TITLE: Easiest way to prove the Rouché–Capelli theorem
QUESTION [5 upvotes]: Rouché–Capelli theorem (Kronecker–Capelli theorem/Rouché–Fontené theorem/Rouché–Frobenius theorem/Frobenius theorem) states that for the non-homogeneous system Ax = b,


$(i)$ $Ax = b$ has a unique solution if and only if $rank[A] = rank[A|b] = n$


$(ii)$ $Ax = b$ is inconsistent (i.e., no solution exists) if and only if $rank[A] <
rank[A|b]$


$(iii)$ $Ax = b$ has infinitely many solutions if and only if $rank[A] = rank[A|b] < n$

How do I derive these conditions ?
My Understanding
$(i)$ $Ax=b$ has a unique solution
$A^{-1}$ exists $\implies \boxed{x=A^{-1}b} \implies |A|\neq 0 \implies rank(A)=n$
If solution exists, then $\vec{A}_1x_1+\vec{A}_2x_2+....+\vec{A}_nx_n=b\implies b$ is a linear combination of the column vectors
$\implies rank[A|b]=rank[A]=n$
$(ii)$ $Ax=b$ is inconsistent (i.e., no solution exists)
$\boxed{|A|x=(adj A)b\implies |A|=0 \quad\&\quad adj A.b\neq 0}$
$|A|=0\implies rank[A]<n$
Is there a way to prove the last two conditions of the Rouché–Capelli theorem ?

REPLY [3 votes]: 2) The system is inconsistent if and only if there aren't $x_1,...,x_n$ such that:
$$A_1x_1+....+A_nx_n=b$$
This happens if and only if we add $b$ as a column to the matrix $A$ the rank increases by 1($b$ is linearly independent).
3) Clearly there are solutions if and only if there are $x_1,...,x_n$ such that:
$$A_1x_1+....+A_nx_n=b$$
This happens if and only if we add $b$ as a column to the matrix $A$ the rank doesn't increase($b$ is linearly dependent). So we have $rank(A)=rank(A|b)$. Now we have to prove when the solutions are infinite. The simplest proof that I know relies on linear algebra, but I don't know if you are familiar with it, so I'll stay on linear systems. If you apply Gaussian Elimination to $A$ you'll obtain a scale matrix with $p$ pivots. The number of pivots of a scale matrix is equal to its rank(and row reduction doesn't modify the rank of A), so $rank(A)=p$. Now the "pivot-columns" will correspond to unknowns and "non-pivot-columns" to parameters
Essentially the set of solutions will depend upon $n-p=n-rank(A)
$ parameters. The solutions are infinite if and only if there are parameters so if and only if $n-rank(A)>0$ so $rank(A)<n$.
Your proof of the first point is wrong if the matrix is allowed to be rectangular. In that case a correct proof would be noticing that a consistent system has an unique solution if and only if there aren't paramenters, so if and only if $n-rank(A)=0$ which implies $n=rank(A)$
[However it's a bit imprecise to talk about "infinitely many solutions" if we don't specify in which field we're working].
Here I write a more advanced(and elegant) proof of the last proposition that relies on linear algebra. If a linear system is consistent than its solution are an affine subspace of the numerical standard affine space, whom direction is the set of solution of the homogeneous associated system. An affine subspace has more than one element if and only if its dimension is non-zero This happens if and only if the dimension of its direction is non-zero. The dimension of the direction is by definition $dim(Ker(A))$. By the rank-nullity theorem:
$$dim(Ker(A))=n-rank(A)$$
We want this dimension to be non-zero, so:
$$n-rank(A)>0$$
$$rank(A)<n$$