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0

Your approach is correct. For better clarity, I suggest proving the relation $$ \overline{\operatorname{Im} T^*} = (\ker T)^\perp $$ separately (it's useful on other occasions too). The proof uses the fact that the closed linear span of $A$ is $(A^\perp)^\perp $: $$ \overline{\operatorname{Im} T^*} = ((\operatorname{Im} T^*)^\perp)^\perp = (\ker T)^\perp ...


0

A corollary to Gershgorin's theorem states: More generally: let $M\in M_{n\times n}(\mathbb{R})$ be such that: \begin{equation} (\forall i\in {1,..,n}) (\sum_{j\neq i} a_{i,j}) < a_{i,i}. \end{equation} However, Gershgorin's theorem implies that each eigenvalue $\lambda$ of $M$ satisfies: \begin{equation} \lambda \in \cup_{i=1}^n Ball_{(\sum_{j\neq i} ...


1

Yes it is. Apply Gershgorin's Theorem.


1

Multiply the x column vector with the given matrix.


0

I think $M_n(F)$ has no proper two-sided ideal, if $F$ is a field. To prove this, you may first try to show that any ideal of $M_n(F)$ must be of the form $M_n(I)$, where $I$ is an ideal of $F$. Since fields have no proper two-sided ideals, then you conclude that there is no non-trivial ideal. Was this what you are asking, or did I misunderstand the ...


1

$V$ is the eigenvector matrix, $D$ is the diagonal matrix. $VD$ is still an eigenvector matrix. Multiplying a diagonal matrix from the right, scales the colums. Columns are eigenvectors. Scaled eigenvectors are still eigenvectors (of the same matrix that $V$ belongs to). $DV$, is not. Multiplying a diagonal matrix from the left, scales the rows. This ...


0

The below is incorrect as of now, but might help someone give a correct proof. Feel free to modify it. If $\det(M)$ is a zero divisor, we have the following: let $a \in A$ be such that $a\det(M) = 0$, and take $x = (a,0,\dots,0)^T$. We have $$ \DeclareMathOperator{\adj}{adj} M[\adj(M)x] = [\det(M)\adj(M)] x = \det(M) Ix = 0 $$ Thus, if $\det(M)$ is a ...


0

When you take a photo you are using a projection operator from a $3D$ space to a $2D$ space.


3

No, you do not necessarily have $F_1 = F_2$. But as you have already shown the coefficiencts of the indices in $(F_1 \setminus F_2) \cup (F_2 \setminus F_1)$ are zero. This combination is unique, up to zero coefficients, cf. 1.


3

No, because this isn't justified: $$\bar{\lambda}\overline{\<u,v\>} = \overline{\<\bar{\lambda}u,v\>} $$ It should be $$\bar{\lambda}\overline{\<u,v\>} =\overline{\lambda \<u,v\>} = \overline{\<\lambda u,v\>} $$


2

No, the third equality does not hold. Note that $$\overline{\lambda}\,\overline{\langle u, v\rangle} = \overline{\lambda\langle u, v\rangle} = \overline{\langle\lambda u, v\rangle}.$$


1

Spectral Theory for a closed densely-defined linear operator $A : \mathcal{D}(A) \subseteq X\rightarrow X$ can be viewed in terms of properties of the resolvent operator $R(\lambda)=(A-\lambda I)^{-1}$. In fact, spectrum is defined in terms of the resolvent. For a selfadjoint operator $A$ on a Hilbert space $X$, the resolvent exists for all ...


2

The correct is 1). I will regard a 2-dimensional subspace as the row space generated by the rows of a rank 2, $2 \times 4$ matrix $\left( \begin{array}{cccc} a & b & c & d \\ d & e & f & g \\ \end{array} \right)$. Since such subspace must not be contained in the span of $e_1,e_2,e_3$ one of the generators, say the first row, must ...


0

Actually you cannot get higher than $u_n=n(n-1)/6$. This is because every triple of positions cover ${3\choose 2}=3$ pairs of positions. If there were more than $u_n$ words of weight three, than some pair would be covered twice, and your argument kicks in. This upper bound can also be achieved - at least for some values of $n$. For example when $n=2^m-1$ ...


1

You might be interested in holomorphic functional calculus. We can extend the notion of Jordan canonical form to compact operators on Banach spaces (which includes Hilbert spaces), but this generalization fails in general for bounded linear operators.


1

You might be looking for the classification of finitely generated modules over a PID. It's the natural generalization of Jordan canonical forms and smith normal forms for operators on nice infinite dimensional spaces. This and related topics were discussed recently in an MO thread. Edit: And by recently I mean five years ago. Time flies.


0

A projection map $f: A \to B$ is something you often need to simplify the space $A$. It is build so that differences which occur in the space $A$ will often no longer occur anymore in the space $B$. Then, if you know enough about the map $f$ and find out a bit about the space $B$ (which is easier as their are less differences) then you may find out ...


1

There is no standard way of presenting a vector space other than by presenting a basis, and so there is no standard way of presenting a linear map between two vector spaces other than a matrix (or something that is equivalent to a matrix, e.g., defining the map on a large enough collection of vectors). So any situation where you are trying to compute ...


2

There are two main cases for your question: 1) The description of the operator is specified in such a way that you would be unable to determine its action on any given basis. e.g. The mapping from $\mathbb{R}^2$ to itself that "is a rotation". There are many such rotations, and given the eigenvalues and eigenvectors, you know the transformation. So there ...


3

Yes, since $\dim (U+W)=\dim U +\dim W - \dim (U\cap W)$. In your case: $\dim (U+W)=8 - \dim (U\cap W)$, and note that $\dim (U+W) \leq 7$.


7

Yes. If their intersection was $\{0\}$, then their sum $U+W$ would be a direct sum, so $\dim (U+W)$ would be equal to $8$, which is impossible, since it is a subspace of $\mathbb{R}^7$.


0

Actually, there is nothing to be computed. I mean that it is obvious that $[(1,2,3)]_B = \left( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right)$ since $(1,2,3)$ is the sum of the three vectors of the base $B$. The notation $(1,1,1)^t$ means as was already notice the transpose of the column.


1

The $t$ means transpose here. So $(1,1,1)^t$ is exactly $$\begin{pmatrix}1\\1\\1\end{pmatrix}$$ And if you calculate correctly, you should get that.


39

This is how I used to imagine projections: If a mouse: gets run over by a steamroller: It will look like this: Now if it gets run over by a steamroller another time, it will still look like this:


2

I think you meant you want to find if the following three functions are linearly independent: $$x^2\cos x, x, \sin x$$ You don't need a matrix to find $\alpha, \beta, \gamma$. You can do substitution, or manipulate them algebraically. For example, the last two equations are $$-\pi\alpha+\beta=0\\ \pi\alpha+\beta=0$$ Adding these two gives you $\beta=0$. ...


1

I'll denote $[a\times b]$ or $[a,b]$ the cross product; $(a,b)$ or $a\cdot b$ the dot product and $(a,b,c)=[a\times b]\cdot c$. We're asked to find $$(CM,CB,BF)=[CM\times CB]\cdot BF=[CB+BM\times CB]\cdot BF= ([CB\times CB]+[BM\times CB],BF)=(0+[BM\times CB],BF)$$ Now, I'd suppose "The point $M$ divides the segment $AB$ on ratio 2" is $\frac{AM}{MB}=2$, as ...


3

I do think the wiki's explanation is pretty good: A projection is a mapping of a set (or other mathematical structure) into a subset (or sub-structure), which is equal to its square for mapping composition (or, in other words, which is idempotent). Even more simply: it is an idempotent homomorphism.


0

let $rank(C) = c, rank(A) = a.$ let us look at the null space of $M.$ we have $$(x, y)^\top \in \ker(M) \iff Ax + By = 0, Cy = 0 $$ pick a particular $y$ which has $\text{number of columns of } C - c$ free variables in it. we can solve $Ax = -By$ which will have $\text{number of columns of } A - a$ free variables in it. therefore the total number of ...


3

If you have an orthonormal basis $\{ e_{k} \}_{k=1}^{N}$ on a finite-dimensional space, such as what you would obtain with Gram-Schmidt, then every vector $x$ is expressed as $$ x = \sum_{k} (x,e_{k})e_{k}. $$ This extends to $L^{2}[0,2\pi]$ using $e_{k} =\frac{1}{\sqrt{2\pi}}e^{ikx}$: $$ f = \sum_{k}(f,e_{k})e_{k} $$ The ...


0

Suppose, $m=rank(A)$, $n=rank(C)$, $p=rank(M)$ The partial matrix $A$ has $m$ linear independent rows and the partial matrix $C$ has $n$ linear independent rows. If you fill up the rows of $A$ with zeros, and take the zeros of the zero-block to $C$, you can easily see that you get $m+n$ linear independent rows of $M$. So, $p\ge m+n$.


3

If you want a practical application of it I have one, it is indeed as they say that you take a higher dimensional object and put it into a lower dimensional one. A real world example is computer games. They are almost all generated in a 3D enviorement where they interact with polygons and all that computer stuff. However our screens are only 2D so all that ...


3

You can imagine derivative as a matrix such as this, as the corresponding values of each element approach zero (limit definition): Or integration as like this(riemann sum): Moreover, fourier transform already has a matrix representation for discrete case https://en.wikipedia.org/wiki/DFT_matrix You need extend this matrix to infinity and shrink the ...


0

It projects the input onto a lower dimensional subset. One example of a projection is the function \begin{align} P:\mathbb{R}^3&\to\mathbb{R}^3\\ (x,y,z)&\mapsto (x,y,0) \end{align} That is $P(x,y,z) = (x,y,0)$. You can visualize this function as it takes an input (a point in $3$D-space) and maps the point to the point below it on the ...


0

I always think of it as mapping an object to its shadow (like a 3D person "projected" onto the plane of the ground a 2D image). You can get different projections by adjusting the location of the light source. Of course this is just 3D-2D "intuition" but it has worked for me...


8

Yes. Equality holds iff $AB = BA$. Hint: Note that $AB - BA$ is skew-Hermitian, and that $$ 2\operatorname{trace}[(AB)^2] - 2\operatorname{trace}(A^2B^2) =\\ \operatorname{trace}(ABAB + BABA -ABBA - BAAB)=\\ \operatorname{trace}[(AB - BA)^2] $$ Note: The inequality assumes that both $\operatorname{trace}[(AB)^2]$ and $\operatorname{trace}(A^2B^2)$ are ...


1

$\newcommand{\Reals}{\mathbf{R}}$In case some theory helps clarify your strategy: Your set $W$ is the union of two planes through the origin: $$ W_{1} = \{(x, y, z) \in \Reals^{3} : x + y = 0\},\qquad W_{2} = \{(x, y, z) \in \Reals^{3} : 2y - z = 0\}, $$ since $(x + y)(2y - z) = 0$ if and only if $x + y = 0$ or $2y - z = 0$. In an arbitrary vector space, a ...


3

Since the matrix is square, it is enough to show that the rows are linearly independent. But since they are orthogonal with respect to an inner product, and each has inner product $1$ with itself, this is a standard exercise in linear algebra (we just need that for each row $v$ we have $\langle v,v\rangle\neq 0$). To elaborate a bit on the exercise, one ...


3

It is not closed under addition. For example, $(0,0,z)$, where $z$ is an arbitrary number, is in $W$. Now find another vector in $W$, for example $(1,1,2)$. Add these two together. You get $(1,1,z+2)$, again, where $z$ is arbitrary. This is clearly not in $W$, except for $z=0$.


2

Since $\text{dim Im}(f-\text{Id})+\text{dim ker}(f-\text{Id})=\text{dim} E$ by dimension theorem, it suffices to show that $\text{Im}(f-\text{Id})\cap \text{ker}(f-\text{Id})=\{0\}$. Let $v\in\text{ker}(f-\text{Id})\cap\text{Im}(f-\text{Id})$. Then $f(v)=v$ and $f(w)-w=v$ for some $w\in E$. We have \begin{align*} f(w)-w&=v\\ f^2(w)-f(w)&=v\\ ...


0

Let $y$ be the profit and $x$ be the number of cakes. Just like an equation of a line , here you have $y_1=30,x_1=25$ and $y_2=100,x_2=90$ , Now to turn that into an equation , you can just find the slope $m$ with the famous formula $$ m = \frac{y_2-y_1}{x_2-x_1}$$ and so $$ m = \frac{100-30}{90-25} = \frac{70}{65} =\frac{14}{13}$$ and now you can get the ...


1

First we prove Pythagoras Theorem for inner product space $$ \left\|\sum \limits_{i=1}^{k}a_i e_i\right\|^2=\sum \limits_{i=1}^{k}|a_i|^2 \|e_i\|^2 \hspace{5 mm} $$ $e_i$ is the orthonormal base. \begin{align} \left\|\sum \limits_{i=1}^{k}a_i e_i\right\|^2&=\langle\sum \limits_{i=1}^{k}a_i e_i,\sum \limits_{i=1}^{k}a_i e_i \rangle \\ &=\sum ...


1

Hints. The answer should be a closed disc of radius $\frac12$ centred at the origin of the Argand plane. Let $x=\pmatrix{u\\ v}\in S^1$ and $y=\pmatrix{e^{i\theta}u\\ v}$. Note that $y$ lies inside $S^1$ too. What is the relationship between $x^\ast Ax$ and $y^\ast Ay$? The numerical range of every matrix is compact and convex. So, in view of item 1, what ...


0

$\left({\begin{array} \ x_1 \\ x_2 \end{array}}\right)^T \left( {\begin{array} \ 0 & 1\\ 0 & 0 \end{array}} \right) \left({\begin{array} \ x_1 \\ x_2 \end{array}}\right),$ so we have to maximize $x_1x_2$ given that $x^Tx = 1.$ Here you can use Lagrange multiplier.


2

if the numerical range is defined as $$\{x^\top Ax : x^\top x = 1\},$$ then with $$z^\top = (z_1, z_2), z^\top Az = (z_1, z_2)(z_2,0)^\top=z_1z_2 \text{ subject to } |z_1|^2 + |z_2|^2 = 1.$$ let $$z_1 = \cos t \, e^{is_1},z_2 = \sin t \, e^{is_2}, s_1, s_2, t \text{ are real.}$$ with that the numerical range of $A$ is $$\left\{\frac12 ...


2

\begin{gather*} \begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\ = \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\ = \begin{bmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3 - a^3\end{bmatrix} \\ = (b^2-a^2)(c^3 - a^3)-(b^3-a^3)(c^2 - a^2) = ...


0

By Pythagoras, $\|v\|^2 = \|\sum_i \langle v,e_i\rangle e_i\|^2 + \|v - \sum_i\langle v,e_i\rangle e_i\|^2$.


1

$\left( \begin{array}{ccc} 1-\lambda & -1 & 2 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right) $ = $\left( \begin{array}{ccc} 2-\lambda & -2+\lambda & 0 \\ -1 & 1-\lambda & 2 \\ 2 & 2 & 2-\lambda \\ \end{array} \right) =(-2+\lambda) \left( \begin{array}{ccc} -1 & 1 & 0 \\ -1 ...


0

Let $V=\{x:f(x)=x\}$. Suppose there is $x\notin V$ such that $f(x)-x\in V$. Let $x^*$ be the projection of $x$ onto $V$. We have $$\operatorname{dist}(f(x),V)\le \|f(x)-x^*\|=\|f(x)-f(x^*)\|\le \|x-x^*\| = \operatorname{dist}(x,V) $$ But since $f(x)-x\in V$, both $f(x)$ and $x$ have equal distance to $V$. Thus, equality holds throughout. In particular, ...


1

in this case of rank one or zero matrix, it is easy to compute the eigenvalues and the corresponding eigenvectors directly. if $A = 0,$ then $AB = 0.$ clearly a singular matrix. therefore suppose $B \neq 0.$ the eigenvalues of $AB$ are $BA, 0$ and the corresponding eigenvectors are $A, B^\perp.$ zero eigenvalue implies the matrix $AB$ is singular.


0

This is simply asking you to find the trace. One way of viewing this is writing your relation (which comes from Elasticity by the way) as $$\boldsymbol{T} = \lambda\theta\boldsymbol{I} + 2\mu\boldsymbol{E}.$$ Then since for any second order tensor $\boldsymbol{A}$, the trace is defined by $$\text{tr}\left(\boldsymbol{A}\right) = ...



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