Problem
Given $Ax=0$, where$A\in R^{m\times n}$，$x\in R^{n}$, What is the solution?
Solution
Obviously the homogenous system always has the solution $x = 0$. But what about other solutions?
- if $m = n$ and $rank(A) = n$, $x = 0$
- if $m = n$ and $rank(A) < n$, there exists at least one free variable, the number of which is $n - rank(A)$
- if $m \neq n$ and $rank(A)$ must not be more than $min(m, n)$.
- if $m < n$, then $rank(A) <= m < n$, so the equations will be undetermined, which means there exists non-trivial solutions.
- if $m > n$, then $rank(A) <= n < m$. For $rank(A) = n$, the system has an unique solution, which is $x = 0$. While $rank(A) < n$, it degenerates to be undetermined with non-trivial solutions.
Some Thoughts
- Q: Can $Ax = 0 \Rightarrow x = 0$ be true?
- A: From the solution part, we could see clearly that it's false.
- Q: For $M(Ax-b) = 0$, how to solve $x$?
- A: From above question, we couldn't get $Ax=b$ directly, except $M^{-1}$ exists(case 1) or case 3.2. So we need to expand the equation into $MAx = Mb$ and solve the form as $ \tilde{A} x= \tilde{b}$.
- Q: For the form $Ax=0$, if it's overdetermined, how about least squares?
- A: No matter whether it is overdetermined, the least squares is meaningless since the result will always be $x = 0$.