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Why is a norm the square root of an inner product?

Jean’s answer covers everything but the one question of “why the square root?” This has a simple answer: otherwise, the norm would not be homogenous. In other (less formal) words, the units would be ...
Malady's user avatar
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Why is a norm the square root of an inner product?

"The norm is the square root of the inner product" is an oversimplification. An inner product takes two vectors, a norm takes just one vector. If $\langle \bullet, \bullet \rangle$ is an ...
Jean Abou Samra's user avatar
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Easy proof of Cauchy-Schwarz

If $||x|| = ||y|| = 1$, Then, $0 \leq \langle x-y,x-y\rangle = \langle x,x\rangle + \langle y,y\rangle - 2 \langle x,y\rangle = 2 - 2 \langle x,y\rangle \implies\langle x,y\rangle \leq 1$. similarly $\...
Afntu's user avatar
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Assume that for every $f\in X^*$, there exists $y \in X$ such that $f(x)=\langle x, y \rangle$ for every $x \in X$. Show that $X$ is a complete space.

This follows easily from the Riesz representation theorem. But for a hands-on proof, indeed you can show $\varphi$ is an isometry, because of the Cauchy-Schwarz: $$|\langle x, a\rangle| \le \|x\|\|a\|\...
Just a user's user avatar
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Let $ H $ be a Hilbert space and let $ A \in \mathcal{B}(H) $. Show that $ H = \ker A \oplus \overline{\operatorname{Im} A^*}. $

A lot of what you wrote is correct, but you appear to be confused as to why $\text{Ker}(A) \subseteq (\overline{\text{Im}(A^{*})})^{\perp}$. For your method of showing that $\text{Ker}(A) \subseteq (\...
Dean Miller's user avatar
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Let $ H $ be a Hilbert space and let $ A \in \mathcal{B}(H) $. Show that $ H = \ker A \oplus \overline{\operatorname{Im} A^*}. $

Notice that $x\in\operatorname{ker}(A)$ iff $$0=\langle Ax,y\rangle =\langle x, A^* y\rangle$$ for all $y\in H$. This equivalent to $x\in\operatorname{ker}(A)$ iff $x\in \big(\operatorname{ran}(A^*)\...
Mittens's user avatar
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Show that $(\operatorname{Im} A)^\perp = 0$ and from this deduce that $A$ is a homeomorphism.

For the first question, in order to conclude that the sequence $(x_{n})_{n\in\mathbb{N}}$ is convergent, you typically need to know what the limit will be. This is not immediately clear. Instead, ...
Dean Miller's user avatar
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1 vote
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Let $H$ be a Hilbert space, $A, B \in \mathcal{B}(H)$, and let $C = A^*A + B^*B$.

As in the proof of (a), we have $$\langle Cx, x\rangle = \langle A^*Ax, x\rangle + \langle B^*Bx, x\rangle = \|Ax\|^2 + \|Bx\|^2$$ In the proof of (a), we used this to show that $Cx = 0$ implies $Ax=...
Brian Moehring's user avatar
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Let $H$ be a Hilbert space and let $P \in \mathcal{B}(H)$ be a projector.

$P$ is a projector, so $P^{2}=P$. $\langle Pz, Pz \rangle = \langle z, Pz\rangle$ follows from the fact that $Pz=z$ for any $x \in Im (P)$. [ $z=Pu$ for some $u$ and $Pz=P^{2}u=Pu=z$]. For the last ...
geetha290krm's user avatar
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Gram-Schmidt process: kernel and image

Some hints: $f$ is a linear map, so the kernel and image should be subspaces. Currently, your answers are not subspaces. What is $f(2w)$? What is $f(-w)$? Drawing a picture when $V=\mathbb{R}^2$ or $...
angryavian's user avatar
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1 vote

Inner product in RKHS

The inner product $\langle\cdot,\cdot\rangle_{\mathcal H}$ can be any scalar product so that $(\mathcal H,\langle\cdot,\cdot\rangle_{\mathcal H})$ is a Hilbert space of functions $\delta_x\colon \...
Konstruktor's user avatar
1 vote

What is wrong with using the $H^1_0$ inner product here?

All that are required to apply these theorems for existence and uniqueness are that the coefficient function is $L^\infty(\Omega)$ and $f\in L^2(\Omega)$, neither of which require any continuity in ...
whpowell96's user avatar
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Linear operator as linear combination of positive operators

Am I supposed to use 1. in order to prove 2. ? Yes. In fact, 1. is the hard part and 2. is the easy part. What you do, is write $$ T=\frac{T+T^*}2+i\,\frac{T-T^*}{2i} $$ and use part 1. on each of ...
Martin Argerami's user avatar
1 vote

Solving vector equations in $ℝ^3$

$ \newcommand\R{\mathbb R} \newcommand\form[1]{\langle#1\rangle} \newcommand\lcontr{\mathbin\rfloor} \newcommand\rcontr{\mathbin\lfloor} $Here is an approach for (1) using geometric algebra that does ...
Nicholas Todoroff's user avatar
1 vote
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Solving vector equations in $ℝ^3$

(i) We can write $$\vec{a} \times[(\vec{r}-\vec{b})\times\vec{a}]=|\vec a|^2\vec r-(\vec a\cdot\vec r)\vec a-|\vec a|^2\vec b+(\vec a\cdot\vec b)\vec a$$ $$\vec{b} \times[(\vec{r}-\vec{c})\times\vec{b}...
mathlove's user avatar
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Solving vector equations in $ℝ^3$

(i) The provided answer is wrong. For instance, if $\vec{a}=(1,1,0)$, $\vec{b}=(0,1,0)$, $\vec{c}=(0,0,1)$, and $\vec{r}=\frac{\vec{a}+\vec{b}+\vec{c}}2=(1/2,1,1/2)$ then Mathcad yields $$\vec{a} \...
Alex Ravsky's user avatar
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Affine function $\mathbb{R}^n \to \mathbb{R}^n$ preserves angles if it maps a ball to a ball

I answer after a long time. I hope this is still useful. $\newcommand\B{\mathcal B}$ Let $$ \phi\;:\;\mathbb R^n \;\longrightarrow\; \mathbb R^n $$ be an affine function such that $$ \phi\big(\B(C_1\,,...
gpassante's user avatar
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Given an inconsistent overdeterminate system AX=b where $A\in M_{m×n}(R)$ and $b\in R^m$ with rank A=n. Find the least square approx. solution of AX=b

Since the question was "Find the least square approximate solution of AX=b", your solution which is the standard least squares result is by definition the correct answer. And indeed you do ...
Jos Bergervoet's user avatar
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Weight function given polynomial basis

The claim does not hold. The roots of two consecutive polynomials must interlace. Still this additional condition is not sufficient. Indeed WLOG we may assume that the polynomials are monic, i.e. that ...
Ryszard Szwarc's user avatar
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Use Inner product $\langle f,g\rangle$ to minimize integral

Since $\{\sin(t),\sin(3t)\}$ is an orthonormal basis you can find the orthogonal projection for that inner product space such that you want to minimize $\int_0^{\pi}(1-f(t))^2\text dt$. So $\langle1,\...
User 123732's user avatar
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$\lambda \in \mathbb{C}$ is an eigenvalue of the operator $A$, then $\text{Re}(\lambda) = 0$ AND $H$ is a complex vector space, then $A = iB$.

For c) as you mentioned, you want to write $A= \text{Re}(A)+i\text{Im}(A)$, where $\text{Re}(A)$ and $\text{Im}(A)$ are self-adjoint. Similarly as for complex numbers, you take $\text{Re}(A)=\tfrac{1}{...
mathemagician99's user avatar
1 vote

$\lambda \in \mathbb{C}$ is an eigenvalue of the operator $A$, then $\text{Re}(\lambda) = 0$ AND $H$ is a complex vector space, then $A = iB$.

(b) Follow your steps in case it is a complex inner product space: $$ \langle Ax, x \rangle = \langle x, A^*x \rangle = \langle x, -Ax \rangle = -\overline{\langle Ax, x \rangle} \\ \langle Ax, x \...
GEdgar's user avatar
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If the operator $A$ is self-adjoint and $\lambda$ and $\mu$ are distinct eigenvalues of $A$, then their corresponding eigenspaces are orthogonal.

(a) You have that $(A^*x,x)=(x,Ax)=\overline{(Ax,x)}=(Ax,x)$. Here the hypothesis that $(Ax,x)$ is real was used in the last equation. Therefore, $((A-A^*)x,x)=0$, for all $x\in H$. Let me call $(x,y)...
ameg's user avatar
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2 votes
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Prove that $(H,${$.,.$}$)$ is a Hilbert space.

I think 1. and 2. are easy to see. For 3. note that $$\{x,y\}=\langle Ux+A^2Ux,Uy\rangle=\langle Ux, Uy\rangle+\langle A^2U x,Uy\rangle=\langle x,y \rangle+\langle AUx,AUy\rangle=\langle y,x\rangle+\...
mathemagician99's user avatar
1 vote
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a Cauchy Schwarz application

Define $x_i^2 = a_i^2+b_i^2, x_i >0$, then Using Multinomial theorem, $$ \left(x_1+x_2+\cdots+x_n\right)^2=\sum_{ \substack{ 0 \le j_1, j_2, \ldots, j_n \le 2\\ j_1+j_2+\cdots +j_n = 2}} \frac{2}{...
Sam's user avatar
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How to understand dot product is the angle's cosine?

Instead of asking "How can one see that a dot product gives the angle's cosine between two vectors?" what about asking "Why is the relationship between a dot product and the angle ...
smichr's user avatar
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5 votes
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True or False: Inner product on $\mathbb{R}^2$ satisfying a specific norm.

The actual parallelogram law is an identity, not an inequality: $$\|v+y\|^2+\|v-y\|^2=2\|v\|^2+ 2\|y\|^2$$ and that is not satisfied in your example, hence the norm is not associated with an inner ...
Lieven's user avatar
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1 vote
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Orthogonal projection is bounded

First you need the fact that $P^*=P$. Indeed, $$ \langle P^*(u+w),u+w\rangle=\langle u+w,P(u+w)\rangle=\langle u+w,u\rangle=\langle u,u\rangle=\langle u,u+w\rangle=\langle P(u+w),u+w\rangle. $$ It ...
Martin Argerami's user avatar

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