let's say that their is a linear transformation T:V --> W. If X1,...,Xr are vectors in V such that T(X1),...,T(Xr) are linearly independent, then is it necessarily true that the vectors X1,...,Xr are linearly independent as well?

Suppose that it's not necessarily so and we imagine your example but with vectors Xi being linearily dependent in V. Then a certain Xi can be written as a linear combination of the others:

 

[math]x_i = \sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } [/math]

 

so then

 

[math]T\left( {x_i } \right) = T\left( {\sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } } \right)[/math]

 

but since T is lineair, we then have

 

[math]T\left( {x_i } \right) = T\left( {\sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } } \right) = \sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j T\left( {x_j } \right)} [/math]

 

and we have written a certain T(Xi) as a linear combination of the other images making them linearly dependent and that's not what we supposed at the start.

 

I hope I didn't make any silly mistakes

Suppose that it's not necessarily so and we imagine your example but with vectors Xi being linearily dependent in V. Then a certain Xi can be written as a linear combination of the others:

 

[math]x_i = \sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } [/math]

 

so then

 

[math]T\left( {x_i } \right) = T\left( {\sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } } \right)[/math]

 

but since T is lineair' date=' we then have

 

[math']T\left( {x_i } \right) = T\left( {\sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j x_j } } \right) = \sum\limits_{\scriptstyle j = 1 \hfill \atop

\scriptstyle j \ne i \hfill}^r {\alpha _j T\left( {x_j } \right)} [/math]

 

and we have written a certain T(Xi) as a linear combination of the other images making them linearly dependent and that's not what we supposed at the start.

 

I hope I didn't make any silly mistakes

Thats a nice proof.

let's say that their is a linear transformation T:V --> W. If X1,...,Xr are vectors in V such that T(X1),...,T(Xr) are linearly independent, then is it necessarily true that the vectors X1,...,Xr are linearly independent as well?

Ya, well you need to be a bit careful about terminology here. A linear operator T acts upon elements of a vector space V, and where T: V-->V it is referred to as a linear transformation. But if T:V-->W, with V and W being different spaces, then T is an operator, not a transformation. But yes, if V and W are of the same dimension, the answer to your question is yes. But they may not be of the same dimension!

 

With that qualification, we can ask: as the maximum number of linearly independent vectors in a vector space V of dimension n is n, is there an operator T:V-->W where W is of dimension n - 1?

 

The answer is yes, there are such operators, so that the n linearly independent vectors in V may not be linearly independent in W under T (because W only admits of n - 1 linearly independent vectors).

 

However, you asked a different question. If the vectors w in W with dim W = n are independent, are the v in V independent if T:V-->W and dim V = n - 1.

 

In general, I suppose, the answer is yes, except that, whereas I can think up an operator which takes an n-space to an n-1-space, I cannot think of one that goes the other way.