Linear Algebra query

[Sarahisme]

hey just a query about this question....

 

for part c where it asks if it is wise to check if w in a lin. combination of the other vectors. well i suppose the answer is no, but if you check if v was a lin. combination of the other, then if u was a lin. combination of the others and then if z was a lin. combination of the others (which i think is the case for this problem) then you would be able to say whether of not the set is linearly dependent or not.

 

what do peoples think?

 

Cheers

 

Sarah

[Sarahisme]

lol i really tried to shink this picture down this time, guess it didnt work eh?

[Dapthar]

what do peoples think?

You're right' date=' it's a waste of time to check, since you have four vectors in [math']\mathbb{R} ^3[/math], thus they must be linearly dependent, since there can't be more than three linearly independent vectors in [math]\mathbb{R} ^3[/math].

[Sarahisme]

lol cool thanks

[Johnny5]

I need to review this myself Sarah, so first let me get this question down. I don't see a part c, so I'm just going to do them all.

 

[math] \mathbf{u} \equiv

\left [

\begin{array}{c}

3\\

2\\

-4\\

\end{array} \right ] [/math]

 

[math] \mathbf{v} \equiv

\left [

\begin{array}{c}

-6\\

1\\

-7\\

\end{array} \right ] [/math]

 

[math] \mathbf{w} \equiv

\left [

\begin{array}{c}

0\\

-5\\

2\\

\end{array} \right ] [/math]

 

[math] \mathbf{z} \equiv

\left [

\begin{array}{c}

3\\

7\\

-5\\

\end{array} \right ] [/math]

 

Question 1: Are the sets {u,v}, {u,w}, {u,z}, {v,w}, {v,z}, {w,z} linearly dependent? Why, or why not?

 

In order to answer this question, we must know the definition of linear dependence.

 

Let us just cover the case where the number of vectors is finite.

 

A set of vectors [math] \mathcal{f} v_1,v_2,v_3,...,v_n \mathcal{g} [/math] is said to be linearly dependent if and only if we can find scalars {r1,r2, r3...rn} such that:

 

[math] \sum_{i=1}^{i=n} r_i \cdot \mathbf{v}_i = \vec {\mathbf{0}} [/math]

 

The symbol on the RHS denotes the zero vector of an arbitrary vector space. This symbol shows up in the axioms of a vector space.

 

Now, there are different 'fields' which the scalars can come from, so we should take this into account. For example, in a given problem, the scalars might come from the real numbers, while in another, the scalars might come from the complext numbers. So we should expect this to vary, from one problem to another. But in all cases, the scalars must come from a field, and therefore obey the field axioms, which means that the scalar 0, and the scalar 1, must be elements of the field.

 

So let us incorporate this into the previous definition of linear dependence.

 

Definition:

 

A set of vectors [math] \mathcal{f} v_1,v_2,v_3,...,v_n \mathcal{g} [/math] chosen from some given vector space, is said to be linearly dependent over some chosen field [math] \mathbb{F} [/math] of scalars, if and only if we can find scalars {r1,r2, r3...rn}, not all zero, ( such that:

 

[math] \sum_{i=1}^{i=n} r_n \cdot \mathbf{v}_i = \vec {\mathbf{0}} [/math]

 

Online definition

 

 

 

Part 1: Are the vectors u,v linearly dependent?

 

u = <3,2,-4>

v = <-6,1,-7>

 

Choose two scalars at random, r1, r2.

 

r1u = r1<3,2,-4> = <3r1,2r1,-4r1>

r2v = r2<6,1,-7> = <-6r2,r2,-7r2>

 

r1u + r2v = <3r1-6r2,2r1+r2,-4r1-7r2>

 

These two vectors will be linearly dependent if there are r1,r2, not both zero, and:

 

3r1-6r2 = 0

2r1+r2 = 0

-4r1-7r2 = 0

 

So now, we have to find two different scalars, r1,r2, which makes the three equations above true simultaneously.

 

The first equation gives us a constraint, which is this:

 

3r1=6r2

 

or equivalently

 

r1=2r2

 

Now, putting this into the second and third equations leads to:

 

2(2r2)+r2 = 0

-4(2r2)-7r2 = 0

 

4(r2)+r2 = 0

-8(r2)-7r2 = 0

 

5r2 = 0

-15r2 = 0

 

The only way for either mathematical statement above to be true, is if r2=0, and using the constraint, it follows that r1 must also be zero.

 

Thus, it is not the case that there are least two scalars r1,r2, not both zero, for which:

 

 

r1u + r2v= 0

 

Hence, by definition, it is not the case that u,v are linearly dependent.

[matt grime]

Johnny, the naturals are not a field so cannot form the underlying scalars of any vector space.

 

I haven't checked what the vectors are but I dont' need to since I can tell that the question is asking this:

 

Suppose that u,v,w,z are 4 vectors in ANY space, for now.

 

 

Suppose further that each pair of vectors, {u,v} {u,w} etc is an independent set.

 

Is the set {u,v,w,z} NECESSARILY independent?

 

The answer is no, as the above example shows: we almost certainly have 4 vectors that are pairwise linearly independent, but since they lie in a 3-d space, they must collectively be dependent.

 

It is not a good idea to check dependence by attempting to write one specific vector as a combination of the other three.

 

Let is show this by taking a deliberatly stupid example, suppose we hve the vectors a,2a,3a and b where a and b are linearly independent (in some space), then we although we know these are obviosuly dependent, we cannot write b as a combination of the vectors a,2a,3a.

[Johnny5]

Johnny, the naturals are not a field so cannot form the underlying scalars of any vector space.

 

That's right Matt.

 

I will go back and change it right now.

[Johnny5]

 

I haven't checked what the vectors are but I dont' need to since I can tell that the question is asking this:

 

Suppose that u' date='v,w,z are 4 vectors in ANY space, for now.

[/quote']

 

Matt, if the number of vectors exceeds the dimension of the vector space, then mustn't they be linearly dependent?

 

In other words, in the case of four vectors in 3D space, one can be written as a linear combination of the other three?

 

In this problem here, the vectors are in a three dimensional space(you can tell from the column vector with three rows), but there are four vectors, hence they are linearly dependent, that is, any one of the vectors can be written as a linear combination of the other three.

[matt grime]

Yes, 4 vectors in 3 dimensions must be linearly dependent, but once more I am completely mystified as to what your point is.