So this here questions got me stuck for a bit:     Determine whether S is a subspace of V, if yes find a basis and its   dimension. Let $S$ be the set of all 3x3 symmetric matrices in $V =  M_{3x3}$   And here's why: I checked the appropriate axioms and it turns out S is in fact a subspace of V, but my main problem is that before this exercise, to find a basis, I'd find the augmented matrix and solve it (RREF) and then determine the solution, but here I'm not sure if I can.  So I have an approach but not sure which if it works:   If I say for e.g.   Assume the matrix $A=\begin{pmatrix}     a &amp;  b &amp; c\\     b &amp; d &amp; e\\      c &amp; e &amp; f\\ \end{pmatrix}$  if det(A) $\neq 0$ => linearly independent => the basis exists.  $k_1\begin{pmatrix}     a\\     b\\      c\\ \end{pmatrix}$ + $k_2\begin{pmatrix}     b\\     d\\      e\\ \end{pmatrix}$ + $k_3\begin{pmatrix}     c\\     e\\      f\\ \end{pmatrix} = 0;  dim(S) = 3$  Please let me know how to go about this. Thank you.
122053
Find a basis for the vector space of symmetric matrices with an order of $n \times n$  This is my thought: by definition of symmetry, $a_{i,j}=a_{j,i}$. Therefore, the basis should consist ${n^2-n} \over 2$ matrices to determine each symmetric pair. In addition, it should also consist $n$ matrices to determine each term in the diagonal.    Therefore, the dimension of the vector space is ${n^2+n} \over 2$.   It's not hard to write down the above mathematically (in case it's true).    Two questions:     Am I right? Is that the desired basis? Is there a more efficent alternative to reprsent the basis?     Thanks!
25590
I didn't want to bloat the title, i'll also add that the basis is in ${\mathbb{R}^n}$. This question seems kind of easy, but its from a test so I assume there is a catch here somewhere.   I tried to prove it by contradiction. Assume by contradiction that there exists no basis that satisfies the condition. But as known, every bilinear form can by identified by a matrix so that $A_{ij} = f(v_i,v_j)$. If there is no basis so $f(v_i,v_j) = -f(v_j,v_i)$ it means there is no skew-symmetric matrix which can identify a bilinear form over $\mathbb{R}$ which is obviously not true( I guess an example can be given here, I think ).  This proof just seems to simple to be true, but I can't seem to figure what could be wrong with it. I'll greatly appreciate any tips, thanks!
48940
We're given: $$ V = Span\left\{ \begin{bmatrix} 2 \\ 2 \\ 2 \\ 1 \end{bmatrix}, \ \begin{bmatrix} 2 \\ 1 \\ 1 \\ 0 \end{bmatrix}, \ \begin{bmatrix} 5 \\ 4 \\ 1 \\ 1 \end{bmatrix}  \right\} \ \ \   \mathrm{and} \ \ \ W = Span \left\{\begin{bmatrix}1\\-3\\2\\1 \end{bmatrix} \right\} $$  and we're asked to find $\mathrm{dim}(V\cap W^{\bot})$  Here's my approach. First, by inspecting the basis of $W$, I managed to construct a basis for $W^{\bot}$, which is the following: $$W^{\bot} = Span \left\{\begin{bmatrix} 3 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}\right\}$$ Then, for each vector $\textbf{w}$ in the basis of $W^{\bot}$, I tried to see if the system $A\textbf{x}=\textbf{w}$ was compatible or not. In this case, I found out that the system was only compatible with two of the vectors in $W^{\bot}$, thus indicating me that $\mathrm{dim}(V \cap W^{\bot} ) = 2$ (which is correct).  What I do not get, however, is  that since the dimension of the intersection is $2$, why aren't two of the vectors in the basis of $V$ orthogonal to the vector which spans $W$ (that was my initial approach, i.e try to see which vectors of the basis of $V$ are orthogonal to the vector that spans $W$).  Also, I was wondering if there was a simpler/quicker way of doing this.
33091
If $V_0$ is the subspace consisting of matrices of the form $C=AB-BA$ for some $A,B$ in a vector space $V$ then $V_0=\{A\in V|\operatorname{Trace}(A)=0\}$.  The problem above is one of the past qualifying exam problems. I can prove that  $V_0\subset\{A\in V|\operatorname{Trace}(A)=0\}$ but I do not know what to do with converse.  Thank you in advance.
28426
Let $V=\{f : S \to \mathbb{R} \}$ be the set of all the functions $f : S \to \mathbb{R}$. Let $S=\{ 1,2,...,n$}, now my train of thought was that the basis of V must be a linearly independent set that Span V so somehow show that there is a set of functions on S that can compose as a linear combination and map to any number in $\mathbb{R}$. Any idea how to get a general answer for basis for dimension of V?
69605
Describe a basis for the vector space of symmetric n x n matrices. What is the dimension of this space?
121057
So I am in an introductory linear algebra(with proofs) class, and I came across the following theorem:  If $S$ is a subspace of a vector space $V(F)$ and is generated by $\{a_1,...,a_n\}$ then any subset $\{b_1,...,b_m\}$ of $S$ with $m&gt;n$ is linearly dependent.   The proof in the textbook is done through induction (as most proofs regarding finite sets are) but I really do not follow it and I feel there is a simpler way of doing it. A hint would be appreciated.
38731
If $A$ and $B$ are two $4\times3$ matrices such that $A\mathbb{v}=B\mathbb{v}$ for all $\mathbb{v}\in\mathbb{R}^3$, then $A=B$?  If it's not true, can you give me an example of it?  Thank you.
21177
Using the Rank-Nullity Theorem, find the dimension and a basis for the subspace W of  $Q_4[x]$ consisting of all the polynomials $\displaystyle a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ such that $\displaystyle a_1 + a_2 + a_3 + a_4 = 0$.  Need some direction on how to start this question. Have some idea if a matrix was used but not a polynomial
36814
I need to prove that if $V$ is a finite dimensional vector space over a field K with a non-degenerate inner-product and $W\subset V$ is a subspace of V, then: $$ (W^\bot)^\bot=W $$ Here is my approach:  If $\langle\cdot,\cdot\rangle$ is the non-degenerate inner product of $V$ and $B={w_1, ... , w_n}$ is a base of $V$ where ${w_1, ... , w_r}$ is a base of $W$ then I showed that  $$ \langle u,v\rangle=[u]^T_BA[v]_B $$ for a symmetric, invertible matrix $A\in\mathbb{R}^{n\times n}$. Then $W^\bot$ is the solution space of $A_rx=0$ where $A_r\in\mathbb{R}^{r\times n}$ is the matrix of the first $r$ lines of $A$. Is all this true?  I tried to exploit this but wasn't able to do so. How to proceed further?
71100
Prove that if $W_1$ is any subspace of a finite-dimensional vector space $V$, then there exists a subspace $W_2$ of $V$ such that $V = W_1 \oplus W_2$   What I have done so far is to note that since $V$ is finite and $W_1$ is a subspace of $V$, we have $\dim(W_1) \leq \dim(V)$. If we have equality, then let $W_2 = \{0\}$, so that we have $V= W_1 \oplus W_2$. So now I need to look at the case when $\dim(W_1) \lt \dim(V)$.  What I have tried for this case is to let $$\beta = \{v_1,v_2,..,v_n\}$$ be a basis for $V$ and $$\gamma=\{u_1,u_2,..,u_m\}$$ a basis for $W_1$. My idea was to extend $\gamma$ to a basis for $V$, so let $$\alpha=\{u_1,u_2,..,u_m,w_1,w_2,..,w_{n-m}\}$$ be the extension of $\gamma$ to $V$, where $w_1,w_2,..,w_{n-m}$ are basis vectors for $W_2$. If I'm doing this right then I would just have to show that $W_1 \cap W_2 = \{0\}$ and $W_1 + W_2= V$   Am I heading in the right direction? Any hints would be greatly 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.
19400
The question is to find in the space $\mathrm{Mat}_{3\times 2}(\mathbb{\mathbb{R}})$ a non-zero matrix that is orthogonal to $$ A= \begin{pmatrix} 1 &amp; 2\\       3 &amp; 4\\       5 &amp; 6\\ \end{pmatrix}.$$  The inner product is $\langle A, B \rangle = tr(A B^T) = tr(B^T A)$.  Any hints or clues on how to start this?
100804
If U is a T-invariant subspace of dim n+1, and if there is no subspace W ⊆≠U such that T(w) ∈ W for all w ∈ W, then the set of vectors {u,T(u),T^2(u),...,T^n(u)}, u ∈ U, u ≠ 0  forms a basis for U.   To show the set of vectors forms a basis. I need to show         The set is a linearly independent set    The span of this set is U.       I am stuck at this point. How can I show the vectors in this set is linearly independent? Also, I am not sure how to show the span(set) = U. I know the dimension of this set is also n, which matches to the dimension of the subspace. Can I claim since all the T-invariant subspaces are in U, the set of vectors span U? Thank you so much!    Edit: added u is non-zero
54217
Consider a vector space $V$ , show that: "$dimV=1\Leftrightarrow V$  has exactly two subspaces". Well I can understand that since $dimV=1$ the only subspaces are the zero and it self, but how can i prove that? How can I prove the equivalence? Thanks in advance.
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?
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