I am trying to prove the set $W=\{(a_n) | \sum a_n^2 &lt;∞\}$ is subspace of vector space $V=\mathbb{R}^∞$ (vector space of all sequences of real numbers)  For this, clearly as $0^2+ 0^2+...=0 &lt;∞$ Hence (0) is in $W$.   Further, if $(a_n), (b_n) \in W$ then we have $\sum a_n^2&lt;∞$ and $\sum b_n^2&lt;∞$.   How can I prove $\sum (a_n+b_n)^2&lt;∞$
121718
So I know that $\sum x_n^2$ converges, and $\sum y_n^2$ converges. ${x_n}$ and ${y_n}$ are both sequences of real numbers. How can I prove that $$\sum(x_n+y_n)^2$$ converges? So far I have $$\sum(x_n+y_n)^2=\sum(x_n)^2 +\sum (y_n)^2 +\sum(2x_ny_n)$$ so I need to prove that $\sum 2x_ny_n$ converges. Is this the right way to approach it? If it is, I am completely lost as to the next step.  I also need to prove that $$\sqrt{\sum(x_n+y_n)^2}\leq \sqrt{\sum(x_n)^2}+\sqrt{\sum(y_n)^2}$$  I assume this will follow pretty easily from a result in the first proof, but I'm stuck on that now, and perhaps it is easier to start with the second one. I have a lot of work I've done on the first problem, but none of it seemed to go anywhere. Any hints?  Thanks!
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?
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.
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!
22048
Put   $$ l^{\infty} = \{ (x_n) \subseteq \mathbb{C} : \forall j \; \;\ \;|x_j| \leq C(x)\} $$  I want to show that $c_0$, the space of all sequences of scalars that converges to $0$ is closed subspace of $l^{\infty}$.  MY atttemt: Take arbitrary sequence $(x_n) \in c_0$. We know $x_n \to 0$. but $0 \in c_0$. Therefore, every sequence in $c_0$ converges to an element in $c_0$. Hence, $c_0$ must be closed  Is this correct? thanks
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!
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 $W_u = 0$.
102350
Let $V$ be a vector space over $\mathbb{C}$ such that $T^2=1_V$. Define  $V_1 = \left\{v\in V |\ T(v) = v \right\}$,  $\ V_2 = \left\{v\in V |\ T(v) = -v\right\}$  prove that $V = V_1 \oplus V_2$  For start, I've shown that $V_1\ ,V2$ are subspace of $V$. However, I'm stuck trying to show that $\forall \ v\in V,\ v = v_1 + v_2$ where $v_1 \in V1, v_2 \in V_2$.  I know at some point I have to make use of  $T^2=1_V$, but cant figure out how. Any help or insight is deeply appreciated.
15674
Let $V$ be a real n dimensional vector space &amp; let $T:V→V$ be a linear transformation satisfying $T^2(v)=-v$ for all $v\in V$. Then how can we show that $n$ is even?  I am completely stuck on it. Can anybody help me please?
87351
Given $V$ a inner product space and $W_1$, $W_2$ subspaces of $V$  Show $(W_1 + W_2 )^\perp = W_1^\perp \cap W_2^\perp$ and $W_1^\perp + W_2^\perp ⊆ (W_1 \cap W_2 )^\perp$.
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.
38632
Prove that if $V$ and $W$ are three-dimensional subspaces of $\Bbb R^{5}$, then $V$ and $W$ must have a non-zero vector in common.  So far I got $V = \{v_{1}, v_{2}, v_{3}\}$ and $W = \{ w_{1}, w_{2}, w_{3}\}$ because they are 3-dimensional subspaces. Then $V \cup W = \{v_{1}, v_{2}, v_{3}, w_{1}, w_{2}, w_{3}\}$  My prof said that now I need to prove that they are linearly independent so this is a contradiction and the statement is wrong because we are in $\Bbb R^{5}$ and we have 6 elements on it.  I don't know how to express them as a linearly independent proof, so I hope anyone can give me a good explanation on it.
56002
Let us assume that every vector in $S_2$ is a linear combination of vectors in $S_1$.  Question: Does that mean that $S_1$ and $S_2$ are bases for the same subspace of $V$?  I know that the answer to this question is yes, subspaces spanned by both $S_1$ and $S_2$ are the same. Let's call them $W_1$ and $W_2$ respectively. How do we prove $W_1=W_2$? Equal subspaces when regarded as sets, must have the same elemnts. How can we show that EVERY vector in one subspace is also in the other subspace?
84836
Could any one just tell me a sketch of the proof of the followings?  $1$. If $n&gt;1$, does every linear transformation on $\mathbb{R}^n$ have an invariant subspaces of dimension $2$?  $2$ Is there a vector space $V$ and a linear transformation $T$ on $V$ such that it has exactly $3$ invariant subspace?  Thank you!
101311
Let $V$ be the vector space made by the $n \times n$ square matrices.  a) Prove that $S=\{A\in V|A^t=A\}$ is a subspace of $V$  b) Prove that $T=\{A\in V|A^t=-A\}$ is a subspace of $V$  c) Prove that $V=S\oplus T$    SOLUTION  a) Let $O$ be the zero matrix, then $O \in S$ because $O^t=O$.  Then, let $A, B \in S$; and $r, k \in \mathbb R$. Hence  $$rA+kB=r(A^t)+k(B^t)=rA^t+kB^t$$ then $rA+kB \in S \implies$ $S$ is a subspace of $V$  b) Let $O$ be the zero matrix, then $O \in T$ because $O^t=O$.  Then, let $A, B \in T$; and $r, k \in \mathbb R$. Hence  $$rA+kB=r(-A^t)+k(-B^t)=-rA^t-kB^t=-(rA^t+kB^t)$$ then $rA+kB \in T \implies$ $T$ is a subspace of $V$  c) I have no idea how to do it.  I HOPE YOU CAN HELP ME VERYFING THE a), b) parts and with the part c).
59318
$W$ and $U$ are subspaces of vector space $V$. Show that $U\cup W$ is not necessarily a subspace of $V$.   I know that one counter-example is enough to show that. What example could fit here?
87564
As an exercise in my textbook, I need to prove that if $V$ is a finite dimensional vector space with dual space $V^*$ over $\mathbb{R}$, then dim$(V)$=dim$(V^*)$.  Let $\omega\in V^*$ and let $\{e_1,...,e_n\}$ be a basis for $V$. Define $e^i\in V^*$ by $e^i(e_j)=\delta_{ij}$. We show that $\{e^1,...,e^n\}$ spans $V^*$. $\omega(v)=\omega(v_1e_1+...+v_ne_n)=v_1\omega(e_1)+...+v_n\omega(e_n).$ If $\omega(e_1)=\lambda_1,...,\omega(e_n)=\lambda_n$, then $\omega(v)=v_1\lambda_1e^1(e_1)+...+v_n\lambda_ne^n(e_n)$=$\lambda_1e^1(v)+...+\lambda_ne^n(v)$.   To show $\{e^1,...,e^n\}$ is linearly independent, suppose that $0=c_1e^1+...+c_ne^n$ is the zero mapping to $\mathbb{R}$. Consider the image of $e_1:$  $0(e_1)=c_1*1+...+c_n*0=c_1$ Hence, $c_1=0$. Repeating the procedure for $e_j$, $2\leq j\leq n$, we see that $c_1=c_2=...=c_n=0$.  Does this proof look correct?  If the proof is correct, I have an additional small question. It seems like this proof is dependent on the fact that $V$ is a real (or complex) vector space, since we define the covectors to have the image set $\{0,1\}$. Is there a proof that works for vectors spaces over general fields?