Tag Info

New answers tagged

0

Another way: Take $\bf x$ and $\bf y$ such that $T{\bf x} = T{\bf y}$. Apply $S$. Then $ST{\bf x} = ST{\bf y}$, and we get ${\bf x} = {\bf y}$. In general, $\ker T \subset \ker ST $. So $T$ is injective, hence bijective (using finite dimension). Now, let ${\bf z} \in \mathcal{V}$. Exists ${\bf x}$ such that $ST{\bf x} = {\bf z}$. Then, ${\bf y} = T{\bf x} ...


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.


0

Well, $(a_3,b_3,c_3)=(1,1,2)$ so those three values are determined. As you said, for the remaining six unknowns $a_i,b_i,c_i$, $i=1,2$, you have three equations, so the system is underdetermined. It is also completely decoupled. \begin{align} a_1+2a_2&=1\\ b_1+2b_2&=3\\ c_1+2c_2&=3 \end{align} Thus, $a_1=1-2a_2$, $b_1=3-2b_2$, and ...


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 ...


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 ...


0

Suppose that $f_1 ,f_2 ,...,f_k$ are linear dependent and let $f_j $ be the polynomial whose degree is the largest then $f_j = c_1f_1+ ...c_{j-1}f_{j-1} + c_{j+1}f_{j+1}+.... +c_{k}f_{k}$ but the degree of right hand side is less thar left hand side and this is contradiction


0

A bounded linear functional is just a special case of bounded linear operator, one for which the codomain is the field of scalars. A linear mapping $f:E\to V$ between normed vector spaces is a bounded linear operator iff: $$ \exists M \gt 0 \text{ s.t. } \forall x \in E \;\; ||f(x)||_V \le M||x||_E $$ The property given by this Question is then the very ...


0

First some definitions before I come to you question: A line represented in Plücker coordinates $L = (m^T, d^T)^T$ can be constructed from two points $a, b \in R^3$ with $m = a \times b$ (we call it momentum) and $d = b - a$ (we call it direction). As Plücker lines are also a homogeneous representation, it is helpful to define a "normalized" version. We do ...


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.


0

Hint: If $A^2 + I_2 = 0$, then what are the eigenvalues of $A$? $A$ will be a satisfactory matrix if and only if it has the correct eigenvalues and is unitarily similar to a diagonal matrix.


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 = ...


0

I'm very curious why you're using the notation $\hat x$ and $p$, which is very reminiscient of the QM notation $\hat x$ (position operator -- in Dirac's notation its associated projection operator would be $|x\rangle\langle x|$) and $\hat p$ (the momentum operator with similar projection operator). But anyway, to your solution: Notice that $P$ is an ...


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 ...


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 ...


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

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 ...


0

First of all, the $c_i$ cannot be all equal to zero as a basis cannot contain the zero vector. After that, you can use the incomplete basis theorem: a linear independent family can be completed to a basis by picking vectors from an existing basis. So use the canonical basis $(e_1,\dots,e_n)$. You'll be able to pick by induction $n-1$ vectors from this ...


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 $y-x_{0}$ ...


0

Hint: use the multiplicative property of determinants, and the connection of the determinant with invertibility.


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.


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 $$ AA^* = UBB^*U^*= UB^*BU^* = A^*A. $$


0

Let $n>0$, let $v_1\dots v_n$ be non-zero generalized eigenvectors to distinct eigenvalues $\lambda_1\dots \lambda_n$. Then the vectors $v_1\dots v_n$ are linearly independent. First I prove this lemma: Lemma: If $\lambda_1\ne\lambda_2$ then $\ker (\lambda_1 I - A)^{k_1}\cap \ker(\lambda_2 I - A)^{k_2}=\{0\}$ for all $k_1\ge1$ and $k_2\ge1$. ...


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 $I-A$ ...


0

Hint: What is the probability that the determinant of that matrix is zero. Note that the determinant is a function of its entries and is thus a continuous random variable. (laplace expansion)


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 ...


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 ...


0

The underlying mapping is $$ f(X)=X, $$ the identity mapping on the vector space $V$ of all matrices. It is linear, hence its derivative at $X$ in direction $\delta X$ is $$ f'(X)\delta X=\delta X, $$ which is $$ f'(X) = f. $$ Note, that both $f$ and $f'(X)$ are linear mappings from $V$ to $V$. The mapping $f'$ is a mapping from $V$ to $L(V,V)$.


0

let us visualize the null space of a matrix in three dimensions. suppose $\vec n = (1,2,3)^T$ is in the null space of $A$. this is a line, can you visualize the plane of vectors $\vec x^T = (x,y,z)$ that is orthogonal to $\vec n.$ the components of $\vec x $ satisfy $0 = \vec x . \vec n = \vec x^T \vec n = x + 2y + 3z.$ the rows of the matrix are all made ...


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 $5$.


0

Consider $\Lambda_i$ inside balls of radius $\epsilon$ around an enumeration of rational vectors in $\mathbb{R}^{n-1}$. Show that for any ball, the probability is zero. By countable additivety, you'll be done.


0

The exact same idea as in the answer to your other question works. That is, now take a Hamel basis of $B(H)$ that extends a Hamel basis of $\mathbb RI$, and you can still get a $\mathbb Q$-linear map (so additive) such that $\mathbb RI\subsetneq \Phi(\mathbb RI)$.


0

What is the result of $P^{-1} \cdot 0$? What is the product of that with $P$?


2

The zero-matrix is diagonal, so it is certainly diagonalizable.


0

You're given all the information you need. If you look at your matrix A, you'll notice that it is in upper triangular form. Recall, that if your matrix A is in upper triangular form, then the eigenvalues of A are simply the diagonal entries. Thankfully, the eigenvectors corresponding to each $\lambda$ value have already been calculated and given to you by ...


1

You have diagonalized the matrix. That is, the expression $PDP^{-1} = A$ is the "diagonalization" of the matrix $A$.


0

You do not swap columns when doing row reduction. The process here is: $$\begin{bmatrix} 0 & 1 & 3 & -1 & 2 &1 \\ 0 & -2 & -6 & 2 & 0 & 2 \\ 0 & 0 & 0 & 1 & -1 & 7 \\ 0 & 2 & 6 & -2 & 2 & 0 \\ 0 & 3 & 9 & -2 & 2 & 7\end{bmatrix}_{\begin{array} \\R_2 \to ...


0

just find the eigen values of $A$ which in your case is $2,-3,1$ and the corresponding eigen vectors the columns of $P$ will be the three eigen vectors


0

[A note on notation: I added vector symbols, since I was nonplussed that you used $a$ for a vector and the adjacent letters $b, c$ for scalars. I trust it will be clear.] [An introductory note: is it clear to you that $\vec{a}$ is the normal vector to the plane, and what normal vectors mean? If not, you should brush up on that, and then come back to ...


1

Let $n$ be even; write $n=2k$. Then $v_n = v_{2k} = \frac{1}{\sqrt{2}} (e_{2k} + e_{2k-1})$ and $v_{n-1} = v_{2k-1} = \frac{1}{\sqrt{2}} (e_{2k-1} - e_{2k})$. Hence $\frac{\sqrt{2}}{2} (v_n + v_{n-1}) = e_{2k-1} = e_{n-1}$. So this allows you to represent any $e_n$ for odd $n$. Use a similar construction to represent $e_n$ for even $n$. Since you can ...


2

Something went wrong here. What they've tried to do -- judging from the results in the first two columns -- is $R_2\Rightarrow R_2+R_1/2$ and $R_3\Rightarrow R_3-R_1/2$, but when I do that, I get $$\begin{bmatrix}4 & -2 & -2\\0 & 4 & 2\\ 0 & 4 & 8 \end{bmatrix}$$ Perhaps the first row was in fact $(4, -2, 2)$. But that doesn't ...


2

Your function $f$ is a linear operator (check this!) between finite dimensional spaces. Assuming your topology on $V$ is induced by some norm, all such operators are continuous. In order to show this, you might like to introduce some norm $\|\cdot\|$ on $V$ then show that there exists $C\in\mathbb{R}$ such that, for any $v\in V$, $$|f(v)|\leq C\|v\|.$$ Since ...


0

The derivative is $$ \eqalign { \frac {\partial f} {\partial v} &= 2\,{\rm sym}\bigg(\frac {\partial f} {\partial A}\bigg)\,v \cr } $$ where ${\rm sym}(X) = {\frac {1} {2}}(X + X^T)$. To derive it, consider the differential $$ \eqalign { df &= \frac {\partial f} {\partial A} : dA \cr } $$ where $X:Y = {\rm tr}(X^TY)$ represents the Frobenius ...


0

The normal vector can be read off directly from the Cartesian equation of a plane, once the equation is understood, the derivation of which is given by the following theorem and the accompanying corollary. Theorem : Vector equation of a plane. Let $\Pi$ be a plane in $\mathbb{R}^{3}$. Let $P_{0}$ be an arbitrary point on $\Pi$. Let $\mathbf{n}$ be ...


0

Notice that you could define this equivalently as follows. Let $f$ be the following function: $$f(a,b)=(a+3b,a-b,2a-b,4b).$$ This forms a bijective linear map from $\mathbb{R^2}$ to $F$. Find a basis for $\mathbb{R}^2$. What is its image under $f$?


3

A vector $(x,y,z)$ is in the plane $P$ if we have $x-y-2z=0$, or in other words, if $$ (1,-1,-2)\cdot(x,y,z) =0 $$ So $P$ consists of exactly those vectors that are orthogonal to $(1,-1,-2)$, just from a rewriting of the equation.


1

If $n = \langle a,b,c\rangle$ is the normal vector to the plane $P(x,y,z)$. Take the vectors $r = \langle x,y,z\rangle$ and $r_0 = \langle x_0,y_0,z_0\rangle$ such that $r - r_0 \in P(x,y,z)$ . The normal vector is orthogonal to every vector in $P(x,y,z)$. So in particular, $$n\dot\ (r-r_0) = 0$$ $$\langle a,b,c\rangle \langle ...


2

An inner automorphism will preserve the spectrum of matrices. If you look at some diagonal matrix like $$\mathrm{Diag}(2,2,\dots,2,2^{-(n-1)})$$ then the transpose of its inverse has a different spectrum, at least for $n>2$. For the $n=2$ case I'm not immediately sure how to proceed. EDIT As @JyrkiLahtonen points out, the statement is false for $n=2$. ...


1

$\begin{matrix} W&= & \{(x_1,x_2,x_3,x_4)| x_1+2x_2+3x_3+4x_4=0,2x_1+2x_2+x_3+3x_4=0, x_i \in \mathbb{R}\}\\ &= &\{(x_1,x_2,x_3,x_4)| x_1=3x_3+x_4,x_2=-\frac{2}{5}x_3-\frac{2}{5}x_4, x_i \in \mathbb{R}\}\\ &= &\{(3x_3+x_4,-\frac{2}{5}x_3-\frac{2}{5}x_4,x_3,x_4)| x_3,x_4 \in \mathbb{R}\}\\ &= ...


0

First, one minor observation. I believe you meant $v_{2n}=\overline{v_{2n-1}}$ (instead of $\overline{v_{2n+1}}$) otherwise $v_1$ is not conjugated of any vector. Now, write $v_{2n-1}=a_{2n-1}-ia_{2_n}$ and $v_{2n}=a_{2n-1}+ia_{2_n}$, where $a_{2n-1},a_{2n}\in\mathbb{R}^n$. Notice tha $a_{2n-1}=\dfrac{v_{2n-1}+v_{2n}}{2}$ and ...


0

Expand the product $$\left(\sum_{i=1}^{100}x_i\right)\left(\sum_{j=1}^{100}y_j\right)$$ and you will get the meaning.



Top 50 recent answers are included