dragonbox algebra

1.0 and 1 (duality:0--1,--+)X + 0 = x, x 1 = XX + 1 = 1, x 0 = 02. idempotentx + x = x, x x = X3. involution(x ') ' = X4. Complementarityx + x ' = 1, x X ' = 05. Commutativex + y = y + X, xY = y X6. Associative(x + y) + z = x + (y + z) = x + y + Z(xY) ·Z = x(YZ) = xY Z7. DistributiveT(Y + Z) = xY + x ZX + (yZ) = (X + Y) · (X +z)8. SimplificationTY + xY ' = x, (x + Y) · (x + Y ') = XX + xY = X, x (x + Y) = X9. Multiplying and factoring(X + Y) ·(X ' + Z) = xZ + X ' · YTY + X ' ·z = (X + z) · (X '

\ \end{pmatrix}\to \begin{pmatrix}0 0 \ 0 A + B \ \ \end{pmatrix}.$$ Because the rank of the matrix does not change under the elementary transformation of the block, the basic formula of rank can be obtained $ $r (A) +r (b) =r (a^2) +r (b) =r\begin{pmatrix} a^2 0 \ 0 B \ \ \end{pmatrix}=r\begin{pmatrix}0 0 \ 0 a+b \ \end{pmatrix}=r (a+b). $$Certificate Act Two (Geometric method-using linear transformation theory) refer to [question 2014a12] and its solution.Certificate Law III (Algebraic

[Problem 2015s01] set \ (M_n (\mathbb{r}) \) is the real linear space of the entire formation of the \ (n\) Order matrix, \ (\varphi\) is a linear transformation on \ (M_n (\mathbb{r}) \), so that for a given \ (A, B\in m_n (\mathbb{r}), or \ (\varphi (AB) =\varphi (a) \varphi (b) \), or \ (\varphi (AB) =\varphi (b) \varphi (a) \) is established. Proof: either \ (\varphi (AB) =\varphi (a) \varphi (b) \) is true for any of the \ (A,b\in m_n (\mathbb{r}) \), or \ (\varphi (AB) =\varphi (b) \varphi

\), and the conclusion of the proof is the formula (13).\[r/\cap i_k\cong r/i_1\times r/i_2\times\cdots\times r/i_n\tag{13}\]First, easy authentication \ (r\to r/i_1\times r/i_2\times\cdots\times r/i_n\) is the homomorphism mapping, if it can be proved that it is full-shot, by the homomorphism fundamental theorem can be concluded. The proof method is the same as the essence in the elementary number theory, we need to construct for each dimension \ (r_k= (\cdots,0,a_k,0\cdots) \). This condition

arithmetic comparer (>,≥,Select * from R,s where r.a=s.c;except (division)Given the relationship R (x, y) and S (y) where x, Y is a set of properties (which can also be a single attribute), y in Y and s in R are attributes (sets) with the same name and can have different property names.But must originate from the same domain set. When solving R÷s, group R by the value of X, and then examine each group, such as Y in a group that contains all Y in S, then take the value of x in that group as a tu

necessary and sufficient condition to prove its existence is \ (| g_k|\) coprime, the full use of the cycle group has just been discussed proof \ (a\) decomposition of each factor is its generation of the elements of the group, the necessity is through the construction of two \ (p-\) Order (refer to the next article) of the product to export contradictions. In addition, if \ (G=g_1\times g_2\) and \ (G_1\leqslant h\), it is easy to prove that there is \ (H=g_1\times (G_2\cap H) \).\[h= (H\cap g

\,}\varphi_b\). Also by (1) know \ (\{ae_1,\cdots,ae_r,be_{r+1},\cdots,be_{r+s}\}\) linear Independent, it can be expanded to \ (u\) a group of base \ (\{ae_1,\cdots,ae_r,be_{r+1},\ cdots,be_{r+s},f_{r+s+1},\cdots,f_m\}\).Finally easy to verify: \ (\varphi_a,\varphi_b\) in \ (v\) a set of base \ (\{e_1,\cdots,e_n\}\) and \ (u\) a set of base \ (\{ae_1,\cdots,ae_r,be_{r+1},\cdots, be_{r+s},f_{r+s+1},\cdots,f_m\}\) The representation matrix is the required matrix. \ (\box\) Fudan University 2014--

obtained by the line vector \ (b\) is a linear combination of \ (a\) line vectors. (3) for the No. 208 page of the Fudan Gaodai review Question 38, the answer can refer to the Fudan high-generation white Paper 121th page Example 4.17, in the fourth Chapter review I also carefully said this proof; Its algebraic proof is also very simple, as long as the \ (ax=0\) and \ (\begin{pmatrix} A \ B \end{pmatrix}x=0\) and \ (bx=0\) The same solution can get \ (b\) line vector of the maximal independent g

, single non-zero vector linearly independent;27) The sufficient and necessary condition for the linear correlation of a vector group is that one of the vectors can be linearly represented by the remaining vectors;28) The sufficient and necessary condition for the linear correlation of two vectors is that their components correspond proportionally;29) The vector group is linearly correlated with the partial vector, then the whole vector group is linearly correlated;30) maximal linear independent

1. Ways to get a column of another column with max/min values:A. Most_bars_country = flags["name"][flags["Bars"].idxmax ()]B. bars_sorted = flags.sort_values ("Bars", ascending=[0])Most_bars_country = bars_sorted["Name"].iloc[0]2. The probability of a certain value in a column:orange_probability = flags[flags["Orange"]==1].shape[0]/flags.shape[0]3. The calculate combination by using factorial:　　　Import Mathdef find_outcome_combinations (N, k): # Calculate The numerator of our formula.Numerator =

Compute Solution of Ax=b (X=XP+XN)Rank rR=M Solutions ExistR=n Solutions UniqueExampleIf we want to solve the equation, what conditions does b1,b2,b3 need to meet? The observation matrix shows that the third line is the first two rows and so the B1+B2=B3Solvability Condition on B:Ax=b is solvable when B was in C (A)If a combination of Rows of a gives zero row, then the same combination of entries of B must give 0Assuming that the above matrix becomes:To find complete solution to ax=b:1.Xp (parti

1. Get the R value and the P value between the dataset:R_fta_pts,p_value = Pearsonr (nba_stats["pts"],nba_stats["FTA"])R_stl_pf,p_value = Pearsonr (nba_stats["STL"],nba_stats["PF"]) # It'll return R value and P value.2. The function of getting convariance form the data set, the Convariance is the value of this measure how much both variables correlated with all other. If one changes to bigger, the other changes to bigger. Which said these, variables is corresponse. Here is the function of gettin

I. Five Representation Methods of Matrix Multiplication 1. General Form 2. Multiply the matrix and column vector 3. Multiply a matrix and a row vector 4. Multiply Matrix Blocks Ii. Matrix Inversion For phalanx, left inverse = Right Inverse Multiply the original matrix by its inverse matrix to obtain the Unit Matrix Several methods to determine whether it is reversible:1. the determinant is 02. the columns of the matrix are linear combinations of the columns of the matrix. 3. When the follow

Label: strong SP 2014 problem on c AMP R BS[Question 2014a01]Try to calculate the value of the following \ (n \) Order determining factor:\ [| A | = \ begin {vmatrix} 1 x_1 (x_1-a) X_1 ^ 2 (x_1-a) \ cdots X_1 ^ {n-1} (x_1-a) \ 1 X_2 (x_2-a) X_2 ^ 2 (x_2-a) \ cdots X_2 ^ {n-1} (x_2-a) \ vdots \ vdots \ 1 X_n (x_n-a) x_n ^ 2 (x_n-a) \ cdots x_n ^ {n-1} (x_n-a) \ end {vmatrix }. \]Tip:\ (A \) is discussed in two cases.Note:In fact, the results of the above two cases can be unified. Why

, \ beta_s $. then $ \ bee \ label {313_7_eq} \ SCRA \ beta_1, \ cdots, \ SCRA \ beta_r \ mbox {linear independence, and it is a group of bases of} \ SCRA w \ mbox }. \ EEE $ \ Bex \ dim W = S + T = \ dim W_0 + \ dim \ SCRA v. \ EEx $ forward certificate \ eqref {313_7_eq }. on the one hand, $ \ beex \ Bea \ quad \ sum _ {I = 1} ^ t K_ I \ SCRA \ beta_ I = 0 \ \ Ra \ SCRA \ sex {\ sum _{ I = 1} ^ t K_ I \ beta_ I} = 0 \ \ Ra \ sum _ {I = 1} ^ t K_ I \ beta_ I \ In w \ cap \ SCRA ^ {- 1} (0) =

[Question 2014a02]Evaluate the value of the following \ (n \) Order Determinant, where \ (a_ I \ NEQ 0 \, (I = 1, 2, \ cdots, n )\): \ [D_n =\begin {vmatrix} 0 a_1 + A_2 \ cdots a_1 + A _ {n-1} a_1 + a_n \ A_2 + A_1 0 \ cdots a_2 + A _ {n-1} A_2 + a_n \ vdots \ vdots \ A _ {n-1} + A_1 _{ n-1} + A_2 \ cdots 0 A _ {n-1} + a_n \ a_n + A_1 a_n + A_2 \ cdots a_n + A _ {n-1} 0 \ end {vmatrix }. \] NoteAt this stage, try not to use the descending formula of the matrix. We recommend t

[Article] the full text of this series of articles on the wall is reprinted. Every language has libraries, besides the big. net libraries, F # has two own: the Core, which is shipped with Visual Studio 2010, and the PowerPack, which is an external library developed by MSR Cambridge and Visual Studio Team. notice that the code quality in PowerPack is actually quite high, it is put outside the Core library because it is evolving fast. once stable, they may be put into the Core. Our concern is mat

RingThe definition of is similar to the interchangeable group, but adds another operation "·" on the basis of the original "+" (note+ And. It is not generally known as addition and multiplication ). In abstract algebra, ResearchRingIsRing Theory. Definition The Set R and binary operations defined on it + and (r, +, ·) constituteRingIf they meet the following requirements: (R, +) forms an exchange group. Its unit is calledZero Element, As '0 '. That

BZOJ 3996 TJOI2015 linear algebra network stream, bzojtjoi2015 Given N then n Matrix B And one 1 limit n Line vector C , Find one 1 limit n 01 matrix A , Make (A × B − C) × Max (A × B − C) × AT = A × B × AT −c × We can consider N Items, each item does not select the corresponding A Where each location is 1 Or 0 Then the row vector C It can be seen as a matrix of the cost

the determinant of the time will give a detailed proof.The other part is the general algorithm used when solving the inverse matrix of 3 order and above.First we give a lemma:Theorem 1: if n x n matrix A is reversible, then for any r^n vector B, the solution to the matrix equation ax = b is only present.Proof: existence, in this matrix equation is multiplied by the inverse matrix of a, then there is x = a^-1 B. Uniqueness, combined with the properties of the inverse matrix uniqueness mentioned