for $u=e_1\wedge e_2 + e_3 \wedge e_4$ we have $W_u = 0$.
121672
I know that probably this question has already been answered, but I'd like to present my attempt of solution.  Let $(E,\|\cdot\|)$ be a normed vector space and let $F\subseteq E$ be a finite-dimensional subspace. Suppose that $\dim_{\mathbb{R}} F=n$.  First of all I proved the following lemma.  Lemma 1 Every finite-dimensional normed vector space is complete.  Then, I noticed that $(F,\|\cdot\|_{F})$ is a finite-dimensional normed vector space, where $\|\cdot\|_{F}$ is the induced norm. So, $F$ is complete by Lemma 1.   Lemma 2 If  $(E,\|\cdot\|)$ is a normed vector space such that $\dim_{\mathbb{R}}{E}=n$, then $E$ is algebraically and topologically isomorphic to $\mathbb{R}^{n}$.  By Lemma 2, I can view $F$ as a complete subspace of $\mathbb{R}^n$ and, since $\mathbb{R}^n$ is itself complete, I can conclude that $F$ is closed. Here I used the fact that every complete subspace of a complete space is closed.  Is there something wrong?
993
Given an arbitrary infinite dimensional Banach space $X$, can we deduce that it's dimension $\dim(X)$ (the cardinality of one of its Hamel bases) is less or equal of the dimension  $\dim(X^{\ast})$ of its dual space (the space of all continuous linear functionals $f:X\to\mathbb{R}$)?
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.
83448
I'm taking a course in Real Analysis and have come across the following question which I initially thought looked quite simple:  Let $X$ be an inner product space over $\mathbb{K}$ ($\mathbb{K} = \mathbb{C}$ or $\mathbb{R}$) Show that x and y are orthogonal if and only if $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$ for all $\mu, \lambda \in \mathbb{K}$.  I have proven the statement in one direction (starting with orthogonality and ending up with $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$). However, I seem to be getting stuck trying to go in the other direction. The following is what I have so far:  $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + 2\bar{\lambda}\mu Re&lt;x,y&gt; + ||\mu y||^2$  Then we must have 2$\bar{\lambda}\mu Re&lt;x,y&gt; = 0$ for the desired equality to hold, and so $Re&lt;x,y&gt; = 0$.   However, for orthogonality I want $&lt;x,y&gt; = 0$ and can't see how I can make this next step.  Any help would be greatly appreciated.
13996
Suppose $W$ is a subspace of the finite-dimensional vector space $V$ and $\dim W = \dim V=n$  Let $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ be a basis for $W$ and $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ be a basis for $V$.  The set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ still spans $W$. Since $W$ is a subspace of $V$, $W\subset V$ and all the $w_i$'s can be written in terms of $\{ \overrightarrow v_{1},\ldots,\overrightarrow v_n\}$.  $\therefore$ We can remove one by one all the $w_i$'s from the set $\{\overrightarrow w_{1},\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ while still having it span $W$, which leaves us with the set $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$. We know it to be linearly independent since it is a basis for $V$. Therefore the basis of $V$ is also a basis of $W$. From there on, it is trivial to prove double inclusion $W\subset V$ and $V\subset W$ which gives us $W=V$.  Is my approach correct?
111819
The other day, my teacher was talking infinite-dimensional vector spaces and complications that arise when trying to find a basis for those. He mentioned that it's been proven that some (or all, do not quite remember) infinite-dimensional vector spaces have a basis (the result uses an Axiom of Choice, if I remember correctly), that is, an infinite list of linearly independent vectors, such that any element in the space can be written as a finite linear combination of them. However, my teacher mentioned that actually finding one is really complicated, and I got a sense that it was basically impossible, which reminded me of Banach-Tarski paradox, where it's technically 'possible' to decompose the sphere in a given paradoxical way, but this cannot be actually exhibited. So my question is, is the basis situation analogous to that, or is it actually possible to explicitly find a basis for infinite-dimensional vector spaces?
72156
Question While looking over the exercise $3.F-34$ in Linear Algebra Done Right, I encountered the following paragraph  Suppose $V$ is finite dimensional. Then $V$ and $V'$ are isomorphic, but finding an isomorphism from $V$ onto $V'$, generally, requires choosing a basis of $V$. In contrast, the isomorphism from $V$ to $V''$ does not require a choice of basis and thus is considered more natural.  and these questions showed up in my mind: $1$. Does the word natural just means that we don't need to choose a basis? I have seen the word canonical is used in the similar manner too. Is there a more precise definition for natural or canonical? $2$. Assuming the answer to question $1$ is Yes, then why there is no natural isomorphism from $V$ onto $V'$? $3$. I think that there is a relation between the answer to question $2$ and the proof of Riesz representation theorem. So, if we cannot find a natural isomorphism between $V$ and $V'$ then it means that we cannot prove Riesz representation theorem without choosing a basis of $V$. Is this true?  Complementary Informations The Isomorphism from $V$ onto $V^{''}$. Suppose $V$ is a finite dimensional vector space. Consider the following map $$ \Lambda(v)(\phi)=\phi(v),  \qquad \forall v \in V, \,\, \forall \phi \in V^{'} $$ then $\Lambda$ is an isomorphism from $V$ onto $V''$. Riesz Representation Theorem. Suppose $V$ is a finite dimensional linear space equipped with an inner product and $\phi$ is a linear functional on $V$. Then there is a unique vector $v_0 \in V$ such that $$\phi(v) = {\langle v,v_0 \rangle}_{V}, \qquad \forall v \in V$$  Other Related Posts I found the following posts related to this question on MSE and MO. , , , .
72455
Let $S$ be a vector space. Suppose that for any positive integer $n$, there exists a linearly independent subset $S_n ⊆ V$ of size $n$.   How do you show that $S$ is not finite dimensional but infinite-dimensional.  I understand that to be finite dimensional then it must have a finite then  there exists a finite set that such at $\operatorname{Span}(T)=S$. So would it be right to prove by contradiction that there is not a finite spanning set that satisfies $\operatorname{Span}(T)=S$, and hence I could argue that the vector space $S$ is infinite dimensional.  How would I show that $S$ does not have a spanning set $T$, such that $\operatorname{Span}(T)=S$?
32901
Given a subspace $U \subset V$ of a vector space $V$, show that there exists a subspace $U' \subset V$, such that $U+U'=V$ and $U \cap U'=\{0_V\}$.  I am aware that $U'$ is the complement of $U$ in $V$.   Would it be correct to prove this over the basis extension theorem?  For every subspace $U$ there exists a basis of $U$, let it be $B_U$. It is true that $B_U \subset V$. According to the basis extension theorem one can extend any set of linearly independant vectors in $V$ to a basis of $V$. Thus one can extend $B_U$ to a basis of $V$. Let this basis be $B_V$. It is true, since the vectors in $B_V$ are linearly independant, that $\langle B_U \rangle \cap \langle B_V \setminus B_U \rangle = \{0_V\}$. It is also obviously true that $\langle B_U \rangle + \langle B_V \setminus B_U \rangle =V $. Let $\langle B_V \setminus B_U \rangle=U'$.   Is this correct or am I missing something?
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 $(X \otimes Y)^* \to X^* \otimes Y^*$?
122695
In the 4th book of Stein and Shakarchi, the statement of the Hahn-Banach theorem is as follows:     Suppose $V_0$ is a linear subspace of $V$ and $p$ is a real sub-linear function on $V$, and that we are given a linear functional $l_0$ on $V_0$ that satisfies $$l_0(v)\leq p(v) ~\forall ~v\in V_0$$   Then $l_0$ can be extended to a linear functional $l$ on $V$ that satisfies    $$l(v) \leq p(v)~ \forall ~ v \in V$$   The proof given is as follows:     We have already proved that if $l_0$ can be extended to a subspace W, then it can also be extended to the subspace $W+ w$ for any vector $w$. We well order all the vectors in $V$ that do not belong to $V_0$, and denote this ordering by $&lt;$. Among these vectors, we call a vector $v$ extendable if the linear functional $l_0$ has an extension of the kind desired to the subspace spanned by $V_0.v$ and all the vectors $&lt;v$. What we want to prove is in effect that all vectors not in $V_0$ are extendable. Assume the contrary, then because of the well-ordering we can find the smallest $v_1$ that is not extendable. Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $&lt;v_1$,then by assumption $l_0$ extends to $V_0'$. The previous step, with $V_0'$ in place of $V_0$ allows us then to extend $l_0$ to the subspace spanned by $V_0'$ and $v_1$, reaching a contradiction. This proves the theorem.   Firstly, how does proving that all vectors not in $V_0$ are extendable imply that an extension exists for the whole space? We know that for any $v \notin V_0$ there exists a valid extension, say $l_v$ for the space spanned by $V_0,v$ and all vectors $&lt;v$. Say we define $l(v):=l_v(v)$. Now we take an arbitrary $w&gt;v$. Couldn't it happen that $l_w(v) \ne l_v(v)$. In this case how can one define $l$? Will it help if we change the definition of "extendable" to something like: A vector $v \notin V_0$ is called extendable if $l_0$ has an extension of the kind desired to the relevant subspace which is equal to the already defined functionals $l_w~ \forall w&lt;v$ in the relevant subspaces? Also, how does one get the statement :"Now if $V_0'$ is the space spanned by $V_0$ and all the vectors $&lt;v_1$,then by assumption $l_0$ extends to $V_0'$."