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prove that $\|f\|=n^{1/2} $

Suppose $M > 0$ satisfies $\|f(x)\| \le M\|x\|$ for all $x\in \Bbb R^n$. Then, $$M \ge \frac{\|f(x)\|}{\|x\|},$$ for every $x\in \Bbb R^n\setminus\{0\}$. Choose $x = (1,1,\ldots, 1)$ to get $$M \ge ...
stoic-santiago's user avatar
0 votes

bv space is a direct sum of $bv_0$ with a one dimensional subspace

You know lots and lots and lots of one-dimensional spaces. Take any nonzero vector $v$ on any complex-vector space, and then $\mathbb Cv=\{\lambda v:\ \lambda\in\mathbb C\}$ is a one-dimensional ...
Martin Argerami's user avatar
0 votes
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bv space is a direct sum of $bv_0$ with a one dimensional subspace

We have to understand how the sequences in $bv$ behave. Note that by the triangle inequality, for $k, n \in \mathbb{N}$ and $(x_n) \in bv$: \begin{align} \lvert x_{k+n} - x_n \rvert \leq \sum_{\ell = ...
Hyperbolic PDE friend's user avatar
0 votes

Proof of Closest Point Theorem in Hilbert Space

All of your arguments are completely correct. Note that it is unnecessary to find a different point: There is no need that a sequence consists of pairwise different points. For instance, the number ...
Martin Väth's user avatar
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0 votes

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
2 votes

Outward pointing normal Tetrahedron

Correct. The outward pointing normal is the normalized cross product $\frac{\vec{CA}\times\vec{AD}}{\|\vec{CA}\times\vec{AD}\|}.$
Lieven's user avatar
  • 764
1 vote
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What is so special about eigenvector that it behaves like a pivot of linear transformation?

I suppose one way to look at it is that it's the vector that has the right combination of $\hat{\imath}$ and $\hat{\jmath}$ such that the stretching and skewing caused by $A$ on each of those basis ...
ConMan's user avatar
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0 votes

Curl of Cross Product of Two Vectors

Here is an example in Physics calculating the Cross product of the $\overset{\rightarrow}{\boldsymbol J}\text{ }$ Current Density for a Spherical shell of radius $r$ with a charge $Q$, in a shell ...
Stephen Elliott's user avatar
1 vote
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Vector space generated by translates

Your deduction on $E$ is correct, so I will answer question 3. It seems it is talking about the vector space of $L^2$ functions defined on some interval $I\subset \mathbb R^n$ with values in the ...
Lorenzo Pompili's user avatar
0 votes

vector space over any field?

You're looking essentially at cardinalities of sets. So yes, the cardinality of a non-trivial vector space must be at least the cardinality of its field. If both are finite, we must have $|V| = |F|^n$ ...
aschepler's user avatar
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1 vote
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To find the dimension of intersection of two vector spaces.

If $\{a,b,c,d\}$ belongs to both $S$ and $T$. Observe that the two given equations will hold. After adding the above two equations you will get: $a+c = 0$ and $b+d = 0$. Now can you guess the ...
Divyanshu Kumar's user avatar
0 votes

Can the $\ker T$ and $\ker T\circ T$ be the same?

If you want to disprove $3$ and $4$ you should try to come up with counterexamples. You are close to proving 1. There is one trivial inclusion, and the other can be proved by contradiction.
Bcpicao's user avatar
  • 918
1 vote

Why $0$ must be an eigenvalue of any Laplacian matrix?

When you compute the determinant, add every other columns/lines in the first one. It's a general observation that if the sum along lines (resp. columns) of a matrix $M$ are constant equal to $\lambda$,...
Ayoub's user avatar
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3 votes
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Why $0$ must be an eigenvalue of any Laplacian matrix?

If the sum of entries along a row of $L$ is $0$ then $$L \cdot (1, 1, \ldots, 1)^T = 0 \cdot (1,1, \ldots, 1)^T$$ Therefore $0$ is an eigenvalue of $L$, and $(1,1,\ldots,1)^T$ is its corresponding ...
Thành Nguyễn's user avatar
4 votes
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Why eigenspace of -1 orthogonal complement of 1?

$A(K_n)$ is a real symmetric matrix, so its eigenspaces are orthogonal to each other. $A(K_n) + I$ is the $n \times n$ matrix of all $1$'s, which has rank $1$ because all its rows are the same. That ...
Robert Israel's user avatar
0 votes

Are all matrices equipped by definition with a matrix-vector product?

You will find that if you trace such ontological dependencies long enough, that what you get a net that has cycles and worse. To get a hierarchy of concepts one has to cut the net to get a spanning ...
Lutz Lehmann's user avatar
1 vote
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Consequences of definition of scalar product

For the first one, I would have just added the line $$\overline{\langle z,x\rangle+\langle z,y\rangle}=\overline{\langle z,x\rangle}+\overline{\langle z,y\rangle}$$ The second one is okay and for the ...
ultralegend5385's user avatar
1 vote
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Dimension of a linear subspace of $2^S$ spanned by $100$ singletons?

The $100$ pairwise disjoint singletons are linearly independent, since a linear combination of any $99$ of them is disjoint from the last one. So the $100$ elements form a basis for their span. ...
Ryan Noonan's user avatar
2 votes
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Writing convention of Courant–Fischer theorem

As an analogy: note that the sum $$ \sum_{i=1}^n \left(\sum_{j=1}^i x_{ij}\right) $$ makes sense, but the sum $$ \sum_{j=1}^i\left(\sum_{i=1}^n x_{ij}\right) $$ does not. Just like the sum from 1 to $...
Ben Grossmann's user avatar
1 vote
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How can we be sure that the triangle inequality works in this case?

If I didn't understand your problem wrongly, then youre trying to find when equality does occur. $|z_1+z_2|=|z_1|+|z_2|$ can only occur when $z_2$ is a scalar multiple of $z_1$. In your case, $$(z + 1 ...
Gwen's user avatar
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0 votes

Writing convention of Courant–Fischer theorem

You have fixed $S=\mathrm{span}\{x_{i_1}, \dots, x_{i_k}\}$, then $\mathrm{dim}(S)=i_k$ and you have either exactly one vector space to minimize (then the entire expression gives you back $\lambda_{...
Severin Schraven's user avatar
6 votes
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Why $X^\perp=\text{span}\space\{x_{k+1}\dots,x_n\}$ true?

Let us recapitulate my comments, before deleting them. Though you didn't mention it, $\{x_1,…,x_n\}$ itself is supposed to be an orthonormal set of eigenvectors for $A$, hence a basis of $\Bbb R^n$. (...
Anne Bauval's user avatar
  • 37.3k
4 votes

Why $X^\perp=\text{span}\space\{x_{k+1}\dots,x_n\}$ true?

Let $V$ be a finite dimensional inner product space, and let $\{v_1,\dots,v_n\}$ be an orthonormal basis for $V$. Let $X=\{v_1,\dots,v_k\}$. We claim that $X^\perp=\operatorname{span}\{v_{k+1},\dots,...
Lorago's user avatar
  • 10k
3 votes
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Different versions of Courant–Fischer theorem

If we replace $A$ by $-A$ and $k$ by $n-k+1$ in the statement of the min-max version of the theorem, we get \begin{align*} \lambda_{n-k+1}(-A) &=\min_{\dim S=n-k+1}\,\max_{x\in S\setminus0}\frac{x^...
user1551's user avatar
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1 vote
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Closure of projecting cone is the tangent cone

Proof of $T(Y, \bar{y}) \subseteq \operatorname{cl} P(Y-\{\bar{y}\})$: If $h \in T(Y, \bar{y})$, then there exists a sequence of nonnegative reals $t_l$ and points $y_l \in Y$ converging to $\bar{y}$ ...
angryavian's user avatar
  • 90.5k
1 vote

$ABACA = 0 \Longrightarrow BAC = 0$ if $A,B,C \ge 0$ are symmetric.

This proof looks correct! Indeed, for an matrix $M$, if $M^T M$, then for any vector $x$, $$ \lVert Mx\rVert^2 = x^T M^T M x = 0, $$ from which any $Mx = 0$ and so $M = 0$. In terms of a concise ...
Damian Pavlyshyn's user avatar
1 vote
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$ABACA = 0 \Longrightarrow BAC = 0$ if $A,B,C \ge 0$ are symmetric.

Note that for real matrices $M$, $M = 0$ if and only if $M^TM = 0$. More generally, for a positive semidefinite matrix $P$, it holds that $M^TPM = 0$ if and only if $\sqrt{P}M = 0$, which holds if and ...
Ben Grossmann's user avatar
1 vote

$ABACA = 0 \Longrightarrow BAC = 0$ if $A,B,C \ge 0$ are symmetric.

Your solution looks essentially correct. Denote the inner product of two vectors $f,g$ by $\langle f,g \rangle.$ For any vector $f$ we have $$ \langle C^{1/2}ABAf,C^{1/2}ABAf\rangle = \langle(ABACA)...
Steen82's user avatar
  • 157
1 vote
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Can I derive vector space-equivalent from direct sum?

I am assuming you are working with finite-dimensional vector spaces only. You are on the right track. Your question is ostensibly about integers $a_i, b_i$ ($i=1,2$) with the properties $a_i \leq ...
Randall's user avatar
  • 19.1k
0 votes
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Find the dimension and basis of the vector of the symmetric polynomials

When you selected $1$ in your basis, you put the condition $a_1=a_2=a_3=a_4=a_5=0$. Then you say for any $a_0$ you can swap $x$ and $y$ and your linear combination does not change. Now let's look at ...
Andrei's user avatar
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1 vote
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Let $T:\Bbb R^5\rightarrow\Bbb R^5$ be a $\Bbb R$ linear transformation.

Assume these three linearly independent vectors as $e_{1}, e_{2}, e_{3}$, then extend this basis of null space of $T$ to get a basis of $\mathbb{R}^{5}$ and say basis vectors are $e_{1}, e_{2}, e_{3}, ...
Afntu's user avatar
  • 1,041
1 vote
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Matrix of an Inner Product and Spectral Theorem

Perhaps things will be clearest when $n = 1$. Let $(V, \langle -, - \rangle))$ be a one-dimensional real inner product space, with basis $(e)$. Suppose that we measure $$ \langle e, e \rangle = 2,$$ ...
Joppy's user avatar
  • 13k
2 votes
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Eigenvalues of a particular linear transformation

Assuming wlog that $k$ is algebraically closed, let $(u_{i,j})_{1\le i\le p, 1\le j\le m_i}$ be a basis of $V_1$ for which the matrix of $T_1$ consists of Jordan blocks with eigenvalues $\lambda_1, \...
Anne Bauval's user avatar
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1 vote

How to identify the basis generated by figures?

You can think of a basis visually as a minimal set of "directions" on a space which "generate" it. When you look at the images, ask yourself these two questions: If I imagine ...
Lourenco Entrudo's user avatar
0 votes

Why is $ R(A^*) \perp N(A)$ true?

$\def\spans{\operatorname{span}}$ $\def\qty#1{\left( #1 \right)}$ $\def\range{\operatorname{range}}$ $\def\nspace{\operatorname{null}}$ First note that if $v \in \nspace A$, $$ Av = \begin{bmatrix} ...
Ted Black's user avatar
  • 750
2 votes
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Why is $ R(A^*) \perp N(A)$ true?

Notation: By $\langle x,y\rangle$ I mean the standard inner product on $\Bbb{C}^{n}$ given by $y^{*}x$. Let $x\in R(A^{*})$ . Then $x=A^{*}(z)$ for some $z\in \Bbb{C}^{n}$ and consider some $y\in N(A)$...
Mr.Gandalf Sauron's user avatar
1 vote

Why is $ R(A^*) \perp N(A)$ true?

If $u \in N(A)$, then $0 = (Au, v) = (u, A^*v)$ for all $v$, so $u \in R(A^*)^{\perp}$. Hence $N(A) \subset R(A^*)^{\perp}$. Since $\dim N(A) = \dim R(A^*)^{\perp}$, we get $N(A) = R(A^*)^{\perp}$.
Kakashi's user avatar
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-3 votes
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Is $\ell^1$ a closed subspace of the normed space $(\ell^2, \||\cdot\||_2)$?

No, all closed subspaces of a Hilbert space are isomorphic to Hilbert space. $\ell^1$ is not isomorphic to Hilbert space. The meaning of the sentence: "This converges in $\ell^2$ norm to the full ...
uniquesolution's user avatar
1 vote
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How is sphere related with Rayleigh quotient?

The key point is to see it as inner products. We have for $x\in\mathbb{R}^n\backslash \{ 0\} $ : $$R_A(x) =\frac{\langle x, Ax\rangle} {\langle x, x\rangle} =\frac{\langle x, Ax\rangle} {||x||^2 }. $$ ...
NancyBoy's user avatar
  • 470
1 vote

If $B$ is a $n \times n$ matrix, then is the column space of $B$ and $BB^{'}$ same?

This works even if $B$ is $m \times n$. We just need to show that, for any $x \in \Bbb{R}^n$, there exists some $y \in \Bbb{R}^m$ such that $Bx = BB'y$. To construct such a $y$, we take a tip from ...
Theo Bendit's user avatar
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1 vote
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Completeness of Y (normed tvs) in the proof of th. 2 in ch. V section 3 of Yosida's Functional Analysis ed 6

My first attempt, as noted by @Jochen, was flawed. Here is a (hopefully) correct solution: Let us denote by $\tau$ the topology on $Y$ induced by the weak topology on $X$ and by $\tau_1$ the norm ...
Pelota's user avatar
  • 843
2 votes
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How orthogonal projection connects with eigen space?

The matrix $P=\begin{bmatrix}0&0\\a&1\end{bmatrix}$ is a projection since $P^2=P$ but it is not orthogonal unless $a=0$ since it is not symmetric. The nullspace of this matrix also depends on ...
CyclotomicField's user avatar
3 votes

Is this a vector space: $k(x,y,z)=(kx,y,z)$?

Your space fails to be a vector space because your multiplication does not distribute over addition: $$(a+b)\cdot(x,y,z)=((a+b)x,y,z)=(ax+bx,y,z)$$ $$a(x,y,z) + b(x,y,z)= (ax,y,z)+(bx,y,z)=(ax+bx,y+y,...
Omar Dennaoui's user avatar
5 votes
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Is this a vector space: $k(x,y,z)=(kx,y,z)$?

You are correct in saying that $V$ is not a vector space, but your friend is right in saying that your argument is wrong. Indeed, within the axiom (which would be more correctly written as) For each ...
Ben Grossmann's user avatar
6 votes
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Is union of orthonormal bases orthonormal?

The answer is no. Consider the matrix $A = \begin{pmatrix} 0 & 0 \\ 1 & 1\end{pmatrix}$. This matrix has eigenvalues $\lambda_1 = 0$, $\lambda_2 = 1$ and corresponding normalized eigenvectors $...
whpowell96's user avatar
  • 5,992
0 votes

Find exercises to practice vector calculus

When I learned vector calculus I used the materials available on MIT OpenCourseWare. They have very insightful exercises on the major subjects covered on an standard course.
Kham Bodrogi's user avatar
0 votes

External direct sum $U_1\oplus U_2$ isomorphic to $U_1+U_2$ does not necessarily imply that $U_1\cap U_2 = \{0\}$

There are two related but distinct questions here: Is the map $U_1 \oplus U_2 \to V$ given by $(u_1, u_2) \mapsto u_1 + u_2$ an isomorphism onto $U_1 + U_2$? Equivalently, is this map injective (...
ronno's user avatar
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3 votes
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Is the nullspace of transpose of any matrix orthogonal to the range of that matrix?

Let $u \in R(A)$ and $v \in N(A^\top)$. By the definition of range, we have $u=Aw$ for some $w \in \mathbb{R}^n$. By the definition of nullspace, we also have $A^\top v = 0$. With these observations,...
angryavian's user avatar
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