Re: linear independence of identity matrix vectors
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Sun, 13 Nov 2005 10:05:34 EST
On Nov 12, 2005 1:48 PM CT, Gary Weselle wrote:
> Narcoleptic Insomniac wrote:
>
> > On Nov 11, 2005 5:06 PM CT, Gary Weselle wrote:
> >
> >
> >>Hello
> >>
> >>Are the standard basis vectors linearly independent?
> >
> >
> > Yes they are. In fact, suppose you are given the
> > collection of vectors {v_1, v_2, ..., v_n} that
> > form a basis for the space W. Then by definition
> > the vectors v_1, v_2, ..., v_n are all linearly
> > independent and W = span{v_1, v_2, ..., v_n}.
> >
> >
> >>I learned that they are but I was thinking about it
> >>and could not think of how could one of them be
> >>gotten using linear combination of the rest.
> >
> >
> > Your intuition is correct; like Russell said, you
> > must have gotten linear dependence confused with
> > linear independence.
> >
> > Regards,
> > Kyle
>
> then linear combination indicated linear dependencies,
> so it seams to me that the following statement is not
> true.
> The span of the vectors v_1, ..., v_n is the set
> of linear combination a_1 v_1 + … + a_n v_n and is
> denoted Sp(v_1,…,v_n)
> Which is a definition of Span from a math book I have.
> If it is true, how can that be?
Yes, this is the definition of "span". The span of a
set of vectors is all possible linear combinations of
said vectors.
It's very important to note that a set of vectors in a
span do not have to be linearly independent. However,
if we happen to have a set of vectors in which some are
dependent on the others, then those dependent vectors
become redundant.
Let's take a concrete example here. Consider the set of
vectors V = {(1,0,0), (0,1,0), (1,-2,3), (0,0,1)} and
B = {(1,0,0), (0,1,0), (0,0,1)}. Note that
(1,-2,3) = 1*(1,0,0) - 2*(0,1,0) + 3*(0,0,1).
This explicitly shows a linear dependence of the vector
(1,-2,3) on the standard basis vectors. However, it's
true that Span{V} = Span{B} = R^3. That is to say, the
collection of all possible linear combinations of V is
identical to the collection of all possible linear
combinations of B. Moreover, this collection is R^3.
Another important thing to note here in this example is
that V is NOT a basis for R^3. From the usual definition
for "basis" that I posted earlier we see that V is not a
basis for R^3 since all the vectors in V are not linearly
independent.
The following technique is useful for determining linear
(in)dependence. Let's take four 3-vectors, say
(1,0,0), (0,2,-1), (1,1,0), and (1,0,1). To check these
for any dependence relations we can set them as columns
in a matrix:
1 0 1 1
0 2 1 0
0 -1 0 1.
If we reduce this matrix we obtain:
1 0 0 -1
0 1 0 -1
0 0 1 2.
Notice how the 3x3 identity is "inside" our reduced
matrix. Each pivot here is telling us that the original
vector that was in that column is linearly independent.
Thus, the set V = {(1,0,0}, (0,2,-1), (1,1,0)} is
linearly independent.
So what happened to the vector (1,0,1)? The cool thing
is that our reduced matrix tells us the *exact* dependence
relationship (1,0,1) has on the others; namely
(1,0,1) = -1*(1,0,0) - 1*(0,2,-1) + 2*(1,1,0)
...as you can verify. Note where the coefficients came
from. The reason I brought this technique up is to
point out what would happen if you began with the
standard basis vectors. The result of this should be
very clear and I should probably stop rambling now.
Regards,
Kyle
.
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- Re: linear independence of identity matrix vectors
- From: Gary Weselle
- Re: linear independence of identity matrix vectors
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