Proof

quackensack · 2007-03-10 12:57:30

hey i take linear algebra, and i have a quiz coming up on monday so i'm studying proofs. i was wondering if someone could show me how to do the following:

suppose that A is an n by n matrix. prove that the rank of A is n if and only if the equation AX=0 has just the trivial solution.

Tigeree · 2007-03-10 13:19:19

i.......never learnt that

Sekky · 2007-03-10 13:23:37

http://en.wikipedia.org/wiki/Rank-nullity_theorem

That should help

Informally, if your kernel is zero (ie the morphism never fails injection), then the rank must be the length of the matrix, because all columns will be linearly independent and therefore the morphism will always be injective.

Tigeree · 2007-03-10 13:34:00

hey wot happend to janefairfax' post

JaneFairfax · 2007-03-10 14:07:35

It was just a query about X; I deleted it as Sekkys post appeared to have answered the question.

quackensack · 2007-03-10 14:24:28

Thanks for all your input, guys!

I was wondering - is there a way to do it without using the nullity theorem? I haven't about the concept of linear dependence in my class (as far as I know, that is).

A friend told me that I can do this proof by putting a matrix A in row-reduced echelon form. If doing this yields a matrix with at least one non-zero entry in each row, then the only solution to AX=0 is trivial. This , in turn, is the definition of rank n if A is an n by n matrix.

I guess where I'm stuck is...how does one row-reduce the matrix A without knowing what it is?

Sekky · 2007-03-10 14:50:32

quackensack wrote:

Thanks for all your input, guys!
I was wondering - is there a way to do it without using the nullity theorem? I haven't about the concept of linear dependence in my class (as far as I know, that is).
A friend told me that I can do this proof by putting a matrix A in row-reduced echelon form. If doing this yields a matrix with at least one non-zero entry in each row, then the only solution to AX=0 is trivial. This , in turn, is the definition of rank n if A is an n by n matrix.
I guess where I'm stuck is...how does one row-reduce the matrix A without knowing what it is?

It's the other way around, if you obtain a row of all zeros when it means one row was a linear combination of the other rows, hence not all rows are linearly independent. If this is true, then there will be infinitely many solutions to AX=0, because it will depend on less rows if you catch my drift. The number of rows of zeros is the kernel of the morphism, the number of non-zero rows is the rank.

quackensack · 2007-03-10 15:31:18

I'm sorry - I'm a little stuck because I haven't learned about linear dependence, kernel, or morphism in my class. Is it possible to show what you mean using a matrix? If it isn't, that's okay. I just think that it would be easier for me to understand it visually.

Ricky · 2007-03-10 15:48:39

I haven't learned about linear dependence, kernel, or morphism

You sure about that? I don't know how you can get to the rank of a matrix without doing the above.

Have you learned about linear transformations?

quackensack · 2007-03-10 15:52:43

Can't one just say:

If AX=0 has the trivial solution, then A is row equivalent to the identity matrix:

(1 0 0)
(0 1 0)
(0 0 1)

The inverse of the identity matrix is itself. And thus it is nonsingular.

Thus, A has rank n.

Because the identity matrix is equal to A, then A has rank n.

Last edited by quackensack (2007-03-10 16:56:31)

Sekky · 2007-03-11 00:43:33

quackensack wrote:

Can't one just say:
If AX=0 has the trivial solution, then A is row equivalent to the identity matrix:
(1 0 0)
(0 1 0)
(0 0 1)
The inverse of the identity matrix is itself. And thus it is nonsingular.
Thus, A has rank n.
Because the identity matrix is equal to A, then A has rank n.

Sure

btw, the solution (or degree of solution) to AX=0 is called the Kernel of the homomorphism.

quackensack · 2007-03-11 02:14:10

i talked to my teacher this morning, and he said that the everything I wrote starting with "The inverse of the identity matrix is itself..." is correct. but he also said that AX=0 always has the trivial solution, but A isn't necessarily row-equivalent to the identity matrix.

i am confused...how do i prove that it is in this case? we haven't learned the invertible matrix theorem, so i'm not sure how else to prove it.

mathsyperson · 2007-03-11 06:57:47

Using the fact that AX=0 has the trivial solution won't help you, as your teacher pointed out.
What you need to use is that it has just the trivial solution. That means that no other matrix X will fit the equation, and you should be able to use that to help you.

Also, keep in mind that it is an 'if and only if' proof, which means that you need to prove that if A has rank n then AX=0 has no non-trivial solutions, and that if AX=0 has no non-trivial solutions then A has rank n.

quackensack · 2007-03-11 08:44:56

mathsyperson wrote:

Using the fact that AX=0 has the trivial solution won't help you, as your teacher pointed out.
What you need to use is that it has just the trivial solution. That means that no other matrix X will fit the equation, and you should be able to use that to help you.
Also, keep in mind that it is an 'if and only if' proof, which means that you need to prove that if A has rank n then AX=0 has no non-trivial solutions, and that if AX=0 has no non-trivial solutions then A has rank n.

Okay, but where I'm stuck is that I'm not sure how to prove that if A has rank n, then AX=0 has no non-trivial solutions.

lightning · 2007-03-12 07:23:15

quackensack wrote:

mathsyperson wrote:
Using the fact that AX=0 has the trivial solution won't help you, as your teacher pointed out.
What you need to use is that it has just the trivial solution. That means that no other matrix X will fit the equation, and you should be able to use that to help you.
Also, keep in mind that it is an 'if and only if' proof, which means that you need to prove that if A has rank n then AX=0 has no non-trivial solutions, and that if AX=0 has no non-trivial solutions then A has rank n.
Okay, but where I'm stuck is that I'm not sure how to prove that if A has rank n, then AX=0 has no non-trivial solutions.

interesting.....any solutions?

Last edited by lightning (2007-03-12 07:23:39)

George,Y · 2007-03-12 18:21:54

rank of the Matrix=rank of the rows of the Matrix=rank of the columns of the Matrix.
This is a theorem, whose proof is rather complex.

Now Ax=0

Now you may understand once solution of x including x_1 that x_1=0, the rank of the 3 columns of A will no longer be 3.

krassi_holmz · 2007-03-12 22:08:35

I haven't learned about linear dependence, kernel, or morphism

Better learn it first.
Linear dependency for vectors basically means, that every has a little "different dimension" form theothers.
Formally, a system {A,B,...,Z} is lineary independent only and only if the equation
aA+bB+cC+...+zZ=0 doesn't have solutions, different from {0,0,...,0} for real a,b,...,z.
For example, if two vectors A,B are collinear, then A=mB, m e R (=/= 0), so
1.A-m.B=0, 1,m =/=0 ==> A and B are dependant.

Math Is Fun Forum

#1 2007-03-10 12:57:30

Proof

#2 2007-03-10 13:19:19

Re: Proof

#3 2007-03-10 13:23:37

Re: Proof

#4 2007-03-10 13:34:00

Re: Proof

#5 2007-03-10 14:07:35

Re: Proof

#6 2007-03-10 14:24:28

Re: Proof

#7 2007-03-10 14:50:32

Re: Proof

#8 2007-03-10 15:31:18

Re: Proof

#9 2007-03-10 15:48:39

Re: Proof

#10 2007-03-10 15:52:43

Re: Proof

#11 2007-03-11 00:43:33

Re: Proof

#12 2007-03-11 02:14:10

Re: Proof

#13 2007-03-11 06:57:47

Re: Proof

#14 2007-03-11 08:44:56

Re: Proof

#15 2007-03-12 07:23:15

Re: Proof

#16 2007-03-12 18:21:54

Re: Proof

#17 2007-03-12 22:08:35

Re: Proof

Board footer