5

Hints: first verify that $\omega^{3}=1$. Then verify that $A^{3}=I$. (It is easy to multiply diagonal matrices). This proves that the dimension is at most $3$. Now verify that $I,A,A^{2}$ are linearly independent.


4

Let's just focus on the first system, and use $e_j'$ to denote the dual basis elements (not standard, but it's used in the problem). We want to find the coordinates of $e_1'=(x_1\ y_1 \ z_2)$ so that $e_1'(e_1)=1,\ e_1'(e_2)=0,$ and $e_1'(e_1)=0$. You can either see this through using the standard basis and its dual basis or doing $e_1'e_j$ as row-column ...


4

What you did is fine. And there is really not much left. The map$$\begin{array}{ccc}\mathbb R(i)&\longrightarrow&\mathbb R^2\\a+bi&\mapsto&(a,b)\end{array}$$is an isomorphism precisely because $a+bi=0\iff a=0\wedge b=0$.


4

HINT If the columns of $A$ are $A_1, A_2, A_3$ your constraint says $$ 1A_1 +2A_2 + 3A_3 = 0 \iff A_1 = -2A_2 - 3A_3, $$ which means that your matrix looks like $$ A = \left[ -2A_2-3A_3,A_2,A_3 \right], $$ so how many free parameters do you really have?


3

What you did is to make a big assumption: every element of $\mathbb{R}(i)$ can be written as $a+bi$. This is part of the question -- that $\{1,i\}$ actually spans $\mathbb{R}(i)$ as an $\mathbb{R}$-vector space. Compared with a similar situtation $\mathbb{Q}(\sqrt[3]2)$. Here $\{1,\sqrt[3]2\}$ is not a basis of $\mathbb{Q}(\sqrt[3]2)$ as a $\mathbb{Q}$-...


3

The theorem you reference assumes the field "k" is either R or C equipped with the standard topology. It doesn't say anything about a vector space over R with R given the discrete topology.


3

Gaussian elimination to RREF is one particular way to solve a system of linear equations into a parametric form for the solution set. The parametric form you get from the book method is not really in a meaningful way "better" than any other form you could choose -- such as your example. In the vast majority of application each of them will be equally good. ...


3

Let $f(x) = (|x_1|+|x_2|)^2$ and note that $f((t,1)) = t^2 + 2|t| +1$, which is not differentiable at $t=0$. Hence $x \mapsto \|x \|_1^2$ is not differentiable.


2

Part (b): Abbreviate $M:=M_{\mathcal{A}}^{\mathcal{A}}(f)$ and using the convention to write $f(v)=M\hat{v}$, where $\hat{v}$ denotes the coordindate represntation of $v$ in terms of basis $\mathcal{A}$, you can determine $M$ by evaluation $f$ on the basis vectors in $\mathcal{A}$, i. e. $f(t^3)=3t^2$, thus the first column vector of $M$ is $(0,3,0,0)^t$, ...


2

$(2x,x)=(2x,2y)$ says that $x=2y$. So the answer is all vectors of the form $(2y,y)$ where $y$ is real number.


2

Don't forget that $\{w_1,\ldots,w_n\}$ is a basis of $W$. Therefore, for any vector space $Z$ and for any vectors $z_1,\ldots,z_n\in Z$, there is one (and only one) linear map $f\colon W\longrightarrow Z$ such that$$(\forall k\in\{1,2,\ldots,n\}):f(w_k)=z_k.$$That's all you need.


2

There are a few properties of differential forms that you didn't mention. First and foremost an important quality they enjoy is being anti-symmetric. Anti-symmetry of course cannot be seen when only dealing with one forms, you have to look at two forms and up. What is anti-symmetry? Well consider the function $$q(x,y)=x-y$$ Now if I exchange x and y I ...


2

Quoting the first line of those notes: Throughout this note $\Bbb K$ will be one of the fields $\Bbb R$ or $\Bbb C$, equipped with the standard topology. All vector spaces mentioned here are over $\Bbb K$. So it doesn't apply to the field in the discrete topology.


2

Not necessarily. Take the $L^\infty$ norm in the plane $$ F(x,y)=\max\{|x|,|y|\} $$ If you square it, you get a non-differentiable function along the lines $x=\pm y$.


2

That's because $h$ is injective and injective maps map linearly independent sets into linearly independent sets. And $h$ is injective because, if $w\in W$ and $h(w)=0$, you can take $v\in V$ such that $g(v)=w$ and then$$0=h(w)=h\bigl(g(v)\bigr)=v$$and therefore $w=g(v)=g(0)=0$.


2

The standard dot product of two complex vectors $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is in fact not $x_1x_2+y_1y_2 + z_1z_2$, but rather $$ \overline{x_1}x_2+\overline{y_1}y_2 + \overline{z_1}z_2 $$ This is necessary to ensure that, for instance, the dot product of any vector with itself is a real (and non-negative) number. Using this will give you ...


2

Your equations give \begin{align*} 3b&=2c\\ a+c&=d\\ \end{align*} Meaning that once $a$ and $c$ are known you can deduce $b$ an $d$. Consequently the matrix $A$ has the following form \begin{pmatrix} a & 2c/3 \\ c & a+c \end{pmatrix} Can you conclude from there ?


2

You can use the nullity-rank theorem with the map $\Phi: \phi^{-1}(U)\to U$ : $\dim(\phi^{-1}(U))=\dim(ker(\Phi))+\dim(im(\Phi))$ but $ker(\Phi)=\ker(\phi)$ because if $v\in Ker(\Phi)$ then $\Phi(v)=\phi(v)=0$ so $v\in Ker(\phi) $while if $v\in Ker(\phi)$ then $\phi(v)=0\in U$ so $v\in \phi^{-1}(U)$ and you have that $\Phi(v)=\phi(v)=0$ then $v\in Ker(\...


2

There is no reason why "everything except $\operatorname{span}\{(1,1)\}$ is in $U$". In fact it is false. Simple example: $$T=\begin{pmatrix}0&1\\1&0\end{pmatrix}$$ Then for $\lambda=1$, $$ T-I=\begin{pmatrix}-1&1\\1&-1\end{pmatrix} $$ and so $U=\operatorname{span}\{(1,-1)\}$ does not contain $(1,2)$ or $(1,3)$, for example. Another ...


1

Matrix $A$ is of the form $\begin{pmatrix} a&b&(-a-2b)/3\\c&d&(-c-2d)/3\\e&f&(-e-2f)/3\\\end{pmatrix}$. Observe that only $6$ free variables have been used.


1

We can write your system as$$\begin{pmatrix} 1 & 3 & 1 \\ 5 & 1 & 1 \\ 1 & 1 & 7 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2\\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 4\\ 2 \end{pmatrix}$$ and that is how you get the matrix that you were talking about. This yields that the first column corresponds to the coefficients of $x_1$ etc.


1

Bilinear pairing $(A,B) \mapsto \mathrm{tr}(AB)$ on $\mathrm{End}(V)$ is non-degenerate, so any functional, invariant or not, admits such a representation. Moreover operator $B$ is uniquely determined. I think the point is that if the functional is invariant, then so is $B$.


1

You are wrong it you claim that uniform convergence is not addressed here. The author of that answer wrote “Next you go on to showing that $f_n$ also converges to $f$ in norm”, which, since the norm here is the $\sup$ norm, is exactly the same thing as asserting that $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$.


1

We have $U= span \{(1,1,0,0), (0,0,1,1)\}.$ Try $W= span \{(1,0,0,0), (0,0,0,1)\}.$


1

Yes, this does hold. Let $T : V \to W$ be linear. One thing you can always do in vector spaces is complement subspaces. That is, given a subspace $X \le V$, you can find $Y \le V$ such that $X \oplus Y = V$. In finite dimensions this is done with bases, but in infinite dimensions, it's done with some kind of axiom of choice argument. Aside: In infinite ...


1

I'm assuming by $V'$ you mean the algebraic dual, i.e., all $\mathbb{F}$-linear maps $V\to\mathbb{F}$. If $V$ is a topological vector spaces over a topological field $\mathbb{F}$, then $V'$ might stand for the continuous dual. I'll use the notation $V^\vee$ for the algebraic dual of $V$. Given any element $\phi\in(\ker T)^0$, define $\psi\in(T(V))^\vee$ ...


1

To briefly address your specific question about "duality" first, there is no "duality" (not sure what is precisely meant by that term here) between complex exponentiation and complex multiplication, unless the arguments of both are real. This is because complex multiplication is commutative; complex exponentiation is not - neither is real exponentiation $(e^{...


1

No such book exists because there is no such object as a space in Mathematics, let alone any sort of gradation between them. The term space is a shortcut for the name of a generic object considered in a given area, so that in Topology a space means a topological space, in Linear algebra it is a vector space, in Functional analysis it may or may not be a ...


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