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## New answers tagged linear-algebra

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For the other direction, if $x$ is orthogonal to $V$, then it is orthogonal to every vector in $V$. You may choose a particular vector $v \in V$, and $x$ will be orthogonal to it. Namely, you can choose each of the basis vectors for $v$. Since $v_i \in V$, and $x$ is orthogonal to every vector in $V$, then $x$ is orthogonal to $v_i$ for all $i$.

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From one high-schooler to another, I would say there are several areas of Linear algebra that can easily be explored. From a purely mathematical stand point here is what I would suggest (in increasing order of difficulty): You can explore the basic matrix [1] operations. These are probably covered in Algebra II or Pre-calculus in your high school. Khan ...

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Write $c := \cos\theta$, $s := \sin\theta$, $\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and $T := T_\mathbf{u}\left(T_\mathbf{v}\right)$, so that we have ... \begin{align} T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf{v})\mathbf{u}\\ R(\mathbf{x}) ... 0 You have the right idea. The k-dimensional subspaces of a diagonalizable linear operator can be found by taking the span of any k eigenvectors. Note that in this particular case, the eigenvectors are \pmatrix{1&0&0}^T,\pmatrix{0&1&0}^T, and \pmatrix{0&0&1}^T. 2 However, if K = \mathbb{Q}, the problem is different, I don't think I can construct A for every m. This is correct. the following result is (I think) well known Let m = p_1^{\alpha_1}\cdots p_k^{\alpha_k} be the prime factorization with p_1<\cdots < p_k. The group \mathrm{GL}_n(\mathbb Q) has an element of order m if and only if ... 0 You should review the rules for taking derivatives, specifically the chain rule. (Either that, or expand all squares in F, but this is more algebraically painful.) The basic idea is that the derivative of the square of some function g should be computed as (g^2)'=2 g g'. For example, the derivative of (3x-7y+5)^2 with respect to y is ... 1 Hint: We can rewrite v - T(v) = (I - T)v $$Where I is the identity matrix. Now, show that$$ T(v - T(v)) = \cdots = -(v - T(v)) $$0 Suppose v-T(v) is not 0_V. Then T(v-T(v))=T(v)-T(T(v))=T(v)-v (since T^2 is the identity operator). In particular, T(v-T(v))=-(v-T(v)), which shows that v-T(v) is an eigenvector (since it is non-zero) with eigenvalue -1. 0 Let S be a sample covariance matrix generated by data matrix X. Then S=\frac{1}{n} XX^T. The dual covariance matrix is given by S^D=\frac{1}{n} X^TX. 0 I don't have enough reputation points to add a comment so, I'll restate Q1 and then offer a partial answer / hint. 1. True or False: If the null space of a 5x4 matrix has dimension 2, then the column space can be isomorphic to a line in \mathbb R^4 . Let's pick it apart and reason through it a bit. Suppose matrix A is a 5 by 4 matrix. In accordance ... 2 A few ideas: (1) Numerical Stuff: Look at various methods of solving linear systems or inverting matrices. Study performance (the number of operations involved), and what sorts of things can go wrong numerically. Show that the naive textbook methods don't work very well in practice. Why "A Math Book Is Not Enough" (Forsythe) (2) Relations to 3D Geometry: ... 2 We had to learn a bit of Matrices in High School. And actually in my opinion a high school student should be able to tackle most parts of matrix algebra and some part of solving system of linear equations. Although I would advice against delving deeply into the the theory, vector spaces and linear transformations etc. Although there are not many books that ... 2 You could probably discuss the row reduction algorithm for solving systems of linear equations. It works quite demonstrably. The deep concept underlying this process is that every matrix has a unique reduced echelon form. You could motivate this through examples. Another related result is that the elementary n\times n matrices giving the row operations ... 2 The mistake is that \dim U_1\ne 1. In particular,$$ U_1=\{(x,y,0):x,y\in F\}=\mathrm{span}\,\{(1,0,0),(0,1,0)\}, $$and hence$$ \dim U_1=2. $$2 The vector space structures on T_aM and T_{f(a)}N are defined so that \psi : T_aM\to \mathbb{R}^n and say \phi : T_{f(a)}N\to \mathbb{R}^m are linear isomorphisms. It is thus enough to show that \phi\circ f_{\ast a}\circ \psi^{-1}: \mathbb R^n\to\mathbb R^m is linear. By definition of the differentiability of f the last map is linear, with matrix ... 1 Your contrapositive is correct, and the thing to be shown is that, given A+\delta is singular, then either A is singular or the norm product is at least 1. Now note that [P or Q] is the same as [(not-P) implies Q], so in proving what you want, you can assume "not-(A is singular)" in the proof, and then proceed to attempt to show the norm product is ... 0 Let A is a required 3x3 matrix and from the problem we have A [w1 w2 w3] = [f(w1) f(w2) f(w3)] Therefore A = [f(w1) f(w2) f(w3)]*[w1 w2 w3]^-1 find [f(w1) f(w2) f(w3)] = (\left[\begin{matrix}0&0&5\\1&-1&11\\0&0&2\end{matrix}\right]) and [w1 w2 w3]^- = ... 2 You can add or remove any row to any other (not the same) row, but in the first step you replace R3 by -R3 (+R1 but you can add). So by negating a row, you negate the determinant. 0 If you have an SVD representation of the matrix, you can use this to compute the product in linear time. I'm not sure how to efficiently compute this representation, though. You can also truncate the SVD of a higher-rank matrix to get a low-rank approximation. 0 You should compute the determinant of the matrix, and if det(A)\neq 0 then the columns of the square matrix are independent and therefore they construct a basis. 1 Let \{v_{i}\}_{i=1}^k be a set of k vectors in \mathbf{R}^n. By "sum of all pairs of inner-products", presumably you mean something like$$ \sum_{i<j} \langle v_{i}, v_{j}\rangle, $$and by "sum of Euclidean distances between all pairs" you mean$$ \sum_{i<j} \|v_{i} - v_{j}\|. $$Consider what happens with two vectors. Since you're asking about ... 2 It is the definition of those continuous functions. It is not a result that can be derived. The function \alpha f is defined to be the function which takes x as input from the domain and spits out the value \alpha f(x) in the codomain. Similarly for the function f+g. Proving that these functions are continuous (and hence belong to the domain V) so ... 3 Note that any eigenvalue \lambda of A satisfies Av = \lambda v \tag{1} for some vector v \ne 0, so that A^2 v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda(\lambda v) = \lambda^2 v, \tag{2} whence 0 = (A^2 + I)v = A^2v + v = (\lambda ^2 + 1)v, \tag{3} and since v \ne 0 this implies \lambda^2 + 1 = 0; \tag{4} but equation (4) has no ... 2 If Av=\lambda v then A^2v=\lambda^2v but on the other side -Iv=-v so we get \lambda^2=-1 which has no real solution. 0 Using the definition of a norm, the distance between two vectors d(A,B)=||A-B||. Source: http://www.math.vanderbilt.edu/~msapir/msapir/feb5.html 1 As I said in my comment, in a convex optimization setting, one would normally not use the derivative/subgradient of the nuclear norm function. It is, after all, nondifferentiable, and as such cannot be used in standard descent approaches (though I suspect some people have probably applied semismooth methods to it). Here are two alternate approaches for ... 1 The short SVD of uu^* is v\sigma v^* with v=\frac{u}{\|u\|} and σ=\|u\|^2. If you compute the bisector w=u+\frac{u_1+0}{|u_1|+0}\|u\|e_1 of u and e_1, then you can get a full unitary matrix V as the reflection matrix I-2\frac{ww^*}{\|w\|^2}, so that uu^*=V\Sigma V^* 0 The basis of all polynomials of degree \le n is {1,x,x^2,...,x^n} no matter of the field of the scalars. It's so because the 'coordinates' of the 'vectors' (read polynomials) are a linear combination of the set {1,x,x^2,...,x^n} where the scalars before any element of that set are from a given field (real, complex or any other). 1 When you go for (G−2I)v=2v,to find v as eigen vector you will no longer be finding eigen vector corresponding to that e-value: since this implies Gv-2v=2v or Gv=4v 2 You must solve (G-\lambda I) = 0. The equation you have written is (G-\lambda I) = \lambda I If you write the correct equations, you will get:$$ \frac{-3v_1}{4} + \frac{\sqrt{3}v_2}{4} = 0\\ \frac{\sqrt{3}v_1}{4} - \frac{v_2}{4} = 0\\ 0 = 0 $$0 Denote$$ A = \begin{bmatrix} 0 & \sqrt{2} & 0 & \sqrt{2} \\ \sqrt{2} & 0 & \sqrt{2} & 0 \end{bmatrix} $$Step 1: Find the spectral decomposition of AA^*. That is, find a diagonal matrix D and unitary matrix U for which$$ UDU^* = AA^* $$To get the diagonal entries of \Sigma, take the square root of the entries of D, so ... 0 Projecting onto v=[1;1;1;1;....] will always product a vector that is a scalar multiple of v, so all the entries of the projection will be equal. Second, the average of a vector x of dimension N is \frac {x \circ v} N (using \circ for dot product). So we wonder if the projection's entries are equal to that. The projection p has two defining ... 0 A similar to B => P with B = P^{⁻1}AP => P with B^T = (P^{⁻1}AP)^T => P with B^T = P^TA^T(P^{⁻1})^T => Q=(P^{-1})^T with B^T = Q^{⁻1}A^TQ => A^T similar to B^T 1$$A\sim B\iff\,\exists\;\text{invertible}\;\;P\;\;s.t.\;\;A=P^{-1}BP\iff A^t=\left(P^{-1}BP\right)^t=P^tB^t\left(P^{-1}\right)^t$$and since \;\left(P^{-1}\right)^t=\left(P^t\right)^{-1}\; we're done 0 Hint: A and B being similar means that A=P^{-1}BP for some P. What happens if you transpose both sides of this equation? 1 As long as you are not obliged to use your formulae for T_{\bf u}({\bf x}) and so on, there is a very easy approach to this. Any reflection in {\Bbb R}^3 is represented by an orthogonal matrix of determinant -1. Any rotation in {\Bbb R}^3 is represented by an orthogonal matrix of determinant +1. Multiply two of the former and you get one of the ... 0 Hint: By induction it can be shown that f(n)=3^n 0 Using properties of adjoints: \langle x, y \rangle = \langle x, (U^* U)y \rangle = \langle Ux, Uy \rangle. By setting x=y we get \|Ux\| = \|x\|. If v is a unit eigenvector corresponding to an eigenvalue \lambda, we have 1 = \|v\| = \|Uv\| = |\lambda| \|v\| = |\lambda|. 1 The first question has virtually been answered by Max. Concerning the second one. Suppose $$Ux = \lambda x \hspace{5mm}\Rightarrow\hspace{5mm} x'U' = \lambda x'$$ so that $$||x||^2 = x'x = x'U'Ux = ||Ux||^2 = ||\lambda x||^2 = |\lambda|^2||x||^2 \hspace{5mm}\Rightarrow\hspace{5mm} |\lambda| = 1$$ 0 Thanks for those kindly person answered or commented on my question. It's helpful. I find 2 ways to solve my problem. 1.The RV coefficient. Take each column of the matrix as an independent realization of a random vector. So, if I want to calculate matrix A_1 and A_2, where A_1 \in R^{n*k},A_2 \in R^{m*k}, m,n \in N^+, I turn this problem into ... 0 assuming that by "r2" you mean \mathbb{R}^{2} I suggest you to take for instance \mathbb{R}\left(1,0\right) and \mathbb{R}\left(0,1\right). 0 You mean like \mathbb R \times \{0\} and \{0\}\times \mathbb R? And \operatorname{Span}\{(1,0,0),(0,1,0),(1,1,0)\}=\mathbb R\times\mathbb R\times \{0\}, but that spanning set is not a basis for that subspace of \mathbb R^3 since it is not linearly independent. These examples are trivial. 0 Doubling each speed is the same as doubling the relative velocity. The relative velocity starts at 15, so the next hour it will be 30, and the next 60. Since we want 90 to be traveled between the both of them, we know it's after 2 hours (45 km) and before 3 (105 km). So after 2 hours, they have 45 km left to go, traveling at 60 km/hr. ... 0 HINT: So, on cumulative effort, they travel together$$5+10=15,2(5+10)=30,2\cdot30$$So, after two hours they travel 15+30=45 KM leaving 90-45=45KM to be covered at a total speed =60KM/hour 1 Your procedures are rather non-standard in terms of finding basis for columnspaces and rowspaces. In particular your second procedure for finding the rowspace is not only unorthodox, but appears to be incorrect. To fully answer all of your questions takes a bit of time, and I apologize in advance for the length. The fact is, you can find both basis in a ... 1 This type of idea reminds of something I stumbled on quite some time ago. I will outline a bit of the connection that I see here, but it might not be exactly what you are looking for. There is an old construction from Tensor analysis that in some sense generalizes the determinant on the level of tensors. It is a monstrous little creature of sorts but it ... 0 Illustrating the example by Shuhao Cao:$$v(x,y) = (x,y) ;\qquad u(x,y) = (-y,x)$$Here \operatorname{div} v\equiv 2 and \operatorname{div} u\equiv 0. The two fields are related by rotation by 90 degrees: red has divergence, blue doesn't. 2 (Edit: The OP has modified the definition of f. The following answer no longer applies.) "It's easy to show that f(A) does not depend on the row of the expansion." Are you sure? Consider A=\pmatrix{1&1\\ 1&-1}. Expand along the first row, we get f(A)=2. Expand along the second row, we get f(A)=0. Your definition of f does depend on ... 2 Let's say more about this linear transformation. First$$A\in \ker t\iff t(A)=A+A^T=0\iff A^T=-A\iff A\in\mathcal{AS_n}(\Bbb R)$$hence 0 is an eigenvalue of t with multiplicity equal to$$\dim\mathcal{AS_n}(\Bbb R)=\frac{n(n-1)}2:=\alpha_n$$and if A\in\mathcal{M_n}(\Bbb R) then$$t(A)=A+A^T\in\mathcal{S_n}(\Bbb R) and by the rank-nullity theorem ...

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No, the kernel is not zero: There are matrices such that $A^T = -A$, called skew-symmetric. Try a matrix with zeros on the main diagonal. To show that the range of $t$ is the collection of symmetric matrices, you need to check two things: For any $A$, $t(A) = A + A^T$ is symmetric. For any matrix $Q$ for which $Q^T = Q$, there is a matrix $A$ such that ...

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