Can 2 matrices performing the same linear transformation be unequal?

Question

If $\Bbb R^m,\Bbb R^n,\Bbb R^o,\Bbb R^p$..., are all different vector spaces, and $A,B,C,D,E$ are transformation matrices.

And there are 2 linear transformations performing the same mapping $$\phi:\Bbb R^n\to \Bbb R^m,\\\gamma:\Bbb R^n\to \Bbb R^m$$ $$\phi(x)=Ax,\\\gamma(x)=Bx$$

Now let there be 3 other transformations: $$\phi_1:\Bbb R^n \to \Bbb R^o,\\\phi_2:\Bbb R^o \to \Bbb R^p,\\\phi_3:\Bbb R^p \to \Bbb R^m$$ And $$\phi_1(x)=Cx,\\\phi_2(x)=Dx,\\\phi_3(x)=Ex$$

Let $$\gamma(x)=\phi_3(\phi_2(\phi_1(x)))\\ \Longrightarrow B=EDC$$

Now if both $\phi$ and $\gamma$ perform the same linear transformation, wrt. the same bases. will $A$ and $B$ be equivalent? Is there a simple way to prove this?

Another way to probably frame this question: 'Is a matrix performing a particular linear transformation always unique?'

Based on the answers I've gotten, if I can equate $Ax=Bx$, can I say $A$ and $B$ are equivalent matrices? Regardless of what they are composite of?

Answer 1

Suppose that $A$ and $B$ are different. Then $B=A+M$.

Since $Ax=Bx$ we have that $Ax=(A+M)x$

Then $Ax=Ax+Mx$

$0=Mx$ for all $x$

$M$ is a zero matrix. So $B=A$

Answer 2

The answer is yes: The matrix representation of a linear transformation with respect to particular bases is unique. To see this, let $\mathcal{B}$ and $\mathcal{C}$ be bases for $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, and $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ a linear transformation. I assume you know that there exists a matrix representation of $T$ with respect to $\mathcal{B}$ and $\mathcal{C}$, so we only show uniqueness. Suppose $M$ and $M'$ are matrices such that for all $x \in \mathbb{R}^n$

$$[T(x)]_{\mathcal{C}} = M[x]_{\mathcal{B}} = M'[x]_{\mathcal{B}}.$$

We get $(M-M')[x]_{\mathcal{B}} = 0$ for all $x$, which means that $M-M' = \mathbf{0} \implies M = M'$.

EDIT based on addendum to the question: Notice that it is important that the relation holds for all $x$ in order to conclude that $M-M' = \mathbf{0}$. If you have $Ax = Bx$ for all $x$, we have $A = B$. You should make sure you understand why that is.