Please I just need an Idea!!!

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fatemeh8989

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hi every body!
A is a square n*n matrix which has not any zero rows.
X is a n*1 matrix which has not any zero entry.
Prove that there is such X matrix such that AX has not any Zero entry.
that is it.
I think I have to prove that the Linear combination of these two matrix entries is not equal to zero.
but I don't know how...? please help me... !!!
 
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To start, let n = 2.

Pick symbols (eg: a, b, c, d, e, f) for A and X.

Write out the multiplication AX.

Show us what you get. :cool:
 
fatemeh8989 said:
I am sorry, X is n*1
Click to expand...

Oh, good. (I missed that.)

Well, I had hoped that you would write out the product matrix AX, using symbols for the entries. Maybe, it would have given you an idea. That's what you're asking for, yes? An idea?

I don't know why I previously suggested using e and f as symbols for the entries in X. Think of A as a coefficient matrix, and think of X as the variable matrix. When you multiply them together, you get the left-hand sides to a system of n equations.

If you look at those expressions, it's clear that all of the coefficients in at least one of them would need to be zero in order for AX to have a zero entry, regardless of n. That's not possible, agree?

\(\displaystyle \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) = \; ?\)


What is your next question about this exercise?


 
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Thanks for your attention.
but it does not give n idea for prove. look,
In that linear system which you mentioned I can put these: x = y = 1 , a = c = 1 , b = d = -1 ;
none of a, b, c, d, x and y are Zero but linear combination of them is zero.
so ... As an Idea i thought about Induction proof.
because it is true for n=1 , [a] = [ab] ; since a, b are not equal to 0 so ab is not equal to zero too. ( Base of Induction )
Assume that for an k*k matrix A ( which has no zero rows) and k*1 matrix X ( which has no zero entries ) AX has not any zero entries.
now we have to show for A (k+1 * k+1 ) and X (k+1 * 1) it is true.
If we add these row and columns in previous matrix A (k*K) and add an none zero entry in X (k*1) and multiply them in a way that we can use the facts that has been assumes before, we can say that the proof has completed.
my problem is in last step, I even used Block multiplication but it has not complete yet i think.
so any suggestion..?
 
fatemeh8989 said:
Thanks for your attention.
but it does not give n idea for prove. look,
In that linear system which you mentioned I can put these: x = y = 1 , a = c = 1 , b = d = -1 ;
none of a, b, c, d, x and y are Zero but linear combination of them is zero.
so ... As an Idea i thought about Induction proof.
because it is true for n=1 , [a] = [ab] ; since a, b are not equal to 0 so ab is not equal to zero too. ( Base of Induction )
Assume that for an k*k matrix A ( which has no zero rows) and k*1 matrix X ( which has no zero entries ) AX has not any zero entries.
now we have to show for A (k+1 * k+1 ) and X (k+1 * 1) it is true.
If we add these row and columns in previous matrix A (k*K) and add an none zero entry in X (k*1) and multiply them in a way that we can use the facts that has been assumes before, we can say that the proof has completed.
my problem is in last step, I even used Block multiplication but it has not complete yet i think.
so any suggestion..?
Click to expand...


I'm confused. Did you not just provide a counter example? How can you expect to prove it then?
 
fatemeh8989 said:
If we add these row and columns in previous matrix A (k*K) and add an none zero entry in X (k*1) and multiply them in a way that we can use the facts that has been assumes before, we can say that the proof has completed.
Click to expand...

I don't think that's a valid argument.



We are told that neither x nor y can be zero.

We are told that a and b cannot both be zero.

Hence, it's impossible for ax + by to equal zero.



The situation is the same higher values of n.



For example, let n = 5:

X contains five variables (x, y, z, u, w)

Row1 of A contains five coefficients (a, b, c, d, e)

We are told that none of the variables equal zero.

We are told that not all of the coefficients in Row1 can be zero.

Hence, it is impossible for ax + by + cz + du + ew to equal zero.
 
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yes you are right about my example... but it was my wrong about writtng the question... we just want to say there is such a matrix - X , which....
sorry..I just find some time at nights to come here and I am sleepy... < blushing>
so...in this condition need to prove that such X exists.
one of my classmates has an idea about Vector Spaces ... ( not complete yet too )
so what do you think now..?
 
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