how to get inverse ?

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종영
종영 2024-8-17
评论： John D'Errico 2024-8-17
a = [1 2 3];
b = [3 2 1 ].';
c = a*b;
aa = c*pinv(b) ;
i want to answer a = aa but i can't
pleas~~~~
  1 个评论
Rik
Rik 2024-8-17
What exactly is your question? Do you want to find out whether a is equal to aa?

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akshatsood
akshatsood 2024-8-17
编辑：akshatsood 2024-8-17
Dear @종영
I understand that you are trying to recover the original vector "a" from matrix multiplication. The issue here is that "c" is a scalar, so when you multiply "c" by pseudo-inverse of "b", you do not necessarily get back the original vector "a".
Explanation: The operation "c * pinv(b)" gives you a vector that tries to approximate "a" under the least-squares solution, but it woould not necessarily equal "a" unless certain conditions are met (e.g., "b" is orthogonal).
Solution: To directly recover "a", you need more information than just "c" and "b". However, if you have control over the process, you can ensure that "b" is orthogonal or use other constraints to make this recovery possible. However, without additional information or constraints, "a" cannot be reconstructed from "c * pinv(b)" operation.
I hope this helps.
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Steven Lord
Steven Lord 2024-8-17
Ran in:
With c == 10, are there other vectors x, such that x*b == 10. In fact, there are infinitely many such vectors. The simplest one is one you will get from pinv, or from lsqminnorm.
The simplest in some sense. But as you said, there are others.
a = [1 2 3];
b = [3 2 1 ].';
c = a*b
c = 10
alsoc = [0 0 10]*b
alsoc = 10
alsoc2 = [0 5 0]*b
alsoc2 = 10
alsoc3 = [3 0 1]*b
alsoc3 = 10
If you didn't know a, how could you rule out that a is [0 0 10], [0 5 0], or [3 0 1] instead of [1 2 3]? Either of the vectors used to create alsoc or alsoc2 could be considered "simpler" as they only have one non-zero element, as does the solution returned by \.
a2 = 10/b
a2 = 1x3
3.3333 0 0
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a2*b
ans = 10
John D'Errico
John D'Errico 2024-8-17
Yes, I guess my use of simplest as a description is arguable. Certainly
a = [10/3, 0, 0]
a = [0 0 10]
a = [0 5 0]
or
a = [10/6, 10/6, 10/6]
Would as easily qualify, and are surely simpler by most definitions of the word. They all work as well as any other.

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