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 ⟶ ℝm 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 ⟶ ℝm 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 jth column is the vector T(ej), where ej is the jth column of the identity matrix in ℝn: $$A=[T(\overrightarrow{e_1})\text{ . . . }T(\overrightarrow{e_n})]$$

Here is my answer:

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

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

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

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

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

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

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

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; $\overrightarrow{x}$≠$\overrightarrow{0}$

∴ TA($\overrightarrow{e_j}$) = TB($\overrightarrow{e_j}$)

A = B

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


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?


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 $B$?

  • $\begingroup$ $\overrightarrow{x}$ = $\overrightarrow{e_j}$? $\endgroup$ – Anton Yershov Feb 11 '18 at 4:48
  • $\begingroup$ Yep, that works. $\endgroup$ – Eric Wofsey Feb 11 '18 at 5:00

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