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5

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.



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