1
$\begingroup$

I remember reading this statement before.

It is as follows.

Transformation is onto if and only if columns are linearly independnet

Transformation is one-to-one if and only if rows are independent

I think it is not right statement because it comes from my unclear memory of reading this

statement before.

But, what I read is quite similar to above statements, but can't recall perfectly.

Can anyone please modify it ?

$\endgroup$
2
$\begingroup$

You've just mixed up your two theorems a bit. Here are the correct statements:

$\textbf{Theorem: }$ Let $T:\mathbb{R}^n \rightarrow \mathbb{R^m}$ be a linear transformation, and let $A$ be the standard matrix of $T$. Then:

  1. $T$ is one-to-one if and only if the columns of $A$ are linearly independent.
  2. $T$ is onto if and only if the columns of $A$ span $R^m$.

I recommend that you look at Is a linear tranformation onto or one-to-one? for the full proof of this theorem, and for additional theorems related to the topic.

$\endgroup$
  • $\begingroup$ For the second statement, How can columns of A span R^m? Matrix of A consists of n columns,, $\endgroup$ – user197971 Dec 10 '14 at 4:12
  • $\begingroup$ As you said, A will consist of n columns. These column vectors will be elements of R^m. Thus, a necessary condition for the columns of A to span R^m is that n is greater than or equal to m. That is, if n<m, then the columns of A certainly do not span R^m, and T is therefore not onto. Does that answer your question? $\endgroup$ – Gecko Dec 10 '14 at 4:50
  • $\begingroup$ certainly thanks! $\endgroup$ – user197971 Dec 10 '14 at 5:23
  • $\begingroup$ You're welcome, glad I could help! $\endgroup$ – Gecko Dec 10 '14 at 6:16

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.