1
$\begingroup$

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.

-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?

$\endgroup$
3
$\begingroup$

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 $B$?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.