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2

consider on this $$ \begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 3 \\ \end{bmatrix} $$


1

No. The simplest example is any upper triangular matrix whose diagonal entries are real: $$\begin{pmatrix} 1 & 1-i & 2 \\ 0 & 2 & 3i \\ 0 & 0 & 3 \end{pmatrix}$$ has complex entries in the non-diagonal spots but its eigenvalues are $1,2,3$ and it is certainly not symmetric!


1

Is the matrix $$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$ symmetric? It has only one positive eigenvalue of multiplicity two.


0

Many have already said it, the problem is unbounded. I want to just give an explicit example: For arbitrary $x,y\ne 0$, without loss of generality norm 1. Extend $x$ and $y$ to orthonormal bases $X$ and $Y$. For $n\in\mathbb N$ define $$ D_n = \operatorname{diag}(n, 1/n, 1, \dotsc, 1) $$ and $$A_n = X D_n Y^T. $$ Then, we have $$|\det(A_n)| = |\det ...


0

This expression attains no maximum or minimum for $n \geq 2$ and non-zero $x,y$. We can prove that this is the case as follows: For any $x,y$ and $C \neq 0$, we can construct an $A$ such that $Ay = Cx$ and $\det(A) = 1$. For this $A$, we have $$ K(A) = x^TAy = x^T(Cx) = C\|x\|^2 $$ Thus, $K(A)$ can be made arbitrarily large or arbitrarily small.


1

This is a bit of a trick question, because it’s impossible to express $X+Y+Z$ in terms of $A$ and $B$. To see this, let $X$ be any real number you like, and set $Y=A-X$ and $Z=B-X$. Then clearly $A=X+Y$ and $B=X+Z$, but $$X+Y+Z=X+(A-X)+(B-X)=A+B-X\;.$$ Since $X$ can be any real number at all, so can $A+B-X$. Thus, the relationship of $X+Y+Z$ to $A$ and $B$ ...


0

When first learning modular arithmetic, you learn how to add, subtract, and multiply congruence classes of integers mod $n$ --- just as you learned basic arithmetic on numbers when you were in elementary school. There is no deep relationship between linear combinations here. A bit later, you begin to look at division. This takes a bit a care, since in ...


0

You need to know $Y+Z$ in terms of $A,B$ in order to find $X+Y+Z$ in terms of $A,B$.


1

The oriented volume spanned by three vectors is a multilinear alternating map of the vectors themselves. That's the starting point. See, e.g., the book by S.Winitzki.


0

I believe I found a proof that the answer is indeed "no", so I'll post it here to avoid misleading future readers. The idea is that no wedge is big enough to contain vectors that are perpendicular to each other, if we aren't allowed to use the $(1,...,1)$ direction. We prove the following equivalent statement: if $V$ is a subspace of $\Bbb R^r$ not ...


3

There is no real solution for the first question $1^x = 2$ Not even a complex number. For the second part: Let $x = a+ib$ and $e^{(2n+1)i\pi} = -1 , n \in \ Z $ so, $2^{a+ib} = 2e^{(2n+1)i\pi}$ $2^{a-1 + ib} = e^{(2n+1)i\pi}$ Take log and you will get, $(a-1+ib)\log2 = (2n+1)i\pi$ Equating real and imaginary part, $a = 1, b = \frac{(2n+1)\pi}{\log2}$ ...


0

Hint. we have $x^TAy=\langle x,Ay\rangle$ and this scalar product has is maxim value wan $Ay$ is parallel to $x$.


0

For your expression we have $$ K(A) = x^t A y = \sum_{i,j} x_i a_{ij} y_j = \sum_{i,j} a_{ij} x_i y_j $$ which is a quadratic form. You can make $K(A)$ arbitrary large by choosing large matrix elements, there is no bound, no maximum. Update: A constraint restricting $A$ to $\lvert \mbox{det }(A) \rvert \le 1$ was added to the problem. This will not ...


0

Let $x = (x_i)_{i=1}^n$ and $y = (y_i)_{i = 1}^n$ be vectors in $\Bbb R^n$. You can think of the situation as $K: \Bbb R^{n \times n}\to \Bbb R$, given by: $$K(a_{11},\cdots,a_{1n},a_{21},\cdots, a_{nn}) = \sum_{i,j=1}^n a_{ij}x_iy_j,$$and now solve the system: $$\frac{\partial K}{\partial a_{ij}} = 0, \quad 1\leq i,j \leq n$$ to find the critical points, ...


0

I am replying to Miller Zhu's comment (2nd comment to question), but the comment box doesn't have enough space. I'm only using the inscribed ellipsoid to get a sense of the principal axes of the polytope. Since you don't have access to the vertices of the polytope, computing the inscribed ellipsoid is the only computationally tractable problem (the ...


0

I that think the key to answering this question is to recognize that Im(A) is a subspace of ℝ^m. It means that for a given b ∈ ℝ^m, there exists a point of Im(A) closest to b; namely, the orthogonal projection of b onto Im(A). We can write this projection as Ax, for some x ∈ ℝ^n, by the definition of Im(A). Thus, we have the result that for any other point ...


1

You should seem $(e_1+e_2,\cdots,e_n)$ and $(\phi(e_1+e_2),\cdots,\phi(e_n))$ . since $\phi(e_1+e_2)\in E^{**}$ then $1=\phi(e_1+e_2)(e_1+e_2)=\phi(e_1)(e_1+e_2)+\phi(e_2)(e_1+e_2)=\cdots=1+1$. this is a contradiction.


1

For a matrix $X$ of rank $r$ I will call the right hand side of $$X = a_1b_1^T + \dots + a_rb_r^T$$ a $decomposition$ of $X$. I will call $a_ib_i^T$ a term in the decomposition. Any rank 1-matrix will be a matrix where a submatrix is made entirely out of ones and the rest of the entries are zeros. A submatrix $B'$ of the matrix $B$ is a matrix where we pick ...


1

Let $Q$ be a real symmetric and idempotent matrix of "dimension" $n \times n$. First, we establish the following: The eigenvalues of $Q$ are either $0$ or $1$. proof. note that if $(\lambda,v)$is an eigenvalue- eigenvector pair of $Q$ we have $\lambda v=Qv= Q^{2} v=Q(Qv)=Q(\lambda v) = \lambda^{2} v$. Since $v$ is nonzero then the result follows ...


1

By distributing {length} you get {length} * {ppf} + {length} * {jacket_price} + {price} + {c1_price} + {c2_price} + {pe_price}


1

Yes your understanding is correct. To apply to the example you gave. You have $u=(1,-4,2,-3), v=(-3,8,-4,6)$. Hence $2u+v=(-1,0,0,0)$ by elementary operation. Finally $(u, v, e_2, e_3)$ is a basis of $\mathbb R^4$


3

Question: Prove that $(666\dots \text{to $n$ digits})^2 + (888\dots \text{to $n$ > digits})=(444\dots \text{to $2n$ digits})$ By dividing by $4$, this is equivalent to \begin{align} (333\dots \text{to $n$ digits})^2 + (222\dots \text{to $n$ digits}) &=(111\dots \text{to $2n$ digits}) \end{align} Let $x=111\dots \text{to $n$ digits}$. Then ...


0

First of all, note that $\lambda+3$ should be in parentheses in your expression for the determinant. Next, observe that, for example, $$(\lambda+4)(\lambda-6)+24=\lambda^2-2\lambda-24+24=\lambda^2-2\lambda.$$ Similarly, you should simplify the quantities within each other pair of parentheses. Next, expand all the products and gather like terms. Then you can ...


2

$(λ+3)((λ +4)(λ -6)+24)+2(-3(λ -6)-24)-1(24-8(λ +4)) $ $=(λ^3+λ^2-30λ-72)+24(λ+3) +2(-3λ-6)-(-8-8λ)$ $=λ^3+λ^2-30λ-72+24λ+72-6λ-12+8+8λ$ $=λ^3+λ^2-4λ-4$ $=(λ+1)(λ^2-4)$Since we can spot $λ=-1$ as a root $=(λ+1)(λ-2)(λ+2)$


0

You can use $\begin{vmatrix}\lambda+3&-2&-2\\-3&\lambda+4&3\\8&-8&\lambda-6\end{vmatrix}=\begin{vmatrix}\lambda+1&-2&-1\\\lambda+1&\lambda+4&3\\0&-8&\lambda-6\end{vmatrix}=(\lambda+1)\begin{vmatrix}1&-2&-1\\1&\lambda+4&3\\0&-8&\lambda-6\end{vmatrix}$ $\hspace{.7 ...


0

I ended up ordering Handbook of Linear Algebra, 2nd Ed by Hogben. Large tome, but great index and binding.


1

A gradient $$ \mbox{grad } \Psi = \nabla \Psi $$ is either zero or orthogonal to the level curves $$ C: \Psi = c = \mbox{const} $$ in particular to the tangent vectors $$ t = \frac{dr}{ds} $$ along $C$. The function $\Psi$ is constant along a level curve $C$, thus $d\Psi = 0$. Using the relation between total differential, gradient and displacement we ...


3

View this as an instruction guide (All steps are literally trivial): 1) Show that $\operatorname{Ker}(f)$ is one-dimensional. 2) Take any $v$ with $f(v) \neq 0$. Deduce from 1) that we have $\operatorname{Ker}(f)=\langle f(v) \rangle$. 3) From 2) we deduce that $v \notin \langle f(v) \rangle$. This is just another formulation of the fact that $v$ and ...


0

The answer in brief: Two matrices are similar if and only if they are similar to the same Jordan canonical form matrix. An equivalent statement, if you are unfamiliar with Jordan form, is that the matrices $A$ and $B$ are similar if and only if for all $\lambda \in \Bbb C$ and positive integers $k$, we have $$ \dim \ker[(A - \lambda I)^k] = \dim \ker[(B - ...


-1

Given $X$, $Y$ and $Z$, composition of homomorphisms can be viewed as a map $$ Hom(Y,Z)\otimes Hom(X,Y) \to Hom(X,Z).$$ This map is surjective if $Y$ has dimension large enough (namely, has dimension at least $\min(\dim X, \dim Z)$). Thus $Hom(X,Z)$ is a quotient of $Hom(Y,Z)\otimes Hom(X,Y)$. However, since we work with vector spaces, every surjection ...


1


4

While the other answers lead you to the precise order of $GL_n(\mathbf{F})$, I'll explain a simple way to see the inequality you asked about. As you note, any $n \times n$ matrix has $n^2$ entries; if these entries are coming from $\mathbf{F}$, then the set of all $n \times n$ matrices has order $q^{n^2}$, since there are $q$ choices for each of the $n^2$ ...


0

Hint: If $A$ and $B$ have the same Jordan form $J$, then $$A = PJP^{-1}$$ $$B = QJQ^{-1}$$ and therefore $$B = Q(P^{-1}AP)Q^{-1} = (PQ^{-1})^{-1}A(PQ^{-1})$$ and $B$ is similar to $A$.


0

Hints: (1) $\text{lcm}(a,N)=\text{lcm}(a,N)$. (2) If $\text{lcm}(a,N)=\text{lcm}(b,N)$, then... (3) If $\text{lcm}(a,N)=\text{lcm}(b,N)$ and $\text{lcm}(b,N)=\text{lcm}(c,N)$, then...


0

Reflexive: $\text{lcm}(a,N) = \text{lcm}(a,N) \Rightarrow aRa$. Symmetric: $aRb \Rightarrow \text{lcm}(a,N) = \text{lcm}(b,N) \Rightarrow \text{lcm}(b,N) = \text{lcm}(a,N)\Rightarrow bRa$. Transitive: $\begin{cases}aRb & \\ bRc &\end{cases} \Rightarrow \begin{cases}\text{lcm}(a,N) = \text{lcm}(b,N)& \\ \text{lcm}(b,N) = \text{lcm}(c,N) & ...


0

Every time starting with $66^2$ adds another 4 to the beginning and a five between the 3 and 6. Also, the number of eights is half the length of $6666...^2$. If this pattern continues, there are (n-1) 4's, a three, (n-1) five's, and finally a six. Because you add n 8's, the number becomes n 4's.


10

Hint: $n$ digits of $6$ is equal to $$ \left(6\left(\frac{10^n-1}{9}\right)\right)^2 = \frac{4}{9}(10^n-1)^2 = \frac{4}{9}(10^{2n}-2\cdot 10^n+1).$$ Apply a similar transformation to the term with $8$'s, then see if you can simplify the sum.


3

In this solution, we only assume that $B$ is diagonalizable (i.e., the eigenvalues $\lambda_i$'s need not be distinct). If $v_1,v_2,\ldots,v_n$ are the eigenvectors of $B$ and $w_1,w_2,\ldots,w_n$ are the left eigenvectors of $B$, where $Bv_i=\lambda_iv_i$ and $w_i^\top B=\lambda_i w_i^\top$ for $i=1,2,\ldots,n$. Then, $$ \begin{align} F\left(v_i ...


5

Let $e_i$ be the eigenvectors of $B$, i.e. $$Be_i=\lambda_ie_i$$ and let $E_{ij}=e_i e_j^T$ be the elementary matrices in this basis, i.e. $$E_{ij}e_k=\delta_{jk}e_i.$$ As it turns out, $F$ is already diagonal in the $E_{ij}$-basis: \begin{align} &BE_{ij}e_k=\delta_{jk}Be_i=\lambda_i\delta_{jk}e_i=\lambda_iE_{ij}e_k\\ ...


0

The asymptotes correspond to the roots of the quadratic polynomial in $x_1,x_2$, as the variables will strive to "achieve $0=4$". By the classical formulas, $$x_1=\frac{6\pm2\sqrt{10}}{4}x_2.$$ The rotated axis are the bisectrices.


0

Hint Quoting Wikipedia : To rotate a figure counterclockwise around the origin by some angle $\theta$ is equivalent to replacing every point with coordinates $(x,y)$ by the point with coordinates $(X,Y)$, where $$x=X \cos (\theta )-Y \sin (\theta )$$ $$y=X \sin (\theta )+Y \cos (\theta )$$ So, transform accordingly your equation $4x^2-12xy-y^2=4$; ...


0

Let $U=\begin{pmatrix}u&v\\0&w\end{pmatrix}$. Then $$UU^T=\begin{pmatrix}u&v\\0&w\end{pmatrix}\begin{pmatrix}u&0\\v&w\end{pmatrix}=\begin{pmatrix}u^2+v^2&vw\\vw&w^2\end{pmatrix}$$ We see that $$w=\sqrt c$$ $$v=\frac b{\sqrt c}$$ $$u=\sqrt{a-\frac{b^2}c}$$ We should also show that all those radicands are nonnegative, but, ...


1

Given the cholesky decomposition $LL^T = DHD$ for some positive definite diagonal matrix $D$, another positive definite diagonal matrix $\tilde D$ and a vector $u$. You can compute the decomposition $$\tilde L \tilde L^T = \tilde D (H+uu^T) \tilde D = \tilde D H\tilde D + (\tilde D u)(\tilde D u)^T$$ as follow: Let $\bar L = \tilde D D^{-1} L$, which ...


0

Those three linear independent vectors span $\mathbb{R}^3$, and the orthogonal complement is $\{0\}$.


1

If $A=(a_{ij})_{i,j}\in GL_n(R)$, and $B=(s_{ij})_{i,j}\in M_n(S)$, then $AB=(\sum_{k}a_{ik}s_{kj})_{i,j}$. Suppose $\sum_{i,j}\alpha_{ij}(\sum_{k}a_{ik}s_{kj})=0$ with $\alpha_{ij}\in R$. Then $\sum_{i,j,k}\alpha_{ij}a_{ik}s_{kj}=0$ and this can be written as $\sum_{j,k}(\sum_i\alpha_{ij}a_{ik})s_{kj}=0$. It follows $\sum_i\alpha_{ij}a_{ik}=0$ for all ...


0

Hint- use the fact that if $W_{1}$ and $W_{2}$ are two subspace of a vector space $V$ then $W_{1}\cap W_{2}$ is the largest subspace of $V$ contained in $W_{1}$ and $W_{2}$ and $W_{1}+W_{2}$ is the smallest subspace containing both $W_{1}$ and $W_{2}$.


1

Look. I'm not a mathematician, but I have a perspective which can explain why the cross product of two vectors is another vector perpendicular to them. It is not a proof but it will help make that idea fimilar. One can understand cross product in this way:imagine a line segment that makes colorful marks wherever it moves on a paper. Now make it move in the ...


2

Let the matrix be denoted by $E$, and the $i,j$ entry by $E_{i,j}$. Then we can write: $$E_{i,j}=\delta_{i,j}$$ If $A$, $B$ are two matrices, the elements of the product are denoted by $$(AB)_{i,j}=A_{i,j}B_{i,j}$$ If you write these two down definitions, it seems to come out quite quickly.


1

Hint: $$ \DeclareMathOperator{\tr}{tr} \tr(A^TJB) = \tr([A^TJB]^T) = \tr(B^TJ^TA) = -\tr(B^TJA) $$


1

A basic of $V$ is $(1,1+1)$. Indeed, every vector $(a,1+a) \in V$ can be written $$(a,1+a)=a*(1,1+1).$$ In particular, $V$ is the vector space of dimension $1$. In the other words, $V \simeq \mathbb{R} $ (note that $V$ is a vector subspace of $\mathbb{R}^2$).



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