MATLAB 中最有效的矩阵求逆
当在 MATLAB 中计算某个方阵 A 的逆时，使用
Ai = inv(A)
% should be the same as:
Ai = A^-1
MATLAB 通常会通知我这不是最有效的求逆方法。 那么什么是更有效率的呢？如果我有一个方程系统，可能会使用 /,\ 运算符。 但有时我需要逆数来进行其他计算。
最有效的反转方式是什么？
When computing the inverse for some square matrix A in MATLAB, using
Ai = inv(A)
% should be the same as:
Ai = A^-1
MATLAB usually notifies me that this is not the most efficient way of inverting.
So what's more efficient? If I have an equation system, using the /,\ operators probably is.
But sometimes I need the inverse for other computations.
What's the most efficient way to invert?
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我建议使用 svd （除非你真的绝对确定你的矩阵没有病态）。然后，根据奇异值，您决定采取进一步的行动。这听起来像是一种“杀伤力过大”的方法，但从长远来看，它会带来回报。
现在，如果您的矩阵 A 实际上是可逆的，那么 A 的伪逆矩阵与 inv(A) 重合，然而，如果您接近“奇点”，您将很容易做出适当的决定，如何继续实际进行
伪逆。当然，这些决定将取决于您的应用。添加一个简单的示例：
I would recommend to use
svd(unless you are really absolute sure that your matrix is not ill-conditioned). Then, based on singular values you make your decisions on further actions to take. This may sound like a 'overkill' approach, but in long run it will pay back.Now if your matrix
Ais actually invertible, then thepseudo inverseofAcoincidence withinv(A), however if you are close to 'singularity' you'll easily make appropriate decision how to proceed to actually make thepseudo inverse. Naturally these decisions will depend on your application.Added a straightforward example:
我认为 LU 分解比反演更有效（如果使用旋转，则可能更稳定）。如果您需要求解多个右侧向量，它尤其有效，因为一旦您进行了 LU 分解，您就可以根据需要对每个向量进行前向反向替换。
我会推荐 LU 分解而不是完全逆分解。如果 MATLAB 就是这么说的，我同意。
更新：3x3 矩阵？如果需要，您可以手动将其以封闭形式反转。只需首先检查行列式以确保它不是奇异的或接近奇异的。
I think LU decomposition is more efficient than than inversion (and potentially more stable if you use pivoting). It works especially well if you need to solve for more than one right hand side vector, because once you have the LU decomposition you can do the forward back substitutions for each one as you need it.
I would recommend LU decomposition over a full inverse. I agree if that's what MATLAB is saying.
UPDATE: 3x3 matrix? You can invert that by hand in closed form if you need it. Just check the determinant first to make sure that it's not singular or nearly singular.
如果你只需要逆，那么就这样做，它在数值上比 inv(A) 更稳定：
If you only need the inverse then just do, it will be numerically more stable than inv(A):
如果没有一种巧妙的方法来完成所有计算而不显式形成逆，那么您必须使用“inv”函数。您当然可以用矩阵和单位矩阵求解线性系统，但这样做没有任何好处。
If there isn't a clever way to do all your calculations without explicitly forming the inverse then you have to use the "inv" function. You could of course solve a linear system with your matrix and the identity matrix, but there is nothing to be gained by doing that.