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A little calculus is helpful here. Suppose to the contrary they are linearly dependent. Then one of the functions, say $|x-i|$, is a linear combination of the others. But all the others are differentiable at $x=i$, so any linear combination of them is differentiable at $x=i$. However, $|x-i|$ is not differentiable at $x=i$.

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Your last sentence is completely correct. Now take a point on the line $y = 0$, say, $(x, 0)$. According to you, this transforms to $(ax + c \cdot 0, d \cdot 0) = (ax, 0)$, which is another point on that same line. So the line (as a set, not as individual points!) remains fixed. Does that help? If you want to have each point of the line $y = 0$ remain ...

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I will let you write the details, this is just a sketch. Even if we are not in an algebraically closed field, we can always conjugate those matrices to their canonical Jordan form (why?). Then try to understand what are the possible canonical Jordan form verifying this condition (easy). Then compute the centralizer of each matrix obtained (this is ...

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Your almost done. If $A\in GL_2(\mathbb F)$ has an eigenvector with eigenvalue $\lambda\in\mathbb F$, then either $A\sim\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ (which gives one conjugacy class per element of $\mathbb F$) or $A\sim\begin{pmatrix}\lambda&0\\0&\lambda'\end{pmatrix}$ (which gives one conjugacy class per unordered(!) ...

2

The textbook chooses to define the action of the dual space as multiplication of row and column vectors. In this approach, an element in $V$ is a column vector, i.e, a matrix of order $n\times 1$, whereas the elements of the dual space $V^{*}$ are row vectors, i.e., matrices of order $1\times n$. So the action of the dual space is given by matrix ...

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Got it. if $AA^T=I_n$ then $A$ has rank $n$ at least, and if $A^TA=I_k$ then $A^T$ has rank $k$ at least. Since $rank(A)=rank(A^T)$ that means $A$ has rank at least $k$ and at least $n$, so they must be equal.

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The general result is that if $\lvert K\rvert\ge n$, $V$ cannot be the union of $n$ proper subspaces (Avoidance lemma for vector spaces). We'll prove that if $V=\displaystyle\bigcup_{i=1}^n W_i$ and the $W_i$s are proper subspaces of $V$, then $\;\lvert K\rvert\le n-1$. We can suppose no subspace is contained in the union of the others. Pick an $u\in ... 2 This looks similar enough to the pair-based definition of the complex numbers that it looks promising to explore how far that analogy will take us. First, it is easy to see that multiplying any pair by$(a,0)$simply scales both of its elements by$a$. If we identify$(a,0)$with the usual real numbers (that is sort of handwavy but bear with me for a ... 2 Let$n$be the degree of the polynomial$p(x) = a_n x^n + \dots +a_1x+a_0$. Since$F$is infinite, there exist$n+1$distinct elements$b_0, b_1, \dots, b_n \in F$. Call$y_k=p(b_k) \in F$for$k=0,\dots, n$. Now, you have $$\left( \begin{matrix} 1 & b_0 & b_0^2 & \dots & b_0^n \\ 1 & b_1 & b_1^2 & \dots & b_1^n \\ \vdots ... 2 By definition a subset V\subset U must fulfill all of the 3 conditions below to be called a subspace of the vector space U: It must contain a zero vector. It must be closed under addition. It must be closed under scalar multiplication. You successfully pointed that the subset in question doesn't fulfill the first condition, hence it isn't a subspace, ... 1 You are right. It is not a subspace, for the reason you mentioned. As for being closed under scalar-multiplication: suppose you multiply a vector in your set by -1. What do you see? Your set \it{is} however closed under addition. Can you prove it? 1 If we simplify the second equation we get$$(a+b)x=\frac34a-\frac{c}4.\tag{A}$$If we simplify the third equation, we get$$(a+c)x=c.\tag{B}$$Now we can simply add these two equations together and use (2a+b+c)x=x on the LHS to get$$x=\frac{3a+3c}4.$$If we multiply this equation by x again, we get$$x^2=\frac34(a+c)x\overset{(B)}=\frac34c,\tag{C}$$... 1 Here is a great link to verify your work. If I were you, I would try demonstrating the W=W_1-W_2 \neq \emptyset by showing an element which is in it. Here is a hint: And further, step 2: Step 3: 1 For p=2, v=(0,1) and w=(1,0) . Then a.v +b.w =0 implies (a,b)=(0,0) for which to hold both a=0 and b=0 must hold. For p=3 , v=(0,0) and w=(1,2). Now , you know that for any vector space , any subset containing the 0 vector is linearly dependant . So the set ... 1 Transformation matrix:$$\left(\begin{matrix}1 & \cos(\theta) & 0 \\ 0 & \sin(\theta) & 0 \\ 0 &0 & 1\end{matrix}\right)$$from (x,y,z) to (q_1,q_2,q_3). Therefore in the opposite direction we have$$\left( \begin{array}{ccc} 1 & -\cot(\theta) & 0 \\ 0 & \csc(\theta) & 0 \\ 0 & 0 & 1 \\ \end{array} ... 1 It is equivalent to show that $$A'(I - B(B'B)^{-1}B')A \geq 0$$ Notice that$H = I - B(B'B)^{-1}B'$is idempotent and symmetric, hence for any$x \in \mathbb{R}^{n}$, $$x'A'(I - B(B'B)^{-1}B')Ax = (HAx)'(HAx) \geq 0.$$ Hence the result follows. 1 If$B$is square (as you've indicated), then the result is obvious: If$B'B$is invertible, then so is$B$. So, we note that $$B(B'B)^{-1}B' = BB^{-1}(B')^{-1}B' = I$$ So, this whole inequality is just $$A'A \leq A'A$$ which is true, but trivial. It seems that there's something missing from your question. 1 Let$\frac{1}{\texttt{1+discount}}=a$. The equation becomes$NVP=-CapEx+\sum_{i=0}^n\texttt{(Revenue-Costs)}\cdot a^i NVP=-CapEx+\texttt{(Revenue-Costs)}\cdot\sum_{i=0}^n a^i \sum_{i=0}^n a^i $is the partial sum of a geometric series.$\sum_{i=0}^n a^i =\frac{1-a^{n+1}}{1-a}$Therefore ... 1 Without loss of generality$CD = C$. By properties of the Kronecker product, the problem is equivalent to $$(B^T \otimes A) vec(X) + X = vec(C)$$ with solution $$vec(X) = \left( B^T \otimes A + I \right)^{-1} vec(C)$$ assuming the inverse exists. Here$vec(A)$is the vector obtained from the matrix$A$by stacking its columns. Let$A$and$B$have ... 1 The tangent plane is defined as: $$x_3 \equiv z \equiv t(x,y) = f(x_0,y_0) + \nabla f(x_0,y_0) \cdot (x-x_0, y-y_0)^T$$ Your derivation is correct, but you missed$ f(x_0,y_0) = 1$. So: $$x_3 \equiv z \equiv t(x,y) = 1 + 2 \cdot (y-1) = 2y - 1$$ and that's what Wolfram Alpha gives as well. 1 The derivatives are correct, but left hand side for equation of tangent plane should be$z-z_0=z-1$. 1 Comparing the coefficients of$i$and$j$on both sides, you can get a set of two equations: $$-A+B=4\\ A+B=-8$$ Is it possible now? 1 Yes, unless$V$is trivial. I'll help you unpack the definitions, but I'll leave the meat of the problem to you. Observation. Let$V$denote a$k$-module. Then$V$is a simple$\text{End}_k V$-module iff:$V$has at least two distinct$k$-submodules, Every non-trivial$k$-submodule of$V$that is closed under the action of$\text{End}_k V$... 1 Hint: The answer is yes (as is often the case for finite dimensional vector spaces). Note that for any$v \in V \setminus \{0\}$, we may select maps$T_1,\dots,T_n \in A$so that$\{T_j(v)\}$forms a basis (or a spanning set, if you prefer). 1 You have two linear equations in$c$and$d$. You know the right-hand side is$(1,0)$. 1 The elements of$V^*$operate on$V$in the obvious way:$(\varphi ,v)\mapsto \varphi (v)$. We ask how can we represent this using the bases for$V$and$V^*$. Given the basis$\mathcal B=\left \{ v_{1},\cdots ,v_{n} \right \}$for$V$, there is the natural basis$\mathcal B^*=\left \{ v_{1}^*,\cdots ,v_{n}^* \right \}$for$V^*$defined by ... 1 Let us fix a basis$\def\b{{\bf b}} \b_1,\dots,\b_n$for$V$. For coordinates of vectors of$V$we use column vectors, i.e.$\pmatrix{x_1\\ \vdots\\ x_n}$represents the vector$x_1\b_1+\dots+x_n\b_n$. Any element of the dual space (i.e. a linear$f:V\to K$) is uniquely determined by its action on the basis, and will simply correspond to the row vector ... 1 Let me reformulate the question. You are essentially asking if the isomorphism$\Lambda^2(V^*) \rightarrow (\Lambda^2 V)^*$where$\varphi^* \wedge \psi^* \mapsto [v \wedge w \mapsto \varphi(v)\psi(w) - \psi(v)\varphi(w)]$is "canonically" induced by the natural isomorphism $$V^* \otimes V^* \rightarrow (V \otimes V)^*$$ where$\varphi \otimes \psi \mapsto ...

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I'm assuming $\mathbb O_n$ means the $n \times n$ matrix $0$ (usually just written as $0$, sometimes $\bf 0$). Actually the binomial theorem does work when $A$ and $B$ commute, which is the case here. The proof is the same as for the case of numbers. Your attempted proof, however, is wrong. You can have $A B = B A = 0$ without $A = B = 0$, e.g. try \$ ...

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