How to prove that the $n$ functions $$ e^{r_{i}t}, te^{r_{i}t},...,t^{k_{i}-1}e^{r_{i}t} $$ $i=1,2,...,s$, where $k_{1}+k_{2}+...k_{s}=n$ and $r_{1},r_{2},...,r_{s}$ are distinct numbers , and are linearly independent on any interval $I$. My try: Let $a_{11}e^{r_{1}t} + a_{12}te^{r_{1}t} +...+ a_{1k_{1}}t^{k_{1}-1}e^{r_{1}t}+a_{21}e^{r_{2}t}+a_{22}te^{r_{2}t}+...+a_{21}e^{r_{2}t} +a_{2k_{2}}t^{k_{2}-1}e^{r_{2}t}+...+a_{s1}e^{r_{s}t}+a_{s2}te^{r_{s}t}+...+a_{sk_{s}}t^{k_{s}-1}e^{r_{s}t} =0. $ Then I differentiate above equation $n-1$ times to get a system of the form $AX=0$, now I need to prove $det(A) \ne 0$, but I am not getting any patterns here. Please help me to figure out this problem.
124299
Let $\lambda_i \in \mathbb C$  $(i=1,2,\cdots,n)$ be $n$ different complex numbers and $p_i \in \mathbb C[t]$ $(i=1,2,\cdots,n)$ polynomials. If we have the relation $${e^{{\lambda _1}t}}{p_1}(t) + {e^{{\lambda _2}t}}{p_2}(t) +  \cdots  + {e^{{\lambda _n}t}}{p_n}(t) = 0$$ for all $t \geq 0$, can we conclude that all $p_i(t)=0$?
107774
I have tried this problem and I'm unsuccessful.  Let $T: M(n,n)\rightarrow M(n,n)$ be the mapping defined by $ T(A) = A^T$. Show that the only eigenvalues of $T$ are $\pm 1$.
98076
Let $T: V \to V$ with the minimal polynomial $z^n$. Prove that there's a vector $v$ such that $v, Tv, T^2v, ..., T^{n-1}v$ are linearly independent.   The way I did it orginally was not allowed. No Jordan Forms are allowed because we did not cover them this semester. I asked my professor for a hint and all he said was we need to use induction. I wish I had something to work with, but I'm so  lost. It's my last problem. Thanks!
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?
89081
Suppose $A$ and $B$ are $n\times n$ matrices over $\mathbb{R}$ such that for $n+1$ distinct $t \in \mathbb{R}$, the matrix $A+tB$ is nilpotent. Prove that $A$ and $B$ are nilpotent.     What I've tried so far:  Define $f(t)=(A+tB)^n$. Then $f(t)$ has polynomials of degree at most n as its entries. Each of these polynomials has n+1 distinct roots, and hence is the constant zero polynomial. This shows that for all $t \in \mathbb{R} \ f(t)=0$, which in particular gives $A^n= f(0) =0$ that shows A is nilpotent. Now if none of the $t_1,...t_{n+1}$ , that make A+tB nilpotent, is zero, we can use $g(t)=(tA+B)^n$ with roots $\frac{1}{t_i}$ similarly and show that B, too, is nilpotent. But I have no idea in case one of the $t_i$ is zero.  Thanks in advance.
7461
Let $A$ be an $n \times n$ matrix and $k$ be a scalar. Prove that $\det(kA)=k^n\det A$.   I really don't know where to start. Can someone give me a hint for this proof?
69198
How to prove a general matrix invertible given by as below?  How to prove that $A^TA+I$ is always invertible for $\forall A \in \mathbb{R}^{n\times n}$?
40757
If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$, thus $\det(A) =0$.    I proved that way, but looks like too simple so I think maybe there is a trick that I missed. Or I proved it right?   Modification: ($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$ ($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)    Now I think I complete the proof.
88366
Here $n$ belongs to natural numbers. Firstly, I proved the relation by putting $n = 1$ . Then, taking $$|\sin(mx)| \leq m|\sin x|$$ true, I had to prove $$|\sin(m + 1)x| \leq (m + 1)|\sin x|$$  Now, here I got stuck. How to prove it?? Please help.
90986
I've been given the following problem as homework:     Prove that there exists a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ such that $T(1,1) = (1,0,2)$ and $T(2,3) = (1,-1,4)$.   Since it just says prove that one exists, I'm guessing I'm not supposed to actually identify the transformation.    One thing I tried is showing that it holds under addition/multiplication in the sense of:  1) $T(x + y) = T(x) + T(y)$  2) $T(cx) = cT(x)$  3) $T((1,1) + (2,3)) = T(1,1) + T(2,3)$? But that's not necessarily true.  I don't know how I'd accomplish this given my limited knowledge.  What's the approach to solving a problem like this?
78290
Problem: Let $A\in M_{n\times n}(\mathbb R)\,$ be a matrix and suppose that a positive number $k\,$ exists such that $A^k = 0\,$ and $A^{k-1} \neq 0$. Suppose that $x=\left[ \begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix} \right]$ is a vector in $\mathbb{R^n}$ such that $A^{k-1} x \neq 0$. Prove that the $k\,$ vectors $\,x,Ax,\dots,A^{k-1}x\,$  are linearly independent.     My attempt: Suppose $x + Ax + \dots + A^{k-1}x = 0$. Multiply both sides with $A^{k-1}$. Then we have $A^{k-1}x + A^k (x + Ax + \dots + A^{k-2}x) = 0 \Leftrightarrow A^{k-1}x = 0 \Leftrightarrow x = 0$   which implies $x + Ax + \dots + A^{k-1}x\,$ is linear independent.    This problem looks quite easy but I want my proof to be checked. Is it correct?
11769
Let $A_n$ denote an $n \times n$ tridiagonal matrix.  $$A_n=\begin{pmatrix}2 &amp; -1 &amp; &amp; &amp; 0 \\ -1 &amp; 2 &amp; -1 &amp; &amp; \\ &amp; \ddots &amp; \ddots &amp; \ddots &amp; \\ &amp; &amp; -1 &amp; 2 &amp; -1 \\ 0 &amp; &amp; &amp; -1 &amp; 2 \end{pmatrix} \quad\text{for }n \ge 2$$  Set $D_n = \det(A_n)$  Prove that $D_n = 2D_{n-1} - D_{n-2}$ for $n \ge 4$.
102659
Given $A\in M_{n \times n}(\mathbb{F})$ and $p_{A}(x)=\det(xI-A)$ why saying that $\det(AI-A)=0$ is not valid?
94189
Suppose $A$ is a $n \times n$ constant matrix. How can I prove $\det(e^A) = e^{\displaystyle \sum_{\lambda_i\in\sigma(A)} \lambda_i}$,  where $\sigma(A)$ is the multiset of eigenvalues of $A$?  The following matlab code shows this is true in $n = 3$ case:  A = rand(3) detA = exp(sum(eig(A))) detmA = det(expm(A))
14924
I would like only a hint to the following exercise:  Let $V$ be a vector space over the field $K$, and $T$, $S$ linear functionals on V such that $Tv=0\Rightarrow Sv=0$. Prove that there exists $r\in K$ such that $S=rT$.  I know how to prove this when $V$ is finite dimensional. I show that if there is no such constant $r$ then $n-2=\operatorname{dim}\textrm{ }(\ker\textrm{ }T\textrm{ }\cap \ker\textrm{ }S)=\operatorname{dim}\ker T=n-1$, a contradiction. But this approach doesn't seem to help at all for the stated problem.
116166
How do you prove that: $$ \begin{pmatrix} 1 &amp; 1\\ 1 &amp; 0 \end{pmatrix}^n =  \begin{pmatrix} F_{n+1} &amp; F_n\\ F_{n} &amp; F_{n-1} \end{pmatrix}$$
47996
I have recently read in a script about statistical methods in a chapter about linear regression that:  Given an $n \times k$-matrix $A$, we have $$A^T A = \sum_{i=1}^{n} a_i^T a_i$$ where $a_i$ denotes the $i$-th row of $A$.  Unfortunately, the author doesn't give a proof of that and I can't figure out one myself. Maybe someone can help me. source:  on page 10