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If you want to show that the two field are equal, it is the same to show that they are included in each other. But by definition of $\Bbb Q(\sqrt2,\sqrt3)$, $\Bbb Q(\sqrt2,\sqrt3) \subset \Bbb Q(\sqrt2 + \sqrt3) \iff \sqrt 2 \in \Bbb Q(\sqrt2 + \sqrt3)$ and $\sqrt 3 \in \Bbb Q(\sqrt2 + \sqrt3)$ And by definition of $\Bbb Q(\sqrt2 + \sqrt3)$, $\Bbb ... 3 We must show that if$c_1 v + c_2 T(v) = 0$, then$c_1=c_2=0$. Now apply$T$to both sides of$c_1 v + c_2 T(v) = 0$to obtain$c_1T(v) + c_2 T^2(v) = T(0) = 0$. But$T^2(v) = 0$, so we get$c_1T(v) = 0 \implies c_1 = 0$since$T(v) \neq 0$. So$c_2T(v) = 0$, but$T(v) \neq 0 \implies c_2=0$as well. 3 For finite-dimensional vector spaces: By simply counting the dimension we see that the image of$B \to C$is isomorphic to the image of$B' \to C'. This enables us to reduce the question to short exact sequences. But short exact sequences split, and for these the claim is obviously true: $$\begin{array}{c} 0 & \rightarrow & A & \rightarrow ... 3 Hints: Well, just define a general line through \;P(x_0,y_0)\; :$$y-y_0=m(x-x_0)$$And now check when a point of this line is at a distance equal to the circle's radius from its center: if the circle is \;(x-h)^2+(y-k)^2=r^2\; , solve$$(x-h)^2+(m(x-x_0)+y_0-k)^2=r^2$$You get many options, but you actually want the ones yielding tangent lines to ... 3 I show below that the eigenvalues of A are exactly the numbers -2\cos\big(j\frac{2\pi}{2n+1}\big) for 1\leq j\leq n. What makes everything work is the identity 2\cos(\theta)\cos(k\theta)=\cos((k-1)\theta)+\cos((k+1)\theta). Unfortunately the eigenvectors are a little complicated to express directly, the best presentation I could find so far was by ... 2 Fun problem. Let the weights of the edges be w_1 \leq w_2 \leq \cdots \leq w_{n-1}. For convenience, define w_0=0 and w_1=1. Let e_1, e_2, ..., e_{n-1} be the edges, in the same order as the weights. Let D be the matrix in the problem. For 1 \leq k \leq n, let D_k be the matrix where (D_k)_{ij}=1 if D_{ij} \geq w_k and (D_k)_{ij}=0 ... 2 I'm assuming that T is an isomorphism of M(n,\mathbb C) as an algebra (I don't think that what you look for is true if you only require T to be a linear isomorphism). The matrices E'_{ij} also satisfy the "matrix-unit relations":$$ E'_{ij}E'_{kl}=T(E_{ij})T(E_{kl})=T(E_{ij}E_{kl})=\delta_{j,k}\,T(E_{il})=\delta_{j,k}\,E'_{il}. $$As ... 2 Yes, because rows two and four are the same on both sides, you can simplify your system to$$ \begin{pmatrix}1 & .1353 & 1\\.3678 & .3678 & 1 \\ .1353 & 1 &1 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2\\ w_3 \end{pmatrix}=\begin{pmatrix} 0\\1\\0 \end{pmatrix} $$If the 3\times 3 matrix has nonzero determinant, then you have a unique ... 2 Are you familiar with quaternions? Clifford algebra offers a good N-dimensional generalization to quaternions. Let me take your example. You have two vectors x and y:$$\begin{align*} x &= 2e_1 + 4e_2 + 5e_3 + 3e_4 + 6 e_5 \\ y &= 6 e_1 + 2 e_2 + 0 e_3 + 1e_4 + 7 e_5\end{align*}$$With quaternions, one would compute a cross product to find ... 1 I solved it like this:$$ \begin{eqnarray*} u^2-v\overline{v}+1 = 0 \wedge (u+\overline{u})v= 0 \Leftrightarrow\\ (v=0 \wedge u^2+1=0) \vee (u+\overline{u} = 0 \wedge u^2-v\overline{v}+1=0) \Leftrightarrow\\ (v=0 \wedge (u=i \vee u=-i)) \vee (\overline{u}=-u \wedge |u|^2+|v|^2=1) \Leftrightarrow \\ \overline{u} = (-u) \wedge (|u|^2+|v|^2=1) ... 1 If you have a valuex$between 0 and$2\pi$, this implies that the segment$[0,x]$represents$\frac{x}{2\pi}$of the whole interval$[0,2\pi]$(as a fraction). Therefore, you want the corresponding point$y\in [1,44100]$to cover the same fraction. This means that you want$y$to satisfy $$\frac{y-1}{44100-1}=\frac{x}{2\pi}.$$ The solution is ... 1 The characteristic polynomial is given by$|A - \lambda I| = 0$, yielding: $$\begin{vmatrix}4 - \lambda &-5 & 2 \\ 5 &-7 - \lambda & 3\\ 6 & -9 & 4 - \lambda \end{vmatrix} = 0$$ Using determinants, this reduces to the characteristic polynomial: $$\lambda^2 - \lambda^3 = 0 \rightarrow -\lambda^2(\lambda - 1) \rightarrow \lambda_1 = ... 1 I will give two proofs of the result, one using minimal polynomials and one without. Proof 1: Let A:V\rightarrow V be an idempotent operator on the n dimensional space V. Suppose that the rank of the operator is r. Then there exists r linearly independent vectors \{\mathbf{u}_1,\ \cdots,\ \mathbf{u}_r\} in the image of A. Suppose that each ... 1 We are given:$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 1 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 \end{pmatrix}$$We find that characteristic polynomial by solving |A - \lambda I| = 0, yielding:$$(\lambda -1)^2 (\lambda^2 + \lambda +1) = 0$$This yields a double and a complex conjugate pair of ... 1 JNF of A should be Q^{-1}AQ, where Q is the matrix having a Jordan basis as its columns. The basis which you found is not a Jordan basis, so it is not a disjoint union of Jordan chains. Different method of getting JNF: Theorem. Let \lambda be an eigenvalue of a matrix A and let J be the JNF of A. Then the number of Jordan blocks of J with ... 1 First divide the numerator by the denominator!$$\frac{3x^3-x+4}{x^3+2x^2+6x}=3+\frac{-6x^2-19x+4}{x^3+2x^2+6x}$$Then note that the denominator factors, at best, to:$$x^3 + 2x^2 + 6x = x(x^2 + 2x + 3)3+\frac{-6x^2-19x+4}{x^3+2x^2+6x} = 3 + \dfrac{A}{x} + \dfrac{Bx + C}{(x^2 + 2x + 3)}$$1 \dfrac{3x^3-x+4}{x^3 + 2x^2 + 6x} = 3+\dfrac{-6x^2-19x+4}{x^3+2x^2+6x}$$ \\ $$\dfrac{-6x^2-19x+4}{x^3+2x^2+6x} = \dfrac{-6x^2-19x+4}{x(x^2+2x+6)} = \dfrac{A}{x} + \dfrac{Bx+C}{x^2+2x+6} You can get the value of A directly by putting x = 0 in this polynomial : \dfrac{-6x^2-19x+4}{x^2+2x+6} . That is the value of A is ... 1$$\frac{(3x^3-x+4)}{x^3 + 2x^2 + 6x}$$First you have to notice that the highest power in the numerator is equal to the highest power in the denominator. So you have to do synthetic division to take out a factor from the numerator: 3x^3+0x^2-x+4 \overline{)x^3 + 2x^2 + 6x+0} You'll end up with: factor + \frac{(...)}{x(x^2 + 2x + 6)} where (...) is ... 1 As vector spaces over {\mathbb R}, {\mathbb R}(X) and {\mathbb R}(X^2) are both infinite dimensional. The index of a field extension corresponds to the quotient of dimensions (over the base field {\mathbb R}) only if the dimensions are finite. Or maybe your confusion lies elsewhere. Suppose you have a vector space V over some field k and an ... 1 General sollution to AX=0 is the kernel Ker(A). To say AX=0 only have zero solution (trivial kernel Ker(A)=\left\{ 0 \right\}) is equivalent to verify the row reduced echelon form of matrix A:$$rref(A)=I$$In your case:$$A = ... 1 What$T$does to the standard basis, gives you a formula for the matrix associated to$T$. The vectors that the basis map to give you the columns. For the basis$\{1,x,x^2\}$I get$T(1)=x^2$,$T(x)=1$,$T(x^2)=-2x$. As vectors these are$(0,0,1)$,$(1,0,0)$, and$(0,-2,0)$. These are the columns of the matrix. Now invert the matrix. The inverse matrix ... 1 I think between your question and your last paragraph you confuse what the index$a$of$V_a$stands for. The examples you show (i.e. the plane$x + y + z = 0$) are all$V_0$, in the corresponding$\Bbb R^n$. Because of scalar multiplication, you need$ax = a$for all$x \in F$if$V_a$is to be a vector space, and that is only statisfied for$a = 0\$. You ...