linear independence question

[robbiek]

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?

[TD]

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

[cosine]

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.

[Xerxes]

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.