I think I need to be using induction in this case. i prove the subspace with dimension 0, then assume there is a subspace of k dimensions and then prove that k+1 holds under addition/scalar multiplication aswell.  any hints/advice?
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
104740
Suppose $v_1,...,v_m$ is linearly independent in $V$ and $w\in V$. Prove that   $$ \dim (\operatorname{span}(v_1+w,...,v_m+w)) \ge  m-1$$   It's an exercise in the book Linear Algebra Done Right.  I'm wondering if I can write $U_1 =\operatorname{span}(v_1,...,v_m)$ and $U_2=\operatorname{span}(w)$ then write $$ \dim(\operatorname{span}(v_1+w,...,v_m+w)) = \dim(U_1+U_2)$$  Would you please help me with this problem, I really want a rigorous proof, thanks.
79831
Let $T_1,T_2$ be two linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$. Let $\{x_1,\cdots , x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1x_i \neq 0$ for every $i=1,2,\cdots, n$ and that $x_i \bot \ker T_2$ for every $i=1,2,\cdots, n$. Which of the following is/are necessarily true?   $T_1$ is invertible. $T_2$ is invertible.  both $T_1,T_2$ are invertible.  neither $T_1$ nor $T_2$ is invertible.   My Approch:  Given dimension of $\mathbb{R}=n$ and  $x_i\perp Ker(T_2)$  So, $\text{Ker}T_2=0$  Row space and null space are perpendicular to each other.  From Rank Nullity Theorem:  let T : V → W be a linear map. Then the rank of T is the dimension of the image of T and the nullity of T is the dimension of the kernel of T, so we have  $\text{dim}(\text{Im}(T))+ \text{dim}(\text{Ker}(T))=\text{dim}(V)$  Applying above result we get,  $\text{dim}(\text{Im}(T))+ 0=n$  Therefore, $T_2$ is one-one.  I could not solve this problem completely. what to do next?
45928
W is a subspace of $\mathbb{R}^n$ iff   The zero vector ∈ W. X + Y ∈ W for any X, Y ∈ W. aX ∈ W for any X ∈ W and a ∈ R.   So, given W = { X : X = [x1...], x1 = 0, x2 = 0, ... xn = 0 } ∈ Rn   The zero vector ∈ W because each X in W is the zero vector by definition, X + Y ∈ W because [0...] + [0...] = [0...] aX ∈ W because a[0...] = [0...]   So if I understand this correctly, the zero vector is itself a subspace of $\mathbb{R}^n$. Is this correct?   In addition, can this be extended to say that for any W = { X : X = [$x_1$...] } ∈ $\mathbb{R}^n$, assuming W is a subspace of $\mathbb{R}^n$, $x_i$ (an arbitrary component of W) can only be a constant if it is 0? (I.e., $x_i$ can't be 1 or 2, but can be 0)
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!
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?
107671
One thing this suggests--at least to me--is that the x-y plane and $\mathbb{R}^2$ are not necessarily equivalent.  For example, I could define the following: $X = \left\{ \begin{bmatrix} x\\y\\z\end{bmatrix} x,y \in \mathbb{R} \land z = 0\right\}$.  Am I wrong to think, one, that this is a subset of $\mathbb{R}^3$?  As I write this it occurs to me that while scalar multiplication is closed under the above rules, addition doesn't pass the smell test for a subspace... so, OK, it's certainly not a subspace.  I would welcome any insight readers of this query can provide.
16826
I cannot understand why $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$.   My reasoning is as follows: Choose any elements $v_1$ and $v_2$ from $\mathbb{R}^2$, add them together you get an element of $\mathbb{R}^2$ and same for scalar multiplication and $0$ vector is an element of $\mathbb{R}^2$. So where did I go wrong?
32156
I want to prove the fact that  If $W\subseteq V$ is a vector subspace of $V$ than   $dim(W)=dim(V)=n \implies W=V$   Without using the fact that if $dim(V)=n$ any set of $n$ indipendent vectors in $V$ is necessarily a basis of $V$, neither the fact that a basis of $\mathscr{L}=(\vec{v_1},...,\vec{v_n})$ is necessarily the set $\big\{ \vec{v_1},...,\vec{v_n} \big\}$. (Since these two facts are consequences of the proof).  By hypotesis I have that $W\subseteq V$ which means that all the vectors in $W$ are in $V$. I must prove the opposite inclusion $V\subseteq W$: all the vectors in $V$ are a linear combination of $n$ vector of a basis, but I don't know how to prove that all the vectors of that basis are contained in $W$.  Can I say that they must belong to $W$ since $W$ has $n$ vectors in its basis too? (But I think in this way I'm assuming the facts listed above).  Can you help me to finish the proof?   Thanks for your help
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 $$U+W=\{(x,x,y,z)\in F^4:x,y,z\in F\}?$$
30473
$V = U_1\oplus U_2~\oplus~...~ \oplus~ U_n~(\dim V &lt; ∞)$ $\implies \dim V = \dim U_1 + \dim U_2 + ... + \dim U_n.$ [Using the result if $B_i$ is a basis of $U_i$ then $\cup_{i=1}^n B_i$ is a basis of $V$]  Then it suffices to show $U_i\cap U_j-\{0\}=\emptyset$ for $i\ne j.$ If not, let $v\in U_i\cap U_j-\{0\}.$ Then \begin{align*} v=&amp;0\,(\in U_1)+0\,(\in U_2)\,+\ldots+0\,(\in U_{i-1})+v\,(\in U_{i})+0\,(\in U_{i+1})+\ldots\\ &amp; +\,0\,(\in U_j)+\ldots+0\,(\in U_{n})\\ =&amp;0\,(\in U_1)+0\,(\in U_2)+\ldots+0\,(\in U_i)+\ldots+0\,(\in U_{j-1})+\,v(\in U_{j})\\ &amp; +\,0\,(\in U_{j+1})+\ldots+0\,(\in U_{n}). \end{align*} Hence $v$ fails to have a unique linear sum of elements of $U_i's.$ Hence etc ...  Am I right?
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!
86749
Let $K$ be a field. Consider the vector space $K^n$ over the field $K$.  Suppose $(a_1,a_2, ... ,a_n) \in K^n$. What is the dimension of the subspace generated by all the permutations of $(a_1,a_2,...,a_n)$?  I think there are 4 different cases    $a_1=a_2=...=a_n=0$ $a_1=a_2=a_3=...=a_n \ne 0$ $a_1+a_2+...+a_n=0,$ $ a_1 \ne a_2$ $a_1+a_2+...+a_n \ne 0$ , $a_1 \ne a_2$
55360
I was looking for a short snazzy proof on the following statement:     n+1 vectors in $\mathbb{R}^n$ cannot be linearly independent   A student of mine asked this today morning and I couldn't come up with a proof solely from the .  From a higher level perspective, I explained that if I put the vectors in a matrix then if the only null space entry is the zero vector, then the vectors are independent but since we have one extra column than row and that the row and column rank are equal, there is no way we can have $n+1$ as the rank of the matrix and hence from Rank-Nullity theorem, the dimension of the Nullspace is at least one which implies that there is a combination of the vectors where not all the scalar multiples in the definition are 0 but yet we get a zero as the linear combination. The student hasn't completely learnt the fundamental subspaces yet so I am not sure he grasped what I was saying.  Is there a cleaner proof?  EDIT: I am stunned how many beautiful answers I got with so much diversity.
78555
I was doing some linear algebra exercies. One I did is this: Prove that the union of two subspaces $U,W$ of $V$ is a subspace of $V$ if and only if one contains the other.   My proof is this: If one is a subset of the other $U \subset W$ then the union is $W$ which by assumption is a subspace. If neither is contained in the other there exists an example where the union is not a subspace: let $V$ be the xy-plane, $U$ the x-axis and $W$ the y-axis. Then the union is not closed with respect to addition.   My question is whether it is valid to prove something by example? How would you prove this in another way?
13084
Let $V$ denote the vector space of polynomials in one variable with coefficients in $\mathbb{R}$, and let $$T(f(x)) = xf(x).$$ Prove that if $W \subset V$ is a subspace such that $T(W)\subset W$, then $V/W$ is finite-dimensional.  First I let $f(x) \in W,$ such that $a_0+a_1x+a_2x^2+...+a_nx^n$, and $T(f(x)) = a_0x+a_1x^2+a_2x^3+...+a_nx^{n+1} \in W$. So I think $W$ is the set contains polynomials such that $a+ax+ax^2+ ... +ax^n +...ax^\infty$. But I am not sure about it, and how it is related to finite-dimension.   Please help me to solve this problem, thank you!!
63650
I was looking for alternate proofs of the theorem that "a vector space $V$ of dimension greater than $1$ over an infinite field $\mathbf{F}$ is not a union of fewer than $|\mathbf{F}|$ proper subspaces" and possible generalizations.  A simple measure-theoretic proof over $\mathbb{R}$ is as follows: By countable additivity the sum of the measures of any collection of subspaces is zero since the measure of each subspace is zero, which is a contradiction.  I would like to look at proofs over arbitrary infinite fields and would like to know if similar statements hold for say modules (finitely generated or otherwise) over infinite rings.