Question about vector independence

[MarcL]

So I was reading my textbook and I confused myself about a theorem

Where if S={v,v₂,...,v_r} and in ℝⁿ then if r>n, then it is linearly dependent

It doesn't make sense to me because if we look at 2 vectors in ℝ³ (lets say u and v)

we have u=(u1,u2,u3) and v=(v1,v2,v3)
So i do k1(u1,u2,u3)+k2(v1,v2,v3)=0
If i use a matrix:

u1 v1 0
u2 v2 0
u3 v3 0

Then it would seem to me as both vectors would be linearly dependent, no?

 

[Stephen Tashi]

So I was reading my textbook and I confused myself about a theorem

Where if S={v,v₂,...,v_r} and in ℝⁿ then if r>n, then it is linearly dependent

It doesn't make sense to me because if we look at 2 vectors in ℝ³ (lets say u and v)

The theorem doesn't say that 2 vectors in R^3 cannot be dependent. It doesn't say that there is dependence "if and only if r > n".

 

[MarcL]

But if I solve for a matrix of a 2x3, I would end up with a free variable no? ( 3rd line, after I REF or RREF the matrix) so that would make it linearly dependent

 

[Number Nine]

But if I solve for a matrix of a 2x3, I would end up with a free variable no? ( 3rd line, after I REF or RREF the matrix) so that would make it linearly dependent

The theorem does not say that 2 vectors in \mathbb{R}^3 must be independent. There is no contradiction.

 

[MarcL]

Well I don't know what I am trying to say is that I see a contradiction, more like " what is wrong in my reasoning" type of thing. I'll give a concrete example I just did, maybe my question will be more clear

Is this set of vector linearly dependent

(8,-1,3),(4,0,1)

8 4 0 --> ( interchange row 2 with row 1, then row 3 with "new" row 2)
-1 0 0
3 1 0

-1 0 0
3 1 0
8 4 0
We end up with 2 rows of leading one, and one row that I can reduce to 0 0 0 ( the third one )easily, so how isn't that linearly dependent?

 

[Stephen Tashi]

so how isn't that linearly dependent?

The two vectors (8,-1,3),(4,0,1) in your example are not linearly dependent.

To determine if two 3-dimensional vectors (u1,u2,u3) and (v1,v2,v3) are dependent, you need to find non-zero solutions x and y to the vector equation:

x ( u1,u2,u3) + y (v1,v2,v3) = 0

Written in matrix form, this can be expressed as:

\begin{pmatrix} u1&v1 \\ u2 & v2\\ u3&v3\end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\0\end{pmatrix}.

Apparently you analyze this equation by writing the "augmented matrix" for it ( where the constants on the right-hand side are put in the matrix, we are dealing with:

\begin{pmatrix} u1 & v1 & 0\\ u2&v2& 0\\u3&v3 & 0\end{pmatrix}

If we row-reduce the matrix so the bottom row is zeroes, what does that tell us?

If it reduces to \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix} , this implies that x = 0, \ y = 0 is the unique solution to the equations, so the two vectors are linearly independent.

If it reduces to something like \begin{pmatrix} 1 & -2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} , this implies there are solutions like x = 2, y = 1 and x = 4, y = 2 etc., so the two vectors are dependent.

 

[MarcL]

Oh I didn't know you guys had a matrix format that I could use... sorry about that

So, trivial solutions are only determined whether or not a x or y ( in this case) variable would have a leading one or not? just seems kinda odd that one row could be 0 without affecting the depedence of the vector

 

[Stephen Tashi]

just seems kinda odd that one row could be 0 without affecting the depedence of the vector

You might be confusing two different techiques for testing independence. Another (and more common technique) for testing the indepdence for a set of n-dimensional vectors is to write the vectors as rows in a matrix and try to row reduce the matrix toward the identity. If you get a row of zeros in that technique you have managed to express the zero vector as a linear combination of other rows, so the vectors are dependent.

If you row reduce \begin{pmatrix} u1 & u2 & u3 \\ v1 & v2 & v3 \end{pmatrix} to something like \begin{pmatrix} 1 &-2 &0 \\ 0 & 0 & 0 \end{pmatrix} then the vectors are dependent.

If you row reduce it something like \begin {pmatrix} 1 & 0 & -2 \\ 0 & 1 & 7 \end{pmatrix} the vectors are independent.

 

[Mark44]

If you have two vectors to consider, it's very easy to determine whether they are linearly independent. If neither one is a constant multiple of the other, the two vectors are linearly independent.

It's a lot harder when you have three or more vectors. Even if no one vector is a multiple of any of the others, the set can still be linearly dependent. For example, consider S = {<1, 1, 0>, <1, -1, 0>, <2, 0, 0>}. No one vector is a multiple of any of the others, but the equation c₁v₁ + c₂v₂ + c₃v₃ = 0, has a solution where not all of the constants are zero.

 

[FactChecker]

if we look at 2 vectors in ℝ³ (lets say u and v)

we have u=(u1,u2,u3) and v=(v1,v2,v3)
So i do k1(u1,u2,u3)+k2(v1,v2,v3)=0

That is not always possible: If u=(1,0,0) and v=(0,1,0) it is not possible to find nonzero k1 and k2 that will give 0.
But it can certainly be true: If u=(1,0,0) and v = (-1,0,0) clearly k1=1 and k2=1 would give 0.
So as long as the number of vectors is less than or equal to the dimension of the space, it can go either way.

However, for R³, it is impossible to find 4 vectors where no set of them are linearly dependent. That is what the theorem says.

 

[Fredrik]

The theorem says that for all positive integers $n$, every subset of $\mathbb R^n$ with cardinality (=number of elements) greater than $n$ is linearly independent. So in particular, it says that every subset of $\mathbb R^3$ with at least four elements is linearly dependent. It says nothing about sets with only two elements, like your $\{u,v\}$.

Some subsets of $\mathbb R^3$ with two elements are linearly dependent, and some are linearly independent. $\{(1,0,0),(2,0,0)\}$ is an example of the former, and $\{(1,0,0),(0,1,0)\}$ is an example of the latter.