Suppose that Ax = b and Cx = b have the same general solution for every b. why is it that then A = C?

Question

If you have any two matrices A and C such that A$\vec{x}$ = $\vec{b}$ and C$\vec{x}$ = $\vec{b}$ have the same general solution for all $\vec{b}$ then why is that A = C.

I can understand this when $\vec{x}$ is always unique, but I don't know how to prove this in general. Does this equation imply that A and C are the same linear transformation? and does this automatically mean that the matrices are equal.

I can understand how the equality of two matrices can be defined as when the matrices have the same linear transformation but I don't understand this computationally. Is there a way to show that for every distinct linear transformation there is only one matrix that corresponds to that linear transformation?

Answer 1

You can reduce to the case where all the solutions are unique. Let $A,C$ be maps $V \to W$, and form the quotient $V/\ker{A} = V/\ker{C}$. Then the induced maps $\bar{A}: V/ \ker{A} \to W$ and $\bar{C}: V/ \ker{C} \to W$ now have the same unique solution for every $\bar{x} = x \mod \ker A$

Answer 2

Let $e_1, \ldots, e_n$ be a basis for $V$, and let $b_i = Ae_i$. Then a solution to $Ax = b_i$ is a solution to $Cx = b_i$. Since $e_i$ is a solution to $Ax = b_i$, we must have $Ce_i = b_i$ as well. Thus $Ae_i = Ce_i$. This implies equality of the matrices of $A$ and $C$.

Answer 3

Your condition implies that $\forall \vec x~((A-C)\vec x = \vec 0)$. Applying this equation to each element of a basis implies $A-C=0$, which implies $A=C$.

Answer 4

Let A $\epsilon \space \mathbb{R}^{m * n}$ Let $e_i$ denote the basic column vectors in $\mathbb{R}^{n * 1}$, for each $1 \le i \le n$.

Let $b_i := Ae_i$, for each $1 \le i \le n$.

By our hypothesis, $e_i$ is a solution of $Cx = b_i$, for each $i$. Hence each $e_i$ satisfies $(A - C)x = O$, which is deduced by subtracting both the matrix equations.

Hence, the dimension of $\mathcal{N}(A - C) \space \textit{[null space]}$ is n, and hence by rank-nullity theorem, we have the $rank(A - C) = 0 \implies (A - C) = 0 \implies A = C$

$\textbf{Hope this helps you.}$