Let $F$ be a field. Then the Euclidean inner product on $F^n$ is defined as $\langle \mathbf{x},\mathbf{y}\rangle=x_1\overline{y_1}+\cdots+x_n\overline{y_n}$ for all $\mathbf{x}=(x_1,~x_2,~\cdots,~x_n),~\mathbf{y}=(y_1,~y_2,~\cdots,~y_n)\in F^n$. Then its induced norm, $\lVert x\rVert:=\sqrt{\langle\mathbf{x}, \mathbf{x}\rangle}$, is a $\ell^2$-norm on $F^n$. From this, I come up with a doubt that can it be possible that we define an inner product on $F^n$ such that its induced norm is $\ell^1$-norm, or $\ell^{\infty}$-norm, or even $\ell^{p}$-norm?
121421
Please help me prove this. I'm not sure how to apply the parallelogram law to the norm.
64906
I know that the following theorem holds true:  Let $S$ be a subset of $\mathbb R^n$, and let $f:S\to \mathbb R^n$ a map such that $d(p,q)=d(f(p),f(q))$ for every $p,q \in S$ (here $d$ is the usual euclidean distance). Suppose that $0 \in S$ and $f(0)=0$. Then $f$ can be extended to an orthogonal transformation of $\mathbb R^n$.  I have two questions:   Can someone give me a reference of a proof of this theorem? I believe that the previous theorem holds true also if we change $\mathbb R^n$ with $\mathbb C^n$ and "orthogonal" with "unitary", with nearly the same proof. Does it holds true also in the $ \ell^2$ space? With the same proof?
41851
(I'll explain what I know first and then I'll ask the questions). Given a finite dimensional vector space $V$, it is often remarked that there is no "natural" isomorphism from $V$ to $V^*$ (I guess this means a basis independent isomorphism?). I understand that one typically constructs an isomorphism $V \to V^*$ by fixing a basis for $V$, call this $B = \{v_1, v_2, \ldots, v_n\}$, and then mapping $v_i \mapsto \delta_i$ where $\delta_i$ is the linear functional given by $\delta_i(v_i) = \delta_{ij}$. Here are my questions:  (1) Why is there no natural isomorphism $V \to V^*$?  (2) If $V$ is a finite dimensional inner product space, we can map each $v \mapsto \langle v, \cdot \rangle$. Is this not "natural"? (I know we can consider sesquilinear forms, but let's keep the discussion to this inner product for simplicity).
83571
Let $(e_n)$ (where $ e_n $ has a 1 in the $n$-th place and zeros otherwise) be unit  standard vectors of $\ell_\infty$.   Why is $(e_n)$ not a basis for $\ell_\infty$?  Thanks.
22490
I was just wondering whether anyone knows an answer to the following:  Suppose that ${\mathbb F}$ is a subfield of a field ${\mathbb G}$ and that $v_1,\ldots ,v_k$ are linearly independent vectors in ${\mathbb F}^n$ (over $\mathbb F$). Is it necessarily true that $v_1,\ldots ,v_k$ are also linearly independent when considered as vectors in ${\mathbb G}^n$ (over $\mathbb G$})?  Any help would be much appreciated. Thanks!
115566
Let $R$ be a Euclidean domain, i.e., a ring with a norm $N : R \rightarrow \mathbb N$ such that for any $a,b\in R$ with $b\not=0$, we may write $a = qb + r$ for some $q,r \in R$ with $N(r) &lt; N(b)$. Let $I$ be a prime ideal of $R$, so $R/I$ is a domain. Must $R/I$ in fact be a Euclidean domain?   I know that this is true if we replace "Euclidean domain" with "PID", and false if we replace it with "UFD", so there is no clear intuition to be gained from similar concepts. My first attempts at a proof was to borrow a construction from analysis and define a norm on $R/I$ by  $N(a + I) = \inf\{N(a+i) : i\in I\}$  but this didn't really get me anywhere.   So, this fact true? If so, how can I prove it? If not, whats a counterexample?
61642
If a norm on a vector space is given by an inner product then we have $\lVert x + y\rVert = \lVert x \rVert + \lVert y \rVert$ only when $x=\lambda y$ for some constant $\lambda$ (this follows from the Cauchy-Schwarz inequality). This does not hold for general (unitary invariant) norms. E.g. when considering generalised Ky Fan norms: $$ \lVert x\rVert_s = \sum_i s_i \lvert x\rvert_i^\downarrow$$ where $s_i\geq 0$ and $\lvert x\rvert^\downarrow$ is $\lvert x\rvert$ rearranged with the components from biggest to smallest. So for example the operator (sup) norm has $s_1 = 1$ and $s_i=0$ for $i&gt;1$, while the trace norm has $s_i=1$ for all $i$. For these norms it is in general possible to find $x$ and $y$ so that the norm is additive for these vectors, while $x$ and $y$ are not multiples of each other.  Now my question is the following. Given a unitarily invariant norm and vectors $x$ and $y$ such that $\lVert x + y\rVert = \lVert x\rVert + \lVert y \rVert$, can we then conclude that $\lVert tx + (1-t)y\rVert = t\lVert x\rVert + (1-t)\lVert y \rVert$ for $0\leq t \leq 1$?   This is true for these generalised Ky Fan norms and for all norms that come from inner products. It is also true for all $p$-norms as this additivity still only holds for $x$ and $y$ that are multiples of each other (even though $p$-norms don't come from an inner product).  My intuition behind this is that this additivity property (when $x$ and $y$ aren't multiples of each other) forces the norm to be "linear" in a certain way, such that the norm has to be very similar to a generalised Ky Fan norm. So as a bonus question: If there are $x$ and $y$ that aren't multiples of each other, such that the norm is additive for these $x$ and $y$, show that the norm is a generalised Ky Fan norm. Note that this can only be true when we assume the norm to be unitary invariant since otherwise we can construct a counter-example of this (take $(X_1,\lVert\cdot\rVert_1)\oplus (X_2,\lVert\cdot\rvert_2)$ with the norm $\sup\{\lVert\cdot\rVert_1,\lVert\cdot\rVert_2\}$ where $\lVert\cdot\rVert_1$ is a Ky Fan norm and $\lVert\cdot\rVert_2$ is some other norm).
77140
This is a question of the book Linear Algebra of Kenneth Hoffman.  Describe explicitly all inner products on $\mathbb{R}$ and on $\mathbb{C}$.  I think that if $&lt;,&gt;$ is one inner product on $\mathbb{R}$ (or $\mathbb{C}$) then $&lt; ,&gt;=K[,]$, where $[,]$ is the standard inner product. Is that correct?
55863
From the  on the dual of a norm:  $X$ and $Y$ are normed spaces, and we associate with each $f\in L(X,Y)$ (the space of bounded linear operators from $X$ to $Y$) the number $$\|f\| = \sup\{|f(x)|:x\in X, \|x\| \leq 1\}.$$  At some point in the proof that $L(X,Y)$ is bounded, it seems like they use the inequality $$\|fx\| \leq \|f\|\cdot\|x\|,$$ where $\|x\|\leq 1$.  I can't see why this should holds.   $\|f|\|$ is the supremum of what $|f(x)|$ can be, given that $\|x\|\leq 1$. So, given that $\|x\| \leq 1$, $|fx|$ should be bounded by $\|f\|$. But why should it be bounded by $\|f\|\cdot\|x\|$?
97226
If B is $n \times n$ real symmetric positive definite matrix, then $(x,y) = x^T By$ defines an inner product on $R^n$. How do you prove that $\|x\|=(x^T B x)^{1/2}$ is a norm on $R^n$?
82578
What is the Orthogonal Projection onto the $ {\ell}_{\infty} $ Unit Ball?  Namely, given $ x \in {\mathbb{R}}^{n} $ what would be:  $$ {\mathcal{P}}_{ { \left\| \cdot \right\| }_{\infty} \leq 1 } \left( x \right) = \arg \min_{{ \left\| y \right\| }_{\infty} \leq 1} \left\{ {\left\| y - x \right\|}_{2}^{2} \right\} $$  I managed to get an answer using the . Yet I would be happy to see if someone can derive the answer directly.  Thank You.
40860
This question is from a set of optional, much harder problems from my first year analysis course, but the subject material is norms on $\mathbb R^K$.  (c) Show that there exists a constant $C &gt; 0$ such that for all $\mathbf x ∈ \mathbb R^k$, $||\mathbf x||\le C||\mathbf x||_\infty$  Hint: Use the standard unit vectors to write $\mathbf x = x_1\mathbf e_1 +. . .+x_n \mathbf e_n$, and use the triangle inequality.  (d) Show that there also exists a constant $c &gt; 0$ such that for all $x ∈ \mathbb R^k$, $||\mathbf x||_\infty \le c||\mathbf x||$  Hint: If it isn’t true, then you can find a sequence $⟨\mathbf x_n⟩$ such that $||\mathbf x_n||$ is bounded but $c_n := ||x_n||_∞ → ∞$. (Why?) Think about the sequence $⟨\mathbf x_n/c_n⟩$. The Bolzano-Weierstrass theorem may be of assistance.  Note that the unspecified norm represents any norm for $\mathbb R^k$. I've done part c), but am stuck on part d). I've had a look for proofs on other questions on MSE, but none of them seem relevant to the hint given, and frequently the terminology is too advanced for me.  I tried considering the case if it isn't true, as suggested, but I can't see how to bound any sequence using that. Any help would be much appreciated.
76686
Today I proved that if $V$ is a normed space with norm $\|\cdot\|$ then I can define a norm on $V \times V$ that induces the same topology as the product topology as follows: $\| (v,w) \|_{V \times V} = \|v\| + \|w\|$.  I think I can do the same for an infinite product $V^{\mathbb N}$ by defining $\|(v_n)\|_{\mathbb N} = \sum_{n=0}^\infty \frac{1}{2^n} \|v_n\|$ and I proved it using the proof of the case $V \times V$ and changing some minor things.  Can you confirm that this result is correct? Thanks.
14182
Let $A$ be a positive definite real $n\times n$ matrix. How can I prove that $$ \int_{\mathbb{R}^n}\mathrm{e}^{-\langle Ax,x\rangle}\text{d}x=\left|\,\det\left(\pi^{-1}{A}\right)\right|^{-1/2}=\pi^{n/2}\lvert\,\det A\rvert^{-1/2}\!, $$ where $\langle\cdot,\cdot\rangle$ denotes the inner product in $\mathbb R^n$, i.e. $\langle x,y\rangle =x^Ty$.
40784
Here's Prob. 8 in the Problems after Sec. 3.2 in Introductory Functional Analysis With Applications by Erwine Kreyszig:      Show that in an inner product space, $x \perp y$ if and only if $\Vert x + \alpha y \Vert \ge \Vert x \Vert$ for all scalars $\alpha$.    If $x \perp y$, then $\langle x, y \rangle = 0$; so for any scalar $\alpha$, we have  $$ \begin{align*} \Vert x + \alpha y \Vert^2 &amp;= \langle x + \alpha y, x + \alpha y \rangle \\ &amp;= \Vert x \Vert^2 + 2 \Re \bar{\alpha} \langle x, y \rangle + \vert \alpha \vert^2 \ \Vert y \Vert^2 \\  &amp;= \Vert x \Vert^2 +  \vert \alpha \vert^2 \ \Vert y \Vert^2 \\  &amp;\ge \Vert x \Vert^2.  \end{align*} $$ So  $$ \Vert x + \alpha y \Vert \geq \Vert x \Vert. $$  Am I right?  Now how to prove the converse? It is my guess that we will have to put in a particular value for $\alpha$.
82505
For the ring $\Bbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}\mid a,b \in \Bbb{Z}\}$ where $d$ is not divisible by the square of a prime, prove that the norm $N(a+b\sqrt{d})= |a^2-db^2|$ satisfies the assertions: $N(x)=0$ if and only if $x=0$; $N(xy)=N(x)N(y)$ for all $x$ and $y$; $x$ is a unit if and only if $N(x)=1$; and if $N(x)$ is prime, then $x$ is irreducible in $\Bbb{Z}[\sqrt{d}]$.
19516
For a real (or complex) inner product space $V$, the inner product can be expressed in terms of the norm as either $$ \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2) $$ or $$ \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2) $$ respectively.  These identities are not hard to verify directly once they are given. I would like to know, is there a systematic way one would derive these from scratch without knowing them beforehand, or are both polarization identities just clever observations?