I am trying to prove the following fact, which seems intuitive to me, yet I cannot find a way to show it.  Consider the set of functions $\{f_1,\ldots,f_n\}$, $f_i:\mathbb{R}\rightarrow\mathbb{R}$. Suppose that this set is linearly independent, i.e. $$\sum_i f_ic_i=0\implies c_i=0\;\;\; \forall i$$  Then there exists numbers $(x_i)_{i=1}^n$ such that the matrix  \begin{bmatrix}     f_1(x_{1})       &amp; f_2(x_{1})   &amp; \dots &amp; \dots &amp; f_n(x_{1})  \\     f_1(x_{2})       &amp; \dots  &amp; \dots  &amp; \dots &amp; \dots  \\      \dots  &amp; \dots  &amp; \dots &amp; \dots &amp; \dots \\     f_1(x_{n})        &amp; \dots  &amp; \dots  &amp; \dots &amp; f_n(x_{n}) \end{bmatrix} has a nonzero determinat.  I have tried to reason by contraposition, assuming that the determinant vanishes for any choice of the $(x_i)_{i=1}^n$ but I didn't get anywhere.
121638
The following proposition is easy to verify:  Given $n$ functions $f_1,...,f_n$, they are linearly independent in an interval $I$ if there exists a set of $n$ points in $I$, namely $x_1,...,x_n$, such that the matrix \begin{pmatrix} f_1(x_1) &amp; \cdots &amp; f_n(x_1) \\ \vdots &amp; \ddots &amp; \vdots \\ f_1(x_n) &amp; \cdots &amp; f_n(x_n) \end{pmatrix}  has full rank.  My question is: can I replace the if by if and only if? That is, if  $n$ functions are linearly independent, is it true that there exists a set of $n$ points such that the above matrix has full rank? I am pretty convinced so, but I am not being able to prove it.
43037
This is a question from Finite-Dimensional Linear Algebra by Mark S. Gockenbach page 57 (exercise 2.5.15). I hope to check my proof.  Let $V$ be a vector space over a field $F$, and suppose $S$ and $T$ are subspaces of $V$ satisfying $S \cap T = \{0\}$. Suppose $\{s_1,s_2, ..., s_k\} \subset S$ and $\{t_1,t_2, ..., t_l\} \subset T$ are both linearly independent sets. Prove that $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$.  My proof) Suppose that $\alpha_1 s_1 +...+ \alpha_k s_k + \alpha_{k+1}t_1 +...+ \alpha_{k+l}t_l = 0$ for some $\{ \alpha_i\}_{i=1}^{k+l} \subset F$ Then $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = -(\alpha_1 s_1 + ... + \alpha_k s_k) \in S$ Since $\{ t_1,...t_l \} \subset T$ and $T$ is a subspace of $V$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_1 \in T $. Since $S \cap T = \{0 \}$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = 0$. Since $\{t_1, ... , t_l \}$ is a linearly independent set, $\alpha_{k+1} = ... = \alpha_{k+l} = 0$. By doing similarly, we can show that $\alpha_1 = ... = \alpha_k = 0$. Therefore, $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$. $\blacksquare$
103477
Let $F(\mathbb{R},\mathbb{R})$ be the vector space of all functions from $\mathbb{R}$ to $\mathbb{R}$. Prove that for all $n\in\mathbb{N}$, the set of vectors $$\left\{sin(x), sin(2x),sin(2^2x),\dots,sin(2^nx)\right\}$$   form a linearly independent subset of $F(\mathbb{R},\mathbb{R})$.  I know that to form a linearly independent subset the set must be closed under addition, closed under scalar multiplication, and contain the neutral element.  Showing that it contains the neutral element is relativley straight forward since when $x=0$ every element also equals $0$. Therefore this subset contains the neutral element. However, I am having a hard time coming up with a way to approach the parts showing that it is closed under scalar multiplication and addition for all $x$. Do you have a piece of advice or a hint on how to approach those part?
87181
Could you help me to show that the functions $\sin(x),\sin(2x),...,\sin(mx)\in V$ are linearly independent, where $V$ is the space of real functions?  Thanks.
18975
For instance, let $v_{1}=(1,-2,3),v_{2}=(0,-1,2)$. The set $\{v_{1},v_{2}\}$ is linearly independent. Enlarging $\{v_{1},v_{2}\}$ to a basis for $\mathbb{R}^3$ I simply form a matrix using $v_{1},v_{2}$ as columns and then augment one of the elementary columns from $I_{3}$ and see whether the resulting echelon form has $3$ pivots, from then I conclude that it spans $\mathbb{R}^3$ and is linearly independent.   But the book I am using gives a solution as: $(a,b,c)$ with $9a-3b-5c\neq 0$. What method was use here and how might I arrive at such an answer? Thanks.
115755
Let $E$ be $\mathbb C$-vector space of dimension $3$. Let $f$ be non zero endomorphism of $E$ such that $f^2=0$. show that there exists a basis $B=\{b_1,b_2,b_3\} $ of $E$ where the matrix of $f$ is given by :  $$\begin{array}{ccc} 0 &amp; 1 &amp; 0 \\  0 &amp; 0 &amp; 0 \\  0 &amp; 0 &amp; 0 \end{array}$$  My try : since $f$ is non zero, there exists $x\in E$ such that $f(x)\not =0$. Take $b_1=f(x)$ and $b_2=x$. Now $b_1,b_2$ can easily shown to be linearly independent. By the exchange lemma we can find $b_3$ in another generating system of $E$ to make $b_1,b_2,b_3$ into a basis of $E$. Now clearly $f(b_1)=f^2(x)=0$ and $f(b_2)=f(x)=b_1$. Now how to find  a $b_3$ such that $f(b_3)=0$ and $b_1,b_2,b_3$ is a basis of $E$? Thank you for your help.   Edit: For the linear independence of $b_1,b_2$, if $\alpha x+\beta f(x)=0$ then we apply $f$ to get $\alpha f(x)=0$ and since $f(x)\not =0$ then $\alpha=0$ it remains that $\beta f(x)=0$ and again this gives that $\beta=0$.
61357
Let $f:\mathbb R^n\to\mathbb R$ differntiable and $f(0)=0$. Prove exist $g_i$ s.t for $x=(x_1,\dots,x_n):f(x)=\sum\limits_{i=1}^nx_i\cdot g_i(x)$. hint:$f(x)=\int\limits_0^1f\prime(tx)dt$.   I dont have any idea how to solve it. maybe we can build $g_i=f(x_i\cdot x)$ but im not sure it may help since I cannot take one time $\int f(x_1\cdot x)dx_1$ and in another $\int f(x_n\cdot x)dx_n$).  how can I build the sequence of functions and furthermore, what is the intuition for this type of questions involves building functions $\varphi:\mathbb R^n\to\mathbb R^m$ based on given condition with sums or diffrentiablity or partial derivatives?
87095
I have the following $N \times N$ matrix. \begin{vmatrix} 0 &amp; 1 &amp; 1 &amp; \ldots &amp; 1 \\  1 &amp; a_1 &amp; 0 &amp; \ldots &amp; 0 \\  1 &amp; 0 &amp; a_2 &amp; \ldots &amp; 0 \\  \vdots &amp; \vdots&amp; &amp;\ddots&amp; \vdots\\ 1 &amp; 0 &amp; 0 &amp; \ldots &amp; a_n \\  \end{vmatrix}  There seems to be a pattern going on for  of this matrix, but I'm not sure how I would find the determinant for the $N \times N$ one.
123064
I am working on a problem from Ideas, Varieties, and Algorithms:     Show that if $f \in \mathbb{C} \left[x_1, \dots, x_n \right]$ vanishes at every point of $\mathbb{Z}^n$, then $f$ is the zero polynomial.   My attempt so far:  This proof is by induction on $n$. Let $f \in \mathbb{C}\left[x\right]$ vanish at every point in $\mathbb{Z}$. Then, it must be identically the zero polynomial, as a polynomial of degree $d$ has at most $d$ roots. Hence, $f$ vanishes at every point at $\mathbb{C}$. Now, assume that $g \in \mathbb{C} \left[x_1, \dots, x_{n-1} \right]$ vanishes at every point in $\mathbb{Z}$. We may collect terms and rewrite $f$ as  $$ f(x_1, \dots, x_n) = \sum_{i=0}^m x^i_n g(x_1, \dots, x_{n-1}), $$ for some functions $g_1,\dots,g_n\in \Bbb C[x_1,\dots,x_{n-1}]$.  How do I proceed from here? I was trying to adapt an earlier proof with no success.
121842
Friends,  Do you know of a neat proof of the linear independence (over $\mathbb{C}$) of the functions $f(t) = e^{at}$ and $g(t)=e^{bt}$ when $a$ and $b$ are linearly independent over $\mathbb{Q}$?
1357
I know there's a theorem in linear algebra that has a list of statements that are equivalent, as follows:   $\det(A) \ne 0$ $Ax = 0$ has only the trivial solution $Ax = b$ is consistent for every matrix $b$ $Ax = b$ has exactly one solution for every $n \times 1$ matrix $b$   (and several other which I do not need for the question that I'd like to ask).  So we've been learning about linear independence, span and bases, and I was wondering the following:  Is statement 2 the same as saying a set of vectors are linearly independent?  Is statement 3 the same as saying the set of vectors span a vector space? (Meaning that every single vector $b$ in the vector space can be made from some combination of $Ax$, where $A$ would represent that vectors and $x$ would be all of the scalar multiples of the vectors?  Is statement 4 the same as saying the set of vectors is a basis? (since that one solution must be the trivial solution, it must be linearly independent, and since it is consistent for every $b$, it spans the vector space?)  It seems pretty obvious that it's all related to each other, but I'm just wondering if they're all exactly the same.
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.
79973
I'm preparing for an exam, and I have come over the following problem: I have the function $f$, where C is a set of complex numbers $$f: C \to C,$$  $$z=a+bi \mapsto f(z)=a+i(a+b)$$  Is the function of the form $f(z) = z + w$, where $w$ does not depend on $z$? Is it of the form $z\cdot w$ (where $w$ does not depend on $z$)?  For the first part, my answer is 'no', seeing as $w=ai$ for the identity to hold, and it is then clear that $w$ depends on $z$. My teacher's solution says the same thing, but that the answer to the question is 'yes'. Am i wrong, or is my teacher wrong?  For the second part of the exercise: I have written $w=c+di$, which gives $z·w=(ac-bd)+ (bc+ad)i$. This gives us two equations, namely: $$ac-bd=a$$ and $$bc+ad=a+b$$  I tried solving this system of linear equations for $c$ and $d$, and ended up with $c=d=1$. However this only holds if $b=0$. I am a bit stuck, and can't really see clearly how to proceed, or how to come to a general conclusion.  Please let me know if something was unclear, or if I should explain more clearly what I have done. Thank you in advance!  EDIT (included my teacher's answer, which I do not understand):  In his solution, my teacher ends up with the same system of linear equations as I did in my post. From there he says that the system has a unique solution if and only if the determinant of the matrix $$ \begin{matrix}a&amp;-b\\a&amp;b\end{matrix}$$ is non-zero. That is $a^2+b^2 \neq 0$.Then he says: But if $a^2+b^2=0 \Rightarrow a=b=0 \Rightarrow f(z) = 0 = 0 · w$ Hence the answer to the question must be 'yes'. If anyone could help me understand what is meant, I would greatly appreciate it. NB: There is a possibility that my teacher has read the task wrong, and therefore answers 'yes' when the answer to both questions is 'no'.
8559
Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible.  I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following $\det(A)=(-1)^{n+1}(n-1)$. To prove this I do induction  n=2 we have the $A=\begin{bmatrix} 0 &amp; 1\\  1 &amp; 0 \end{bmatrix}$ $\det(A)=-1$ and my formula gives me the same thing (-1)(2-1)=-1  Now assume if for $n \times n$ and $\det(A)=(-1)^{n+1}(n-1)$  Now to show for  a matrix B of size $n+1 \times n+1$. I am not sure I was thinking to take the determinant of the $n \times n$ minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.
24024
I want to prove that the matrix $$\begin{pmatrix} 1   &amp;\cfrac{1}{2} &amp;\cfrac{1}{3} &amp;\cdots &amp;\cfrac{1}{n} \\ \cfrac{1}{2} &amp;\cfrac{1}{3} &amp;\cfrac{1}{4} &amp;\cdots &amp;\cfrac{1}{n+1} \\ &amp;\vdots &amp;&amp;\ddots &amp;\vdots \\ \cfrac{1}{n} &amp;\cfrac{1}{n+1} &amp;\cfrac{1}{n+2} &amp;\cdots &amp;\cfrac{1}{2n-1} \end{pmatrix}$$ is positive definite. Using mathematical induction, I only need to show that its determinant is positive. But I can't find the way out.
94994
Let $M$ be a (real) $n \times n$ matrix. For $1 \leq i, j \leq n$ we denote by $M_{ij}$ the $(n-1) \times (n-1)$ matrix that we get when the $i$th row and $j$th column of $M$ are removed. Now, consider fixed $i$ and $j$ with $i\neq j$. Let $N$ be the $(n-2) \times (n-2)$ matrix that we get when removing both the $i$th and $j$th row and the $i$th and $j$th column from $M$. Then the following identity holds: $$ \det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}. $$  We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of $n!$ terms (for instance, terms containing $a_{ji}a_{ij}$ are counted both $1$ time in the LHS and $1$ time in the RHS).  We are looking for a slick proof that does not involve writing out the determinant.  Any suggestions or approaches to this problem are appreciated!
55527
If $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, and $(\beta_1,\beta_2,\beta_3) = (\alpha_1,\alpha_2,\alpha_3)C$. Prove that if $\mid C\mid \neq 0$, then $\beta_1,\beta_2,\beta_3$ are linearly independent.   This statement seems to be trivial, if $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, then $\det{(\alpha_1,\alpha_2,\alpha_3)} \neq0$, and $\det{(\alpha_1,\alpha_2,\alpha_3) C} = \det{(\alpha_1,\alpha_2,\alpha_3)}*\det{}C\neq 0$, which implies $\det(\beta_1,\beta_2,\beta_3) \neq0$, and hence the columns are linearly independent.  I'm not sure how to prove it rigorously, or if I'm missing any steps here. Would really appreciate any help! Thank you!