linear algebra – Matrix multiplication rearrangement

I want to calculate matrix multiplication $ ABA $, where $ A $ Y $ B $ They are real and orthogonal matrices. In fact, they are specifically $ 3 times3 $ rotation matrices However, it is much easier if I can reverse the order of $ BA $ somehow, because I can perform multiplication much more easily.

I know that matrix multiplication is not commutative, however, I ask why both $ A $ Y $ B $ they are orthogonal matrices and, hopefully, there may be some trick to use their orthogonality to reorder the product.

I tried to solve this, but I got stuck here:

ABA = A ((BA) -1) -1 – = A (A -1 B -1) -1

Is there any way to proceed from here?

Functional analysis: proof of the spectral theorem multiplication operator form for normal operators without limits

I'm looking for proof that

Leave $ H $ be a separable Hilbert space and $ N $ an unlimited normal operator in $ H $. Then there is a finite measurement space $ (X, scr {M}, mu) $, a unit operator $ U: H a L ^ 2 ( mu) $and a measurable complex function $ f $ such that: (i) for $ x in H $, $ x in D (N) $ iff $ f , Ux in L ^ 2 ( mu) $; (ii) for $ phi in U (D (A)) $, $ UNU ^ {- 1} phi = f phi $. In other words, $ N $ becomes multiplication by $ f $.

Yes $ N $ is bounded, this can be found in A functional analysis course by G. B. Folland; however, this book does not address unlimited operators. Yes $ N $ It is unlimited but self-attached, this can be found in Methods of modern mathematical physics I: functional analysis by M. Reed and B. Simon; however, this book does not seem to care about unlimited normal operators.

Thanks in advance!

Are there efficient probabilistic multiplication algorithms that use O (n log n) doors?

Recently, Harvey and Hoeven published a document that demonstrates that multiplication of integers can be performed using a maximum of O operations (n ​​log n). This algorithm is theoretically interesting, but in practice it is completely silly because you only begin to see advantages in numbers with an absurd number of digits.

But suppose we just wanted a probabilistic multiplication circuit, which returned the incorrect result with probability as maximum epsilon. Then, perhaps certain shortcuts could be taken, to avoid the most inconvenient parts of multiplying two numbers.

For a fixed acceptable failure rate, epsilon, is there a multiplication algorithm O (n log n) that achieves this failure rate without being terribly inefficient in practice?

What is the temporal complexity of a binary multiplication using the Karatsuba algorithm?

My apologies if the question sounds naive, but I am trying to understand the idea of ​​the complexity of time.

In general, it is said that the multiplication of Karatsuba has a temporal complexity of O(n^1.5...).
The algorithm assumes that addition and subtraction take approximately O(1) every. However, for binary addition and subtraction, I don't think it is O(1). If I'm not mistaken, a typical addition or subtraction of two binary numbers takes O(n) hour.

What will be the total time complexity of the next program that multiplies two binary numbers using Karatsuba Something which in turn performs binary addition and subtraction?

long multKaratsuba(long num1, long num2) {
 if ((num1>=0 && num1<=1) && (num2>=0 && num2<=1)) {
   return num1*num2;

 int length1 = String.valueOf(num1).length(); //takes O(n)? Not sure
 int length2 = String.valueOf(num2).length(); //takes O(n)? Not sure

 int max = length1 > length2 ? length1 : length2;
 int halfMax = max/2;

 // x = xHigh + xLow
 long num1High = findHigh(num1, halfMax); // takes O(1)
 long num1Low = findLow(num1, halfMax); // takes O(1)

 // y = yHigh + yLow 
 long num2High = findHigh(num2, halfMax); // takes O(1)
 long num2Low = findLow(num2, halfMax); // takes O(1)

 // a = (xHigh*yHigh)
 long a = multKaratsuba(num1High, num2High);

 // b = (xLow*yLow)
 long b = multKaratsuba(num1Low, num2Low);

 //c = (xHigh + xLow)*(yHigh + yLow) - (a + b);
 long cX = add(xHigh,xLow); // this ideally takes O(n) time
 long cY = add(yHigh,yLow); // this ideally takes O(n) time
 long cXY = multKaratsuba(cX, cY);
 long cAB = add(a, b) // this ideally takes O(n) time
 long c = subtract(cXY, cAB) // this ideally takes O(n) time

 // res = a*(10^(2*m)) + c*(10^m) + b
 long resA = a * (long) Math.pow(10, (2*halfMax)); // takes O(1)
 long resC = c * (long) Math.pow(10, halfMax); // takes O(1)
 long resAC = add(resA, resC); // takes O(n)
 long res = add(resAC, b); // takes O(n)

 return res;

unit: round to the nearest multiplication of 0.75 in C #

I was wondering how I could round a number to the nearest multiple of o.75. The context is that I have a block and I want its position to be rounded to the nearest 0.75 mult. Ex 0, 0.75, 1.5, 2.25, 3, 3.25, 4.5, 5.25, 6 …
How can I do that ? (c # in the unit)

Fa analysis Functional: does an operator that retains signs subtract $ L ^ 2 $ a multiplication

Leave $ T: L ^ 2 ( mu) a L ^ 2 ( mu) $ Be a linear and continuous operator, where. $ L ^ 2 ( mu) $ It is the real) $ L ^ 2 $-space to some measurement space $ ( Omega, Sigma, mu) $.

$ T $ it is assumed that it retains signs in the sense that
v (x) cdot (Tv) (x) ge0

for $ mu $-almost every $ x en Omega $ and all $ v in L ^ 2 ( mu) $.

This implies that $ T $ Is it a multiplication? That is, does it exist? $ phi en L ^ infty ( mu) $ such that $ Tv = u cdot v $?

It could show the following property:
chi_ {A ^ c} cdot (L chi_A) = 0

$ mu $-Almost everywhere for all characteristic functions. $ chi_A $ of $ A en Sigma $. This would prove the question for $ mathbb R ^ n $ or $ l ^ 2 ( mathbb N) $. I could not prove the question in the general case.

integration – integration of the multiplication of two functions by parts

I am having difficulties with the computational analytical solution for the convolution of two functions by parts:

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The above form can be simplified to the following (where instead of f1, f2, consider the respectful functions of the multiple arguments that are omitted shortly):

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The problem I have is to find out the limits for each component of the resulting function in parts, how to address this problem?

Thanks in advance!

Gwnum library to accelerate the multiplication in mathematics.

Is it possible to use the Gwnum library, to accelerate the multiplication in mathematics.

Algorithms – Karatsuba multiplication rule to divide a number into two parts

In the Karatsuba algorithm to multiply two numbers, we divide each number into two. For example:

x= 1234
y= 2456

Then a = 12, b = 34, c = 24, d = 56

What happens if the digits in each number are not even or equal? What is the rule by dividing it into two parts?


 x = 12345
 y = 2478


 x = 12456778
 y = 241

Please help.

Python integer multiplication algorithm of Karatsuba

This code is not passing all the test cases, can anyone help? I just pass the direct test and then lose precision.

import mathematics
import unittest

IntegerMultiplier class:

def multiplies (yo, x, y):
if x <10 o and <10:
returns x * y

x = str (x)
y = str (y)

m_max = min (len (x), len (y))
x = x.rjust (m_max, & # 39; 0 & # 39;)
y = y.rjust (m_max, & # 39; 0 & # 39;)

m = math.floor (m_max / 2)

x_high = int (x[:m])
x_low = int (x[m:])

y_high = int (and[:m])
y_low = int (and[m:])

z1 = self.multiply (x_high, y_high)
z2 = self.multiply (x_low, y_low)
z3 = self.multiply ((x_low + x_high), (y_low + y_high))
z4 = z3 - z1 - z2

returns z1 * (10 ** m_max) + z4 * (10 ** m) + z2

TestIntegerMultiplier class (unittest.TestCase):

def test_normal_cases (self):
intergerMultiplier = IntegerMultiplier ()

case1 = intergerMultiplier.multiply (1234, 5678)
self.assertEqual (case1, 7006652)

yes __name__ == & # 39; __ main __ & # 39;
unittest.main ()
`` `