Matrix equations proof

Question

I want to show a few equivalences for the quadratic matrix $A \in K^{n\times n}$:

1. The equation $Ax=b$ has no solution for at least one $b \in K^n$.
2. The equation $Ax=b$ has multiple solutions for at least one $b \in K^n$.
3. The equation $Ax=b$ has exactly one solution for no $b \in K^n$.

I already thought about using the Rank–nullity theorem: $$\operatorname{dim}(\operatorname{im} (A)) + \operatorname{dim} (\ker (A)) = \operatorname{dim} (A)$$ I know the dimension of $A$, which is $n$. Do I know the dimension of $b$? Is it $n$ as well? I think if I know the dimension of the kernel I can say something about the amount of solutions.

I'm not allowed to use the term "determinant".

Thanks in advance!

Answer 1

First recall that $Ax=0$ has always at least one solution $x=0$. Here $0$ denotes the $n$-dimensional vector $(0,0,\dots,0)\in K^n$. So

• $(1.)\implies (2.)$ If there exists such a $b$, then $b\notin \operatorname{im}(A).$ Hence $\dim(\operatorname{im}(A))\le n-1$, hence $\dim(\ker (A))\ge 1$ and hence $Ax=0$ has multiple solutions.
• $(2.)\implies (3.)$ Let $Ax=b_1$ have multiple solutions, i.e. let $b_1$ be such that $Ax_1=b_1$ and $Ax_2=b_1$ where $x_1\neq x_2$. Assume that there exists $b_0$ be such that $Ax=b_0$ has a unique solution $x=x_0$. But then $$A\left(x_0+\frac{x_1-x_2}{2}\right)=b_0$$ So $x_0+\frac{x_1-x_2}{2}\neq x_0$ is a second solution for $Ax=b_0$ which is a contradiction.
• $(3.)\implies (2.)$ This immediate for $b=0$.
• $(2.)\implies (1.)$ Again dimension formula.

Of course you can go directly from $(3.)$ to $(1.)$ if you take $b=0$.