I understand when dimension of null space is zero. Then Ax=0 has only trivial solution (Since only zero vector is included).
But what if the dimension is not zero? What does it mean? Is there any difference between a null space with dimension of 1 and 2?
At present I only get to know when the dimension of null space is not zero, and x* is a particular solution to the equation Ax=b, then any vector of the form x*+h is also a solution for any h belongs to the null space (Which means there could be innumerable solutions). Could someone help me to understand the relationship between innumerable solutions and the non-zero dimension of null space？
[Math] How to understand the dimension of null space
linear algebra
Best Answer
Let $A$ be an $m$ by $n$ matrix over a field $F$. Let us say for instance that $F$ is $\mathbb{R}$ or $\mathbb{C}$, but the following discussion holds over any field $F$. Let us study the solution set of $Ax=b$, where $x$ is $n$ by $1$ ($x$ is unknown) and $b$ is $m$ by $1$ ($b$ is knowns). $A$, $x$ and $b$ all have entries in $F$.
Case $1$: it may happen that $Ax = b$ has no solution.
Case $2$: $Ax=b$ has at least one solution, say $x^*$. Define $h = x - x^*$. Then $x$ is a solution of $Ax=b$ if and only if $h$ is a solution of $Ah=0$. Indeed, to prove the forward direction, assume that $x$ is a solution of $Ax=b$. Then
$$Ah = A(x-x^*) = Ax - Ax^* = b - b = 0.$$
Conversely, assume that $h$ is a solution of $Ah = 0$. Then $x = x^* + h$, so that
$$Ax = A(x^* + h) = Ax^* + Ah = b + 0 = b.$$
Hence $x$ is a solution of $Ax=b$. Define a map $f$ from the set of solutions $x$ of $Ax=b$ to the set of solutions $h$ of $Ah = 0$, which maps $x$ to $h=x-x^*$. It is easy to see that $f$ is a bijection, from the above discussion. Also $f^{-1}$ maps $h$ to $x = x^* + h$. Hence there is a one-to-one correspondence between the solution set of $Ax=b$ and the solution set of $Ah=0$.
So if $Ax=b$ has at least one solution, and the solution set of $Ah=0$ has positive dimension, and assuming the field $F$ is infinite (which would be the case if $F=\mathbb{R}$ or $F=\mathbb{C}$ for instance), then this implies there are infinitely many solutions $h$ of $Ah=0$, and therefore there are infinitely many solutions $x$ of $Ax=b$.