使用 Python 将矩阵简化为阶梯形式练习
Matrix reduction to echelon form exercise using Python
我正在定义变量以将给定矩阵转换为其阶梯形式，练习要求我定义不同的变量以检查矩阵是否为奇异矩阵，并将其转换为其阶梯形式，这是锻炼和我到目前为止所做的事情：
import numpy as np
# Our function will go through the matrix replacing each row in order turning it into echelon form.
# If at any point it fails because it can't put a 1 in the leading diagonal,
# we will return the value True, otherwise, we will return False.
# There is no need to edit this function.
def isSingular(A) :
B = np.array(A, dtype=np.float_) # Make B as a copy of A, since we're going to alter it's values.
try:
fixRowZero(B)
fixRowOne(B)
fixRowTwo(B)
fixRowThree(B)
except MatrixIsSingular:
return True
return False
# This next line defines our error flag. For when things go wrong if the matrix is singular.
# There is no need to edit this line.
class MatrixIsSingular(Exception): pass
# For Row Zero, all we require is the first element is equal to 1.
# We'll divide the row by the value of A[0, 0].
# This will get us in trouble though if A[0, 0] equals 0, so first we'll test for that,
# and if this is true, we'll add one of the lower rows to the first one before the division.
# We'll repeat the test going down each lower row until we can do the division.
# There is no need to edit this function.
def fixRowZero(A) :
if A[0,0] == 0 :
A[0] = A[0] + A[1]
if A[0,0] == 0 :
A[0] = A[0] + A[2]
if A[0,0] == 0 :
A[0] = A[0] + A[3]
if A[0,0] == 0 :
raise MatrixIsSingular()
A[0] = A[0] / A[0,0]
return A
# First we'll set the sub-diagonal elements to zero, i.e. A[1,0].
# Next we want the diagonal element to be equal to one.
# We'll divide the row by the value of A[1, 1].
# Again, we need to test if this is zero.
# If so, we'll add a lower row and repeat setting the sub-diagonal elements to zero.
# There is no need to edit this function.
def fixRowOne(A) :
A[1] = A[1] - A[1,0] * A[0]
if A[1,1] == 0 :
A[1] = A[1] + A[2]
A[1] = A[1] - A[1,0] * A[0]
if A[1,1] == 0 :
A[1] = A[1] + A[3]
A[1] = A[1] - A[1,0] * A[0]
if A[1,1] == 0 :
raise MatrixIsSingular()
A[1] = A[1] / A[1,1]
return A
# This is the first function that you should complete.
# Follow the instructions inside the function at each comment.
def fixRowTwo(A) :
# Insert code below to set the sub-diagonal elements of row two to zero (there are two of them).
A[2] = (A[2] - A[2,0]) * (A[2] - A[2,1])
# Next we'll test that the diagonal element is not zero.
if A[2,2] == 0 :
# Insert code below that adds a lower row to row 2.
A[2] = A[2] + A[3]
# Now repeat your code which sets the sub-diagonal elements to zero.
A[2] = (A[2] - A[2,0]) * (A[2] - A[2,1])
if A[2,2] == 0 :
raise MatrixIsSingular()
# Finally set the diagonal element to one by dividing the whole row by that element.
A[2] = A[2]/A[2,2]
return A
# You should also complete this function
# Follow the instructions inside the function at each comment.
def fixRowThree(A) :
# Insert code below to set the sub-diagonal elements of row three to zero.
A[3] = (A[3] - A[3,0]) * (A[3] - A[3,1]) * (A[3] - A[3,2])
# Complete the if statement to test if the diagonal element is zero.
if A[3,3] == 0:
raise MatrixIsSingular()
# Transform the row to set the diagonal element to one.
A[3] = A[3]/A[3,3]
return A
我开始使用 fixRowZero(A)、fixRowOne(A) 和 fixRowTwo(A)，但是当我使用 fixRowThree(A) 时，我收到以下错误消息：
---------------------------------------------------------------------------
MatrixIsSingular Traceback (most recent call last)
<ipython-input-58-d25352be9b2c> in <module>
----> 1 fixRowThree(A)
<ipython-input-57-4faff915d9e0> in fixRowThree(A)
82 # Complete the if statement to test if the diagonal element is zero.
83 if A[3,3] == 0:
---> 84 raise MatrixIsSingular()
85 # Transform the row to set the diagonal element to one.
86 A[3] = A[3]/A[3,3]
MatrixIsSingular:
我做错了什么？当我传递一个矩阵来检查它是否是单数时，它会相应地给 returns 我 True 或 False，所以它似乎工作正常，但是当我使用 fixRowThree() 函数时 returns 我这个错误留言
我可能来晚了回答这个问题。但是，让我们为可能遇到此练习的人澄清一些事情。
依我看，你没有理解手头的任务：
# set the sub-diagonal elements of row three to zero.
如果你return到函数fixRowOne，你会看到这个表达式：
A[1] = A[1] - A[1,0] * A[0]
这个表达式有什么作用？
# First we'll set the sub-diagonal elements to zero, i.e. A[1,0].
你所要做的就是遵循这个表达方式。即使您的 FixSecondRow 函数没有错误，它也是错误的，因为它没有遵循相同的模式。也许，您的代码在示例测试中很幸运。
第二行有两个子对角线元素：A[2,0]和A[2,1]。次对角线元素是位于所述行的对角线元素之前的元素。你必须做同样的过程来取消每一个。因为在前面的其他行中，对角线元素已经是1，我们可以用它们通过矩阵变换很容易地消除子对角线：
A[2] = A[2] - A[2,0]* A[0]
A[2] = A[2] - A[2,1]* A[1]
对于第三行，您必须遵循相同的模式。
我希望这个解释能消除任何试图理解梯形和矩阵变换的人的困惑。
我没有 post 完整解决方案的原因是此练习是 Coursera 课程的一部分：机器学习数学：线性代数。我有一个正确的代码通过了作业，但它将 counter-productive 到 post 它。
我正在定义变量以将给定矩阵转换为其阶梯形式，练习要求我定义不同的变量以检查矩阵是否为奇异矩阵，并将其转换为其阶梯形式，这是锻炼和我到目前为止所做的事情：
import numpy as np
# Our function will go through the matrix replacing each row in order turning it into echelon form.
# If at any point it fails because it can't put a 1 in the leading diagonal,
# we will return the value True, otherwise, we will return False.
# There is no need to edit this function.
def isSingular(A) :
B = np.array(A, dtype=np.float_) # Make B as a copy of A, since we're going to alter it's values.
try:
fixRowZero(B)
fixRowOne(B)
fixRowTwo(B)
fixRowThree(B)
except MatrixIsSingular:
return True
return False
# This next line defines our error flag. For when things go wrong if the matrix is singular.
# There is no need to edit this line.
class MatrixIsSingular(Exception): pass
# For Row Zero, all we require is the first element is equal to 1.
# We'll divide the row by the value of A[0, 0].
# This will get us in trouble though if A[0, 0] equals 0, so first we'll test for that,
# and if this is true, we'll add one of the lower rows to the first one before the division.
# We'll repeat the test going down each lower row until we can do the division.
# There is no need to edit this function.
def fixRowZero(A) :
if A[0,0] == 0 :
A[0] = A[0] + A[1]
if A[0,0] == 0 :
A[0] = A[0] + A[2]
if A[0,0] == 0 :
A[0] = A[0] + A[3]
if A[0,0] == 0 :
raise MatrixIsSingular()
A[0] = A[0] / A[0,0]
return A
# First we'll set the sub-diagonal elements to zero, i.e. A[1,0].
# Next we want the diagonal element to be equal to one.
# We'll divide the row by the value of A[1, 1].
# Again, we need to test if this is zero.
# If so, we'll add a lower row and repeat setting the sub-diagonal elements to zero.
# There is no need to edit this function.
def fixRowOne(A) :
A[1] = A[1] - A[1,0] * A[0]
if A[1,1] == 0 :
A[1] = A[1] + A[2]
A[1] = A[1] - A[1,0] * A[0]
if A[1,1] == 0 :
A[1] = A[1] + A[3]
A[1] = A[1] - A[1,0] * A[0]
if A[1,1] == 0 :
raise MatrixIsSingular()
A[1] = A[1] / A[1,1]
return A
# This is the first function that you should complete.
# Follow the instructions inside the function at each comment.
def fixRowTwo(A) :
# Insert code below to set the sub-diagonal elements of row two to zero (there are two of them).
A[2] = (A[2] - A[2,0]) * (A[2] - A[2,1])
# Next we'll test that the diagonal element is not zero.
if A[2,2] == 0 :
# Insert code below that adds a lower row to row 2.
A[2] = A[2] + A[3]
# Now repeat your code which sets the sub-diagonal elements to zero.
A[2] = (A[2] - A[2,0]) * (A[2] - A[2,1])
if A[2,2] == 0 :
raise MatrixIsSingular()
# Finally set the diagonal element to one by dividing the whole row by that element.
A[2] = A[2]/A[2,2]
return A
# You should also complete this function
# Follow the instructions inside the function at each comment.
def fixRowThree(A) :
# Insert code below to set the sub-diagonal elements of row three to zero.
A[3] = (A[3] - A[3,0]) * (A[3] - A[3,1]) * (A[3] - A[3,2])
# Complete the if statement to test if the diagonal element is zero.
if A[3,3] == 0:
raise MatrixIsSingular()
# Transform the row to set the diagonal element to one.
A[3] = A[3]/A[3,3]
return A
我开始使用 fixRowZero(A)、fixRowOne(A) 和 fixRowTwo(A)，但是当我使用 fixRowThree(A) 时，我收到以下错误消息：
---------------------------------------------------------------------------
MatrixIsSingular Traceback (most recent call last)
<ipython-input-58-d25352be9b2c> in <module>
----> 1 fixRowThree(A)
<ipython-input-57-4faff915d9e0> in fixRowThree(A)
82 # Complete the if statement to test if the diagonal element is zero.
83 if A[3,3] == 0:
---> 84 raise MatrixIsSingular()
85 # Transform the row to set the diagonal element to one.
86 A[3] = A[3]/A[3,3]
MatrixIsSingular:
我做错了什么？当我传递一个矩阵来检查它是否是单数时，它会相应地给 returns 我 True 或 False，所以它似乎工作正常，但是当我使用 fixRowThree() 函数时 returns 我这个错误留言
我可能来晚了回答这个问题。但是，让我们为可能遇到此练习的人澄清一些事情。
依我看，你没有理解手头的任务：
# set the sub-diagonal elements of row three to zero.
如果你return到函数fixRowOne，你会看到这个表达式：
A[1] = A[1] - A[1,0] * A[0]
这个表达式有什么作用？
# First we'll set the sub-diagonal elements to zero, i.e. A[1,0].
你所要做的就是遵循这个表达方式。即使您的 FixSecondRow 函数没有错误，它也是错误的，因为它没有遵循相同的模式。也许，您的代码在示例测试中很幸运。
第二行有两个子对角线元素：A[2,0]和A[2,1]。次对角线元素是位于所述行的对角线元素之前的元素。你必须做同样的过程来取消每一个。因为在前面的其他行中，对角线元素已经是1，我们可以用它们通过矩阵变换很容易地消除子对角线：
A[2] = A[2] - A[2,0]* A[0]
A[2] = A[2] - A[2,1]* A[1]
对于第三行，您必须遵循相同的模式。
我希望这个解释能消除任何试图理解梯形和矩阵变换的人的困惑。
我没有 post 完整解决方案的原因是此练习是 Coursera 课程的一部分：机器学习数学：线性代数。我有一个正确的代码通过了作业，但它将 counter-productive 到 post 它。