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You're right-- this sort of question is studied a lot. As you have defined things, you're looking at a semiring, instead of a ring because there are no additive inverses to the direct sum operation. Of course distributivity goes through, since $L \otimes (K \oplus J) \cong L \otimes K \oplus L \otimes J$ via $l \otimes(j \oplus k) \mapsto l \otimes j ... 3 HINT: If$V\subsetneqq U$, there must be some vector$u\in U\setminus V$. Show that $$\{u,\langle 1,0,1,0\rangle,\langle 0,1,0,1\rangle,\langle 1,1,0,0\rangle\}$$ must be linearly independent and hence that$\dim U\ge 4$. Then use the same argument to show that if$U\subsetneqq\Bbb R^4$,$\Bbb R^4$must have dimension at least$5$. 3 What about using the$\infty$-norm? That is $$\|A\|_\infty = \sup_{x: \|x\|_\infty=1} \|Ax\|_\infty.$$ Take a vector$x$. Then $$\|Px\|_\infty \le \max_{i}\left|\sum_j p_{ij} x_j\right| \le \max_{i}\sum_j p_{ij} (\max_k |x_k|) \le\|x\|_\infty.$$ Denote$z:=Px. Then $$\|P^T\Xi^2 z\|_\infty = \max_i \left|\sum_j p_{ji}\xi_j^2 z_j\right| \le\max_i ... 3 The sum of the entries on each row is always 1, so \left(1,\begin{bmatrix} 1\\1\\1\end{bmatrix}\right) is an eigenpair. A further useful observation is that, due to the first column, (it's easy to see that) \left(2,\begin{bmatrix} 0\\0\\1\end{bmatrix}\right) is an eigenpair. The trace yields the remaining eigenvalue and consequently the ... 3 Hint: Every 2\times 2 skew-symmetric matrix has the form \begin{pmatrix}0&t\\-t&0\end{pmatrix}, so all you need to do is find those t that have the desired property. Writing out the elements of A^2+I explicitly would lead you a long way. 3 solve the recurrence relations D_n = D_{n-1} - D_{n-2} with the initial condition D_1 = 1 \mbox{ and} D_2 = 0. try D_n = \lambda^n. the indicial equation is \lambda^2 - \lambda + 1 = 0 whose roots are \lambda = {1 \pm i\sqrt 3 \over 2}. sso D_n = k (\cos(n\pi/3 + \phi). requiring D_2 = 0 gives \phi = -\pi/6 and D_! = 1 shows k = ... 2 The zero-matrix is diagonal, so it is certainly diagonalizable. 2 Vector spaces with \oplus and \otimes do not form a semiring, since associativity etc. do not hold - the laws only hold up to isomorphism. These isomorphisms fit together in a certain way, and what we get is called a 2-semiring or 2-rig. Just like a semiring is a "fusion" of two monoids (one being commutative), a 2-semiring is a "fusion" of two ... 2 Ok, so suppose you want to solve \begin{bmatrix} 1 & 2 & 5 \\ 2 & 0 & 9 \\ 0 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 4 \\ 8 \\ 7 \\ \end{bmatrix}. As you know, we are trying to find the vector \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} that makes this equation work. Well, if you ... 2 A system$$Ax=b\tag{1}$$of equations in unknowns x_1, x_2, \ldots, x_n implicitly defines the subset$$S:=\{x\in{\mathbb R}^n\>|\>Ax=b\}\quad\subset{\mathbb R}^n\ .$$"Implicitly" means that for any given x\in{\mathbb R}^n it is easy to test whether it is an element of S or not (just compute Ax and check whether this is =b); but you ... 1 In differential geometry, there is the following "well known" Theorem 1: Let f:E\supset U \rightarrow F be a smooth submersion (just as in your question). Let b \in f(U). Then the set S:= f^{-1}(\{b\}) is in fact a smooth submanifold of E with \dim S = \dim E - \dim F. The proof of Theorem 1 considers representations of f in local coordinates ... 1 There are productive ways to think about the solutions of the linear system Ax = b geometrically. There is already a link above on this. However most applications of linear systems are not geometric. They come up everywhere we have any kind of mathematical modeling: physics, chemistry, biology, medicine, epidemiology, computer science, all types of ... 1 Given a matrix equation$$A {\bf x} = {\bf y},$$where A is m \times n, \bf x is n \times 1 (and so we write {\bf x} \in \mathbb{R}^n), and \bf y is m \times 1 ({\bf x} \in \mathbb{R}^m), there are several interpretations, including: If we think of the matrix A as the map \mathbb{R}^n \to \mathbb{R}^m defined by {\bf u} \mapsto A{\bf ... 1 Pick any point x_{0} on the hyperplane H=\{ x : \langle x, a \rangle = b \}. Then$$ H-x_{0} = \{ x-x_{0} : x \in H \} $$is a subspace because \langle (x-x_{0}),a \rangle = 0. So your hyperplane is a translation of a subspace in a particular vector direction. If you want to project y onto H, that's the same as projecting y-x_{0} ... 1 You have diagonalized the matrix. That is, the expression PDP^{-1} = A is the "diagonalization" of the matrix A. 1 Let A, B given with B=U^*AU with UU^*=U^*U=I. Then B^*=(U^*AU)^*=U^*A^*U and$$ BB^*=(U^*AU)(U^*A^*U)=U^*AA^*U, \quad B^*B =U^*A^*UU^*AU = U^*A^*AU . $$Let A be normal. Then by the above calculations$$ BB^*=U^*AA^*U=U^*A^*AU=B^*B. $$If B is normal then$$ AA^* = UBB^*U^*= UB^*BU^* = A^*A. 1 We have ST is invertible if and only if \det(ST)=\det(S)\det(T)\ne0 so \det(S)\ne0 and \det(T)\ne0 and then S and T are both invertible. 1 I think (I can't actually remember at the moment) that the dragging factor is certainly proportional to the speed. So, the last equation you wrote yields two ordinary differential equations: \begin{align*} &\frac{d^2}{dt^2}r_x(t) + \frac{c}{m}\frac{d}{dt}r_x(t) = 0\\ &\frac{d^2}{dt^2}r_y(t) + \frac{c}{m}\frac{d}{dt}r_y(t) + g = 0 \end{align*} To ... 1 Let \lambda_{\max}<1. Then \rho(A)=\lambda_{\max}=1-\tau, \tau\in(0,1) and \|A^k\|< (1-\tau/2)^k for k sufficiently large. Thus the Neumann series \sum_{k=0}^\infty A^k = (I-A)^{-1} $$converges. As A is non-negative, A^k is non-negative, and by the series representation (I-A)^{-1} is non-negative. If \lambda_{\max}=1 then I-A ... 1$$A^tJA=J\implies A^t\left(JAJ^{-1}\right)=JJ^{-1}=I$$Now watch closely at \;\left(JAJ^{-1}\right)^t\; . Added: Using associativity of matrix multiplication, you can also begin with$$A^tJA=J\implies J^{-1}(A^tJA)=(J^{-1}A^tJ)A=J^{-1}J=I\$ and etc.