Question 1031968
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It is not necessarily true.

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I just said it, and I am repeating it one more time and again.


>>> <U>IT IS NOT NECESSARILY TRUE</U>. <<<


It is true only in the case when the operator L is non-degenerated (has the zero kernel). 
Which is not always the case for linear transformations.


The proof of the other tutor is wrong, unfortunately.


<pre>
A contr-example is:


Take 3 linearly independent vectors in {{{R^3}}}.

Let the operator L be the projection {{{R^3}}} on {{{R^2}}}.

Every three vectors in {{{R^2}}} are dependent.

So are dependent in {{{R^2}}} the projections of the original vectors from {{R^3}}}.

But the original vectors were chosen as linearly independent. 

It is on the level of elementary knowledge of linear algebra.
</pre>

Again: the fact that the images are linearly dependent DOES NOT IMPLY that the pre-images are necessarily linearly dependent.