independant proof

**transgalactic** · June 22nd 2011, 01:52 PM

this is a two part question:
v1..vn is a basis
of R^n
and there is a vector b which belongs to R^n and b differs the sero vector
A.
proof that {v1-vn,v2-vn,...,vn-1 - vn}

by definition
in order to prove that a group is independant
we need to show the the only way for

a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

is a1=..=an=0 all the coefficient hs to be zero
so it rang a bell "i need to get a trivial solution "

but trivilal solution is in Ax=0 system could be should if |A| differs zero.

but A is a square matrices by difinition.
how to construct from this single equation a square matrices?
??

the second question:
if v1..vn are solutions to Ax=b system then
rho(A)=1
??
rho(A) is the dimention of row or column space

if v1..vn are solving this system
then the dimention of the solution space is n dim(P(A))=n
and from the formula where n=dim(P(A))+rho(A) we get n=n+rho(A)
so i got
rho(A)=0

but i am asked to prove that rho(A)=1

where is my mistake

**tonio** · June 22nd 2011, 06:53 PM

Originally Posted by transgalactic

this is a two part question:
v1..vn is a basis
of R^n
and there is a vector b which belongs to R^n and b differs the sero vector
A.
proof that {v1-vn,v2-vn,...,vn-1 - vn}

by definition
in order to prove that a group is independant
we need to show the the only way for

a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

is a1=..=an=0 all the coefficient hs to be zero
so it rang a bell "i need to get a trivial solution "

but trivilal solution is in Ax=0 system could be should if |A| differs zero.

but A is a square matrices by difinition.
how to construct from this single equation a square matrices?
??

the second question:
if v1..vn are solutions to Ax=b system then
rho(A)=1
??
rho(A) is the dimention of row or column space

if v1..vn are solving this system
then the dimention of the solution space is n dim(P(A))=n
and from the formula where n=dim(P(A))+rho(A) we get n=n+rho(A)
so i got
rho(A)=0

but i am asked to prove that rho(A)=1

where is my mistake

Make yourself a favour and do use LaTeX for your questions (already more than 1,000!). It's extremely hard to read them as they are.

Tonio

**topspin1617** · June 22nd 2011, 07:54 PM

Originally Posted by tonio

Make yourself a favour and do use LaTeX for your questions (already more than 1,000!). It's extremely hard to read them as they are.

Tonio

In addition to spell check. Maybe grammar check as well, if such a thing is possible.

**transgalactic** · June 22nd 2011, 11:27 PM

at first i have what is given fro both question.

then i write the firrst question

and explain how i tried to solve it

same thing for the second question

**topspin1617** · June 22nd 2011, 11:32 PM

Originally Posted by transgalactic

at first i have what is given fro both question.

then i write the firrst question

and explain how i tried to solve it

same thing for the second question

Yes, that's all great, but that doesn't change the fact that your work is EXTREMELY difficult to read.

**transgalactic** · June 22nd 2011, 11:53 PM

i will try to write in a more understandable way.

i am given:
v1..vn is a basis in $R^n$
$b\epsilon R^{n}$
$b\neq\{0\}$

first question:
proof that {v1-vn,v2-vn,...,vn-1 - vn} is an independant group

how i tried to solve the first question:

by definition:
in order to prove that a group is independant,
we need to show the the only way for the bottom equation to exist

a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

is if a1=..=an=0 all the coefficient has to be zero.
so it rang a bell "i need to get a trivial solution "

but trivilal solution is in Ax=0 system could be should if |A| differs zero.

but A is a square matrices by definition.
how to construct from this single equation a square matrices?
??

second question:

if v1..vn are solutions to Ax=b system then
$\rho(A)$ =1
??

how i tried to solve the second question:
$\rho(A)$ is the dimention of row or column space

if v1..vn are solving this system
then the dimention of the solution space is n, dim(P(A))=n
and from the formula where n=dim(P(A))+ $\rho(A)$ we get n=n+ $\rho(A)$
so i got
$\rho(A)$ =0

but i am asked to prove that $\rho(A)$ =1

where is my mistake
?

**topspin1617** · June 23rd 2011, 01:57 AM

Well... for #2, the rank of the matrix certainly cannot be 0, unless it is the zero matrix. The problem is that the system is nonhomogeneous, so you can't directly apply that formula.

Show first that $v_1-v_2,v_2-v_3,v_1-v_3\in \mathrm{null}(A)$ . Then show that any two of those form a basis for $\mathrm{null}(A)$ .

For #1, you don't really need matrices. Just distribute everything out, collect like terms, and see what happens.

**tonio** · June 23rd 2011, 03:02 AM

Originally Posted by transgalactic

i will try to write in a more understandable way.

i am given:
v1..vn is a basis in $R^n$
$b\epsilon R^{n}$
$b\neq\{0\}$

first question:
proof that {v1-vn,v2-vn,...,vn-1 - vn} is an independant group

how i tried to solve the first question:

by definition:
in order to prove that a group is independant,
we need to show the the only way for the bottom equation to exist

a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

is if a1=..=an=0 all the coefficient has to be zero.
so it rang a bell "i need to get a trivial solution "

but trivilal solution is in Ax=0 system could be should if |A| differs zero.

but A is a square matrices by definition.
how to construct from this single equation a square matrices?
??

Several things:

1) Stop calling "group" to sets of vectors. This is incorrect and confusing since groups are very different things.

2) If all you have to do is to prove that $\{v_1-v_n,\,v_2-v_n,\,...,\,v_{n-1}-v_n\}$ is an independent set , then do as follows:

$a_1(v_1-v_n)+a_2(v_2-v_n)+...+a_{n-1}(v_{n-1}-v_n)=0\Longrightarrow a_1v_1+a_2v_2+...+a_{n-1}v_{n-1}+(a_1+...+a_{n-1})v_0$ . and now

use that [tex]\[v_1,...,v_n\}[\tex] is an indep. set and deduce directly that $a_1=a_2=...=a_{n-1}=0$ .

Tonio

Pd. What's your mother tongue?

second question:

if v1..vn are solutions to Ax=b system then
$\rho(A)$ =1
??

how i tried to solve the second question:
$\rho(A)$ is the dimention of row or column space

if v1..vn are solving this system
then the dimention of the solution space is n, dim(P(A))=n
and from the formula where n=dim(P(A))+ $\rho(A)$ we get n=n+ $\rho(A)$
so i got
$\rho(A)$ =0

but i am asked to prove that $\rho(A)$ =1

where is my mistake
?

.

**transgalactic** · June 23rd 2011, 05:03 AM

tonio:
my mother tongue is russian but i am a student in israel so my course is in hebrew.
thanks i got the idea

topspin1617:
so in the second question i cant use this formula
n=dim(P(A))+ $\rho(A)$
because P(A) represents the solution set of the homogeneus system.

but here i got non homogeneus system.
so what to do in the second question?

**tonio** · June 23rd 2011, 12:14 PM

Originally Posted by transgalactic

tonio:
my mother tongue is russian but i am a student in israel so my course is in hebrew.

Ну, я понимаю, и я надеюсь, у вас есть удачи в учебе.

באיזו אוניברסיטה אתה לומד? אני בוגר מהאוניברסיטה העברית בירושלים

thanks i got the idea

topspin1617:
so in the second question i cant use this formula
n=dim(P(A))+ $\rho(A)$
because P(A) represents the solution set of the homogeneus system.

but here i got non homogeneus system.
so what to do in the second question?

I'm answering in english for the sake of other participants: indeed, you can't use that

formula because the set of all solutions to a non-homogeneous system is never a subspace.

The second question is not that easy: if $Ax=b\, ,\,b\neq 0$ , then either $\rho(A)=n$ and

there's a unique solution, or else $\rho(A)<n$ and there're more than one solution (in

fact, infinite solutions if we're working over an infinite field).

Now, what the question says, I think, is that the basis $\{v_1,...,v_n\}$ is a solution to the non homogenoeus

system, and from here we must deduce $\rho(A)=1$

Now we must remember that any solution to a non-homogeneous system is of the form

$\alpha+\xi\, , \,\alpha=\mbox{ a particular solution to the non-homog. system and } \xi=\mbox { a solution of the asociated homogeneous sytem}$ ,

so we can choose say $v_n$ as a particular solution and then we get that

$\forall \,\,1\leq k <n\, , \,v_k=v_n+\xi_k \iff \{\xi_1,...,\xi_{n-1}\}=\{v_1-v_n,...,v_{n-1}-v_n\}$ , and this last set has

dimension n - 1 by the first question, so...end the argument.

Tonio

**transgalactic** · June 23rd 2011, 02:58 PM

Ну, я понимаю, и я надеюсь, у вас есть удачи в учебе.

באיזו אוניברסיטה אתה לומד? אני בוגר מהאוניברסיטה העברית בירושלים

wow what are the odds for that to happen
thanks i hope i get this succses in this course.

i study at Ben gurion university

**transgalactic** · June 23rd 2011, 03:27 PM

i though of a similar solution:
v1..vn are solutions to the non homogeneus system Ax=b,
if we subtract two solutions for the non homogeneus system then we get a solution to the homogeneus system AX=0.

so from subtracting pairs from the n solutions to the non homogeneus we get n-1 solutions to the homogeneus system.

so dim(P(A))=n-1
now we apply the formula
n=dim(P(A))+ $\rho (A)$
n=n-1+ $\rho (A)$
$\rho (A)$ =1

is this solution correct?

**topspin1617** · June 24th 2011, 11:02 PM

Originally Posted by transgalactic

i though of a similar solution:
v1..vn are solutions to the non homogeneus system Ax=b,
if we subtract two solutions for the non homogeneus system then we get a solution to the homogeneus system AX=0.

so from subtracting pairs from the n solutions to the non homogeneus we get n-1 solutions to the homogeneus system.

Okay up to here.

so dim(P(A))=n-1
now we apply the formula
n=dim(P(A))+ $\rho (A)$
n=n-1+ $\rho (A)$
$\rho (A)$ =1

is this solution correct?

So we have $n-1$ solutions to the system; i.e., $n-1$ elements of $\mathrm{null}(A)$ . This does NOT automatically force $\mathrm{dim}\,\mathrm{null}(A)=n-1$ , however.

tonio's solution works. I also have a solution, which looks different (though is probably the same idea anyway).

We know that $Av_1=\cdots =Av_n=b$ . As you have already noted, each of $v_1-v_n,\ldots,v_{n-1}-v_n$ is an element of $\mathrm{null}(A)$ .

We know these are linearly independent by the first part. Let's show that they also span $\mathrm{null}(A)$ .

Let $v\in \mathrm{null}(A)$ be arbitrary. Since $v\in V$ and $\{v_1,\ldots,v_n\}$ is a basis for $V$ , we can write $v=a_1v_1+\cdots +a_nv_n$ , for some $a_1,\ldots,a_n\in \mathbb{R}$ . So we have

$0=Av=A(a_1v_1+\cdots a_nv_n)=a_1Av_1+\cdots +a_nAv_n=a_1b+\cdots +a_nb=(a_1+\cdots +a_n)b$ .

Now, assume $b\neq 0$ (otherwise $\mathrm{dim}\,\mathrm{null}(A)=n\Rightarrow A=0$ ). This forces the scalar $a_1+\cdots a_n=0$ (the only way a scalar times a vector can be zero is if either the scalar or the vector is already zero). Solving for $a_n$ , we have $a_n=-(a_1+\cdots +a_{n-1})$ . Substituting this into the expression for $v$ ,

$v=a_1v_1+\cdots +a_{n-1}v_{n-1}-(a_1+\cdots +a_{n-1})v_n=a_1(v_1-v_n)+\cdots +a_{n-1}(v_{n-1}-v_n)$ .

Thus, $v\in \mathrm{span}(v_1-v_n,\ldots,v_{n-1}-v_n)$ , which is what we wanted to show.

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