# Linearly dependent sets

1. Feb 14, 2010

### Mola

If A is a 3x3 Matrix and {v1, v2, v3} is a linearly dependent set of vectors in R^3, then {Av1, Av2, Av3} is also a linearly dependent set???

Is this true? Can someone please explain why or why not??

What I think: I think it is true because I read that a linear transformation preserves the operations of vector addition and scalar multiplication.

2. Feb 14, 2010

### rasmhop

If v1,v2,v3 are linearly dependent you can find constants a1, a2, a3 not all 0 such that
$$a_1v_1 + a_2v_2 + a_3v_3 = 0$$
Now left-multiplying this by A you get:
$$A(a_1v_1+a_2v_2+a_3v_3) = A0=0$$
Now use your rules for matrix arithmetic to derive:
$$a_1(Av_1)+a_2(Av_2)+a_3(Av_3)=0$$
(HINT: Ak = kA for constants k, and A(v+w) = Av+Aw for vectors v, w where the expression makes sense).

3. Feb 14, 2010

### Mola

That makes sense. So if we have a1(Av1) + a2(Av2) + a3(Av3) = 0, then at least one of the constants could be zero and that will definitely result to a linearly dependent set.
Thanks.

That leads me to a related theory: Let's assume we are talking about {v1, v2, v3} being a linearly INDEPENDENT set now. If we multiply the vectors by the matrix A, how does it affect the independece? Would it make a differerence if the matrix A is invertible?

4. Feb 14, 2010

### rasmhop

This is actually a quite interesting little question (well in my opinion anyway). First for fixed A, v1,v2,v3 note that if we take the contrapositive of your initial result we get:
If Av1, Av2, Av3 are linearly independent, then v1,v2,v3 are linearly independent.
so for linear independence it goes backwards. For an arbitrary matrix A we can not prove your new statement since we can just let A be the 0 matrix. However if A is invertible, then we can just go backwards by noting that if,
$$a_1Av_1+a_2Av_2+a_3Av_3 = 0$$
Then we can left-multiply by $A^{-1}$ to get,
$$a_1v_1+a_2v_2+a_3v_3 = 0$$
so if v1,v2,v3 are linearly independent and A is invertible, then Av1, Av2, Av3 are linearly independent.

5. Feb 15, 2010

### Mola

Thanks rasmhop... I did think "A" being an invertible matrix could make a difference but I didn't know how to prove it.
That was a very good help from you.