Let $E$ be a $\mathbb{K}$-vector space. I have seen that every subspace $F \subset E$ has an (algebraic) complement $F'$  ($F+F'=E$ and $F \cap F'=\{0\}$).   One proof (using that every vector space admits a basis) is found , and another using directly Zorn's lemma in .  What I would like to know is if it is possible to prove that statement without using the axiom of choice.   One idea that came to me was this: in the finite case, the complement of a subspace  $F \subset E$ is isomorphic to the quotient space $E/F$. Is this true in the general case? If there were a way to embed $E/F$ in $E$ in such a way that it is the complement of $F$, we could prove the statement.  On the other hand, a way to prove that this is not possible to achieve could be to deduce that every vector space admits a basis (without using AC) from the fact that every subspace admits a complement. But I can't see how this could be done.  Thanks in advance for any help
939
Let say we live in the category of vector spaces over $\mathbb{R}$ or $\mathbb{C}.$ Here are three sentences:   Axiom of choice Every vector space has a base. For every vector space $V$ and its subspace $E\subset V$ there is a subspace $F\subset V$ such that $V=E\oplus F.$    I know how to prove that (1)->(2)->(3). How about the inverese? Do (2)->(1) and (3)->(2) hold?  If this is not the case, then is there some weaker version of AC which imply (3)?
64894
I am reading  that finite-dimensional subspaces of normed vector spaces have closed direct sum complements. This is the proof:      Let $\{e_1, ..., e_n\}$ be a basis for $\mathcal M$. Every $x \in   \mathcal M$ has then a unique representation $$x = \alpha_1(x)e_1 +...+  \alpha_n(x)e_n.$$ Each $\alpha_i$ is a continuous linear functional on    $\mathcal M$ (a linear map from  finite dimensional space is always    continuous) which extends to a member of $\mathcal X^*$,  by the    Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let    $\mathcal N$ be the intersection of the null spaces of these exten­sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.   I follow it until the last sentence. Clearly $\mathcal N$ is closed, as it is the intersection of closed sets (the inverse of $0$ under any continuous function, and in particular any linear functional on a vector space, is closed). Also it is easy to see that $\mathcal M \cap \mathcal N= \{0\}$, since if some $m\in\mathcal M$ is in every null space, it must have the zero coefficient for every basis vector, and so it must be zero.  My question: Why do we have $\mathcal X =\mathcal M + \mathcal N$? This seems to only hold if the extension given by Hahn-Banach is zero off of $\mathcal M$, but I see no reason why this should be the case.
121038
Let $V$ be a vector space over a field $F$ (not necessarily algebraically closed), and $T:V\rightarrow W$ be a $F$-linear map. If $c(T)$ and $m(T)$ denote the characteristic and minimal polynomial respectively, then I want to show that $c(T)$ and $m(T)$ have the same set of irreducible factors.  I know that the claim is true if $F$ is algebraically closed, because then the two polynomials have the same roots. The typical reasoning one gives when $F$ is not algebraically closed is: Take an algebraic closure of $F$, say $K$, and look at $m(T)$ and $c(T)$ in $K$ to conclude the proof. But as far as I know, the proof that any root of $c(T)$ is a root of $m(T)$ requires the fact that $V$ is a $K$-vector space. But we only know that $V$ is an $F$-vector space. I can't understand how to proceed after this.
55173
There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that  \begin{align} \{e_K |~ K \subset G~ conjugacy~ class \} \end{align} is a basis for $Z(\mathbb{Z}[G])$. Serre even writes "one immediately checks that the $e_K$ form a basis", but even reading the other post  I don't see that...  My approach to this is:  Be $x \in Z(\mathbb{Z}[G])$, which is equivalent to: $x = gxg^{-1}$ for all $g \in G$. Writing  $x = \sum\limits_{h \in G} \lambda_h h$, we get $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$. I don't really know how to go on at this point. I mean, I know that when $a,b \in K$ (same conjugacy class) you get $\sum\limits_{s \in K} \mu_s s = \mu\sum\limits_{s \in K} s$, since \begin{align} a = hbh^{-1} \Rightarrow \mu_a = \mu_b ~\forall a,b\in K \end{align} So all elements in the same conjugacy class have the same scalar.  But from here I don't really know how I can generate the $x = \sum\limits_{h \in G} \lambda_h ghg^{-1}$ with the $e_K$ ...  I would be really thankful, if someone could give me quite a basic proof that even I understand...
20931
Prove that any two bases of a finite dimensional vector space must have the same number of elements.  By considering the following two bases  $$S_1 = \{ \alpha_1, \alpha_2 , \ldots, \alpha_n \},$$  $$S_2 = \{ \beta_1, \beta_2, \ldots, \beta_m \},$$  how do I show that $m = n$?  Hints to get started. Thanks very much.
45563
I encountered this question in a document I found on a  google search, it bugged me because my perception keeps telling me I'm wrong no matter what I do.     Let $U$, $W$ and $Z$ be subspaces of a vector space $V$ and the following conditions are satisfied:         $U + Z = W +Z$     $U \cap Z = W\cap Z$     $U \subset W$           Prove that $U=W$   So I tried proving by contradiction that $U$ is not equal to $W$,  by saying there exists some vector $w \in W \setminus U$ and  if $w \in W \cap Z$ then $w \in U \cap Z$ which means $w \in  U$. At this momment my mind is shouting  "This isn't it", and I think I should be using a different approach.
50219
Prove if nonempty $K \subset \mathbb{R}$ is compact, then both sup $K$ and inf $K$ exist and are elements of $K$.   I manage to prove the existence of sup $K$ by using the Axiom of Completeness. But I failed to prove that $\sup K \in K$. Can anyone guide me ?
25471
I know that the proof that every vector space has a basis uses the Axiom of Choice, or Zorn's Lemma. If we consider an axiom system without the Axiom of Choice, are there vector spaces that provably have no basis?
53896
I need help proving  this stunning result.  Let me state it this way.  Let  $\mathbb L$ be a field and $\mathbb K$ be a field with $\mathbb K \subset \mathbb L$  Let $A$ be the set of algebraic numbers ie $A=\{a \in \mathbb L, \exists P \in \mathbb K [X], P(a)=0 \}$  I managed to prove that $A$ is a ring with the following reasoning:  If $a,b \in A$ then $ \mathbb K[a]$ is a ring.  And $ \mathbb K \subset \mathbb K[a] \subset \mathbb K[a][b] $  Some finite-dimension argument(ask details if needed) proves that $\mathbb K[a][b]$ has finite dimension as a vector space over the field $\mathbb K$  Reminding that $\mathbb K[a+b] \subset \mathbb K[a][b]$ and $\mathbb K[ab] \subset \mathbb K[a][b]$ proves that they both have finite dimension, hence $a$ and $b$ algebraic.  How should I proceed with inverses ?
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.
92721
Question: Let $E$ be a normed space. Let $G$ be a closed subspace of $E$ and let $F$ be a finite dimensional subspace of $E$. Show that $F+G$ is a subspace of $E$ and is closed.  I'm having trouble in showing $F+G$ to be closed. I know that $F$ is itself closed and complete, as it is a finite dimensional subspace of a normed space, and that if $F$ were compact that $G+F$ would be closed. I also know that the closed unit ball of any finite dimensional normed space is compact.  I tried two methods. One was to take a convergent sequence $(x_n)_{n \in \mathbb N}$ in $G+F$. Then we can write $x_n = f_n + g_n$ where $f_n$ and $g_n$ are sequences in $F$ and $G$ respectively, and I attempted to find a way to force the individual components $f_n$ and $g_n$ inside the unit ball which would enable me to say that they had convergent subsequences. I couldn't see how to do this, however.   The other thought was to try and show that $F$ is compact, but I don't see a way to do this as I can't imagine it to be simply true without some other conditions on $F$.  Are one of these methods the right way to go? Or should I go another direction? I would appreciate any help I can get, although I would prefer not to be presented with a full proof so that I can do some work for myself. Thanks!
64351
Assuming the Axiom of Choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$, the vector space of real sequences?
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?
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
81032
My attempt at a solution:  Proof  Given a vector space $V$, let $S={v_1,...,v_n}$ be a basis for $V$. Consequently $\dim(V)=n$. Now given $W$, a subspace of $V$, it will be shown that $\dim(W)\le \dim V$   First, consider the case where $W=${0}. Then $dim(W)=0 \le\ n$,   Subsequently, consider the case for $W=V$. Then it follows that $dim(W)=dim(V)=n$.  Finally, consider the case where $W \subset V$ and without loss of generality assume $v_n \notin W$. Let $T={v_1,...,v_k}$ where $k\lt n$ be a basis for the subspace W. Now, since S is a basis for V, for every $v_i \in S$,  $c_1v_1+...+c_nv_n=0$ has only the trivial solution, thus none of the vectors in S can be expressed as linear combinations of the other vectors in S. It follows that the set T is not a basis for V as $v_n$ cannot be generated from T. The $dim(W)$ is given by the number of vectors in the basis. Therefore, $dim(W)=k \lt n$  Therefore, it has been shown that if $W \subseteq V$ then $dim(W) \le dim(V)$
120266
Suppose $V$ and $W$ are subspaces of a finite-dimensional vector space $U$.      (a) Show that $W \subset V$ if and only if $V^0 \subset W^0$.      (b) Show that $(V \cap W)^0 = V^0 + W^0$. (Hint: Show $\supset$ holds, then use lots of fomulas for dimension.)   I am trying to show this question. My understanding of annihilators is that for a vector space $V$ over $K$, with $S$ being a subset, the annihilator of $S$ is the subspace $S^0$ of linear functions $f$ in $V^*$ so that $f(s)=0$ for every $s$ in $S$. However, I don't know how to use this definition to prove (a) and (b). The hint for (b) suggests I should use formulas such as   $$\dim(V + W) = \dim V + \dim W − \dim(V \cap W)$$  but I don't know how to reach this stage.
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?