How to verify the uniqueness of a standard matrix?

Question

I think I have answered this question sufficiently, I just want to know if I am correct, or if I missed anything.

Here is the question:

Verify the uniqueness of A in Theorem 10. Let T : ℝⁿ ⟶ ℝ^m be a linear transformation such that T($\overrightarrow{x}$) = B$\overrightarrow{x}$ for some m × n matrix B. Show that if A is the standard matrix for T, then A = B. [Hint: Show that A and B have the same columns.]

Here is Theorem 10:

Let T : ℝⁿ ⟶ ℝ^m be a linear transformation. Then there exists a unique matrix A such that $$T(\overrightarrow{x}) = A\overrightarrow{x} \text{ for all $\overrightarrow{x}$ in } ℝ^n$$ In fact, A is the m × n matrix whose j_th column is the vector T(e_j), where e_j is the j_th column of the identity matrix in ℝⁿ: $$A=[T(\overrightarrow{e_1})\text{ . . . }T(\overrightarrow{e_n})]$$

Here is my answer:

A = [ T_A($\overrightarrow{e_1}$) . . . T_A($\overrightarrow{e_n}$) ]

B = [ T_B($\overrightarrow{e_1}$) . . . T_B($\overrightarrow{e_n}$) ]

assuming A$\overrightarrow{x}$ = T = B$\overrightarrow{x}$

A$\overrightarrow{x}$ = B$\overrightarrow{x}$

A$\overrightarrow{x}$ - B$\overrightarrow{x} = \overrightarrow{0}$

( [ T_A($\overrightarrow{e_1}$) . . . T_A($\overrightarrow{e_n}$) ] - [ T_B($\overrightarrow{e_1}$) . . . T_B($\overrightarrow{e_n}$) ] )$\overrightarrow{x}$ = $\overrightarrow{0}$

[ T_A($\overrightarrow{e_1}$) - T_B($\overrightarrow{e_1}$) . . . T_A($\overrightarrow{e_n}$) - T_B($\overrightarrow{e_n}$) ]$\overrightarrow{x}$ = $\overrightarrow{0}$

Here is where I feel like I might be jumping to conclusions without proper reasoning

[ $\overrightarrow{0_1}$ . . . $\overrightarrow{0_n}$ ]$\overrightarrow{x}$ = $\overrightarrow{0}$ $\forall$ $\overrightarrow{x}$ $\in$ ℝⁿ; $\overrightarrow{x}$≠$\overrightarrow{0}$

∴ T_A($\overrightarrow{e_j}$) = T_B($\overrightarrow{e_j}$)

A = B

This is my first attempt at stating any kind of proof, and any help revising it would be much appreciated.

-Edit-

Using Eric Wofsey's reasoning that

Ax=Bx for any vector x

then can I just say that

A$\overrightarrow{e_j}$ = B$\overrightarrow{e_j}$

$\overrightarrow{A_j}$ = $\overrightarrow{B_j}$

A = B

is it really this simple?

Answer 1

The step you are concerned about is indeed incorrect. Basically, your argument is that $(A-B)x=0$ for any vector $x$, and the zero matrix also satisfies $0_{m\times n}x=0$ for any vector $x$, therefore $A-B=0_{m\times n}$. But this logic is wrong: how do you know there can't be two different matrices which, when multiplied by any vector, give $0$?

Instead, you're going to need to use the fact that $Ax=Bx$ for any vector $x$. This means you can choose $x$ to be any specific vector you want. Can you think of any specific choice of $x$ for which the equation $Ax=Bx$ would tell you some useful information about the entries of $A$ and $B$?