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## Hot answers tagged linear-algebra

4

To show that the set of periodic functions $\mathbb R\to\mathbb R$ is not a vector space, you need to show that the sum of two periodic functions might not be periodic. Let $f(x)$ be periodic with period $\alpha$, let $g(x)$ be periodic with period $\beta$, and let $h(x)=f(x)+g(x)$. Suppose $\beta/\alpha$ is rational and can be written as $\beta/\alpha=r/s$ ...

4

We assume the classical requirements for success of a conjugate gradient method, i.e. $A$ is symmetric positive definite. Use of a pre-conditioner with the modifications discussed below is not precluded. A modified conjugate gradient method which deals with multiple right hand sides is found in the literature, called block conjugate gradient, going back to ...

3

Any matrix can be transformed to a matrix in upper triangular form. The rank of a matrix in upper-triangular form has a rank equal to the number of non-zero entries on the main diagonal. Any sub-matrix has a subset of the main diagonal as its main diagonal. It would be there cannot me more non-zero entries on the main diagonal of the sub matrix.

2

I'm not sure if there is any elegant solution, but your conjecture can be easily proved by a continuity argument. Let $A=\pmatrix{X&Y}$ and $B=\pmatrix{Z\\ W}$, where $X$ and $Z$ are $n\times n$. Suppose that $X$ and $Z$ are invertible. Then the followings are equivalent: (for convenience, I write $g$ instead of $\gamma$) \begin{align} ...

2

The indices are different because the sums are independent from one another. If one intends to multiply the sums, then this distinction is a critical one. Suppose that we were to multiply the summations $\sum_{n=1}^2a_n$ and $\sum_{n=1}^2b_n$ and naively failed to make this distinction. Then, we would have incorrectly \begin{align} ... 2 Conjugate gradient (and similar iterative methods) don't even know that there is a matrix involved, so it's hard to see how they could give us any information about its structure. Here's an idea (which I have never tried myself) ... After you have solved a few different versions of the problem with different values of b, you have a few known values of ... 2 Yes, there are bijections, simply because of cardinality. But they're not at all smooth, and of little help with visualization. Certainly there are no linear bijections. (If you assume there is one, using the rank function you can quickly prove that 0 = 1.) Space-filling curves are continuous surjections from the unit interval or the reals \Bbb R to ... 2 P(x)=ax^3+bx^2+cx+d\begin{aligned} p(1)&=a+b+c+d \\ p(-1)&=-a+b-c+d \end{aligned} $$so p(1)-p(-1)=0$$ p(1)-p(-1)=a+b+c+d -(a+b-c+d)=2a+2c$$2-d plane like x-y axis your basis does work \{x^2-1,x^3-x\} feel that grader would not like how you show it is a subspace. Do not use lambda it means to grader e-value. I seen a peer get ... 2 Sorry, but your line of thought is quite wrong. Your part B is the worst one. Apart from using x_1 and x_2, which is already bad, you can't prove the set is closed under addition by showing that the sum of two particular element is in the set. You need to show that for any choice of polynomials p,q\in W, the sum p+q\in W. Similarly for the closure ... 2 Let w \in Span(u,v) lets check whether w also a solution to Ax=0. w \in Span(u,v) \ so:$$w = au + bv \ s.t \ a,b \in F  than \ Aw = A(au+bv) = Aau + Abv = aAu + bAv = a*0 + b*0 = 0 + 0 = 0.$$so , w also a solution. 1 Your thought process is correct, you are also right in finding a base that has 2 elements: the space W is 2-dimensional. You can show this by writing the general element p = ax^3+bx^2+cx+d and using p(1)=0=p(-1) on it, to get a system of linear equations that yield a = -c, b = -d and thus rewrite the space as \{ax^3+bx^2-ax-b\} (abusing notation a ... 1 The idea is that if f is a-periodic and g is b-periodic, and \frac{a}{b}\in \mathbb{Q}, then it's easy to see that f+g is periodic : if \frac{a}{b} = \frac{p}{q} with p,q\in \mathbb{N} then put c = qa=pb. Since f is a-periodic, it's also c-periodic. Likewise, g is c-periodic since it's b-periodic. Thus f+g is c-periodic. ... 1 That's correct. If Ax=b has a unique solution for any b, then the columns of A are linearly independent. If they were linearly dependent then, on the one hand, the columns of A wouldn't span all of \mathbb{R}^n and there would be a solution for every b; on the other hand, for those unlikely b for which there is a solution, the solution would ... 1 This is not an answer, and don't take it as such. From a theoretical point of view, you have A=\sum_{k=1}^{n}\lambda_k P_k where the P_k are orthogonal projections. If you start iterating in A, or perform various functions of A, the problem is that f(A)=\sum_{k=1}^{n}f(\lambda_k)P_k. You can get down below the level of a projection. You can form ... 1 Your assertions about the Jordan form of T are correct. That is, the Jordan form is$$ \left[\begin{array}{rrrrr} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5 \end{array}\right] From here we see that \begin{align*} ... 1 Since T^2=T, T^2-T = 0. Therefore T satifies the polynomial p(x) = x^2-x = x(x-1). The minimal polynomial of T must divide p(x). Let m(x) be the minimal polynomial. Then m(x) = x or m(x) = x-1 or m(x) = p(x). In any of these cases, the eigenvalues of T are the roots of m(x) are among 0 and 1. So taking the sum of the eigenvalues ... 1 Hint Denote e_1,e_2 the two given vectors and complete the family (e_1,e_2) on an arbitrary basis (e_1,e_2,e_3,e_4,e_5) of \Bbb R^5. Now use the Gram-Schmidt process to construct an orthogonal basis (v_1,\ldots,v_5) and then U^\perp=\operatorname{span}(v_3,v_4,v_5). 1 Hint: Instead of using Gram-Schmidt, you can write the conditions that a vector v is orthogonal to the two given vectors as a system of equations. Then, by reducing the system of equations, you have the answer you're looking for. 1 \mathcal A(n) and \mathcal S(n) are stable by transposition. Let's denote by f_A and f_S the restrictions of f to \mathcal A(n) and \mathcal S(n) respectively.Transposition is the direct sum f_A\oplus f_S=(-\operatorname{Id}_{\mathcal A(n)})\oplus \operatorname{Id}_{\mathcal S(n)}, hence\det f=\det(-\operatorname{Id}_{\mathcal ...

1

For example in $\mathbb{R}^3$ the set $B=\{(1,1,1),(1,1,0),(1,0,0)\}$ it is a basis for this vector space, and therefore is a generator system, but as you say in the comments, adding any other vector of the vector space, this set will remain generator system because it contains a base, for example the set $B'=B$ \$\cup\{{(2,-4,1)}\}, is clearly not ... 1 Orthornormality refers to the basis e_i. When a basis is orthonormal it means the inner product between any two elements of the basis e_i,e_j is \langle e_i, e_j \rangle = \delta_{ij} (see kronecker delta). More generally, two vectors u,v are orthogonal if \langle u, v \rangle = 0. The normality part comes from elements of the basis having norm ... 1 That's because it is notationally bad and depending on how you read it, it may even give different results. For example, consider simple vectors in \mathbb{R}^{3} given by v=a_{1}e_{1}+a_{2}e_{2}+a_{3}e_{3}=\sum_{n=1}^{3}a_{n}e_{n}. Then if we use same index, \begin{aligned} ... 1 The two vectors that you provided are already orthogonal to each other, that is their inner product is 0. You just need to divide each of them by their norm. In general, check out Gram-Schmidt process. 1 Use the Gram-Schmidt procedure. This procedure is very well known, so I'll just link you to a treatment of it that will help you: https://www.math.purdue.edu/academic/files/courses/2010spring/MA26200/4-12.pdf 1 Presumably, this is with respect to the "standard" inner product. Perform a single Gram-Schmidt step. Ifa=\begin{bmatrix} 1\\2\\3\\-1\\2\end{bmatrix}\text{ and } b=\begin{bmatrix}1\\0\\-1\\0\\1\end{bmatrix}$$then$$a^*=a\text{ and } b^*=b-\frac{a^*\cdot b}{a^*\cdot a^*}a^*$$or$$a^*=\begin{bmatrix} 1\\2\\3\\-1\\2\end{bmatrix}\text{ and } ...

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