Homework Help: Showing a set is linear dependent without calculations

1. Oct 6, 2012

potetochippu

1. The problem statement, all variables and given/known data

Without any calculation, explain why [(1 3 0)' (2 2 1)' (-1 0 1)' (1 -3 1)' ] is a linear dependent set.

3. The attempt at a solution

I used the theorem which states that the number of vectors in any linearly dependent set is greater than or equal to its dimension. But I cannot use any calculations to find the sets basis and consequently its dimension.

I am finding linear algebra challenging due to the theorems presented. There are too many theorems and some of them are difficult to comprehend! Any tips for this topic will be greatly appreciated as well. Thanks!

2. Oct 6, 2012

HallsofIvy

This is incorrect. For example, the set of just two vectors {(1 3 0), (2 6 0)} is dependent. What is correct is the "converse", that if a set of vectors has more vectors than the dimension of the vector space then they are dependent, which has hypothesis and conclusion reversed from what you have: "if p then q" is NOT the same as "if q then p".

A "set" does not have a basis! These vectors all have three components so are in R3. Do you know what the dimension of R3 is?

You seem to have great difficulty just learning the basic definitions. Mathematics, not just Linear Algebra, requires precision. You need to learn the precise words of definitions and theorems.

I'm not sure what you mean by "too many theorems". Theorems are what mathematics is all about. If you do not like mathematics, which is what you seem to be saying, then I do not understand why you are taking Linear Algebra.

3. Oct 7, 2012

potetochippu

Ok. I believe I have the correct answer now after carefully rereading the definitions and theorems.

The set is in R^3. The dimension of the space R^3 is 3. The numbers of vectors in a linearly independent set of a vector space is less than or equal to the dimension of the vector space. And since the number of vectors in this set is greater than the dimension of its vector space it cannot be linearly independent.

Thanks for the advice about precision, I really need to keep that in mind.