为什么在Matlab和Octave中inv()和pinv()的输出不相等? [英] Why is the output of inv() and pinv() not equal in Matlab and Octave?

问题描述

我注意到，如果A是一个NxN矩阵，并且它具有逆矩阵.但是inv()和pinv()函数的输出不同. -我的环境是Win7x64 SP1，Matlab R2012a，Cygwin Octave 3.6.4，FreeMat 4.2

I have noticed that if A is a NxN matrix and it has the inverse matrix. But what the inv() and pinv() function output is different. - My environment is Win7x64 SP1, Matlab R2012a, Cygwin Octave 3.6.4, FreeMat 4.2

看看Octave中的示例:

Have a look at the examples from Octave:

A = rand(3,3)
A =
0.185987   0.192125   0.046346
0.140710   0.351007   0.236889
0.155899   0.107302   0.300623

pinv(A) == inv(A)
ans =
0 0 0
0 0 0
0 0 0

• 通过在Matlab中运行相同的命令，获得与所有ans相同的结果.
• It's all the same ans result by running the same command above in Matlab.
• 然后我计算inv(A)*A或A*inv(A)，结果是Octave和Matlab中的3x3恒等矩阵.
• A*pinv(A)和pinv(A)*A的结果是Matlab和FreeMat中的恒等3x3矩阵.
• A*pinv(A)的结果是八度中的单位3x3矩阵.
• pinv(A)*A的结果是八度中的 not 身份3x3矩阵.
• And I calculate inv(A)*A or A*inv(A), the result is identity 3x3 matrix in both Octave and Matlab.
• The result of A*pinv(A) and pinv(A)*A are identity 3x3 matrix in Matlab and FreeMat.
• The result of A*pinv(A) is identity 3x3 matrix in Octave.
• The result of pinv(A)*A is not identity 3x3 matrix in Octave.

我不知道为什么inv(A) != pinv(A)的原因，我已经考虑了矩阵中元素的详细信息.似乎是导致此问题的浮动精度问题.

I don't know the reason why inv(A) != pinv(A), I have considered the details of the element in the matrix. It seems to be the floating accuracy problem which causes this problem.

点后的10个数字可能会有所不同，如下所示:

The 10+ digits after the dot point may be different like this:

inv(A)(1,1)中针对

的

• 6.65858991579923298331777914427220821380615200000000元素 pinv(A)(1,1)

6.65858991579923209513935944414697587490081800000000元素

推荐答案

这个问题已经很老了，但无论如何我都会回答，因为它在某些Google搜索中几乎排在最前面.

This question is quite old, but I'll answer it anyway because it appears almost on top in some google searches.

在我的示例中，将使用magic(N)函数，该函数返回一个N×N的魔术方块.

I'll use for my example the magic(N) function which returns an N-by-N magic square.

我将创建一个3x3的幻方M3，取伪逆PI_M3并将它们相乘:

I'll create a 3x3 magic square M3, take the pseudoinverse PI_M3 and multiply them:

   prompt_$ M3 = magic(3) , PI_M3 = pinv(M3) , M3 * PI_M3

  M3 =

    8   1   6
    3   5   7
    4   9   2

  PI_M3 =

     0.147222  -0.144444   0.063889
    -0.061111   0.022222   0.105556
    -0.019444   0.188889  -0.102778

  ans =

     1.0000e+00  -1.2212e-14   6.3283e-15
     5.5511e-17   1.0000e+00  -2.2204e-16
    -5.9952e-15   1.2268e-14   1.0000e+00

正如您所看到的，答案是单位矩阵，除了一些舍入错误. 我将使用4x4的魔术方块重复该操作:

As you can see the answer is the identity matrix save for some rounding errors. I'll repeat the operation with a 4x4 magic square:

   prompt_$ M4 = magic(4) , PI_M4 = pinv(M4) , M4 * PI_M4

 
  M4 =

     16    2    3   13
      5   11   10    8
      9    7    6   12
      4   14   15    1

  PI_M4 =

     0.1011029  -0.0738971  -0.0613971   0.0636029
    -0.0363971   0.0386029   0.0261029   0.0011029
     0.0136029  -0.0113971  -0.0238971   0.0511029
    -0.0488971   0.0761029   0.0886029  -0.0863971

  ans =

     0.950000  -0.150000   0.150000   0.050000
    -0.150000   0.550000   0.450000   0.150000
     0.150000   0.450000   0.550000  -0.150000
     0.050000   0.150000  -0.150000   0.950000

结果不是单位矩阵，这表示4x4幻方不具有逆. 我可以通过尝试使用Moore-Penrose伪逆的规则之一来验证这一点:

The result is not the identity matrix, it means that the 4x4 magic square does not have an inverse. I can verify this by trying one of the rules of the Moore-Penrose pseudoinverse:

   prompt_$ M4 * PI_M4 * M4

  
ans =

   16.00000    2.00000    3.00000   13.00000
    5.00000   11.00000   10.00000    8.00000
    9.00000    7.00000    6.00000   12.00000
    4.00000   14.00000   15.00000    1.00000

满足规则A * B * A =A.这表明pinv在可用时返回逆矩阵，而在不可用时返回伪逆.这就是为什么在某些情况下您会得到很小的差异，而在一些舍入误差上却会得到较大差异的原因. 为了说明这一点，我将获得两个魔力象限的逆并将其从伪逆中减去:

The rule A*B*A = A is satisfied. This shows that pinv returns the inverse matrix when it is available and the pseudoinverse when the inverse is not available. This is the reason why in some situations you get a small difference, just some rounding errors, and in other situations you get a bigger difference. To show it I'll get the inverse of both magic quadrants and subtract them from the pseudoinverse:

   prompt_$ I_M3 = inv(M3) , I_M4 = inv(M4) , DIFF_M3 = PI_M3 - I_M3, DIFF_M4 = PI_M4 - I_M4

  I_M3 =

     0.147222  -0.144444   0.063889
    -0.061111   0.022222   0.105556
    -0.019444   0.188889  -0.102778

  warning: inverse: matrix singular to machine precision, rcond = 1.30614e-17
  I_M4 =

     9.3825e+13   2.8147e+14  -2.8147e+14  -9.3825e+13
     2.8147e+14   8.4442e+14  -8.4442e+14  -2.8147e+14
    -2.8147e+14  -8.4442e+14   8.4442e+14   2.8147e+14
    -9.3825e+13  -2.8147e+14   2.8147e+14   9.3825e+13

  DIFF_M3 =

     4.7184e-16  -1.0270e-15   5.5511e-16
    -9.9226e-16   2.0470e-15  -1.0825e-15
     5.2042e-16  -1.0270e-15   4.9960e-16

  DIFF_M4 =

    -9.3825e+13  -2.8147e+14   2.8147e+14   9.3825e+13
    -2.8147e+14  -8.4442e+14   8.4442e+14   2.8147e+14
     2.8147e+14   8.4442e+14  -8.4442e+14  -2.8147e+14
     9.3825e+13   2.8147e+14  -2.8147e+14  -9.3825e+13

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