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Proof of linear independence and dependence

  • Thread starter andorrak
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  • #1
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1. Homework Statement

There are two proofs:

Let X and Y be two matrices such that the product XY is defined. Show that if the columns of Y are linearly dependent, then so are the columns of the matrix XY.


Let X and Y be two matrices such that the product XY is defined. Show that if the columns of the matrix XY are linearly independent, then so are the columns of Y


2. Homework Equations
N/A


3. The Attempt at a Solution

Solution for first. i do not know

Solution for second one perhaps?

If XY are assumed to be the identity matrix. Thus we know I = (A^-1)(A)

Therefore, A=X and A^-1=Y. Then we know X^-1=Y. Then by the invertible matrix theorem, the equation Ax=0 has only the trivial solution and must be linearly independent?

Those are my two cents, can anyone help me?
 

Answers and Replies

  • #2
lanedance
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for the first part start with
[tex]
Y = \begin{pmatrix} y_1 & .. & y_i .. & y_n \end{pmatrix}
[/tex]

[tex] X = \begin{pmatrix} x_1^T \\ .. \\ x_j^T \\.. \\ x_m^T \end{pmatrix}
[/tex]

now assume some y_i = a.y_r + b.ys (ie a linearly dependent column vectro for some r,s) and consider the action of the multiplication (each element in XY will correspond to a dot product between the row vector of X and the column vectros of Y)

so what is x^T dot ( a.y_r + b.ys )?
 
Last edited:
  • #3
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Um could you elaborate? I have no idea what you are saying. sorry
 
  • #4
lanedance
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i re-wrote above for clarity & expanded a little
 
  • #5
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Sorry to sound nagging.

But I still have no idea what you are saying. Or what significance it brings. I am only on chapter 2 of the linear algebra book by Lay. I assumed it had something to do with the invertible matrix theorem
 
  • #6
lanedance
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try expaninding this (for arbitrary vectors)

x^T dot ( a.y_r + b.ys )?
 
  • #7
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what is the dot product? if you could answer it systematically it would be very helpful
 
  • #9
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O thats just matrix multiplication. but i do not understand your notation, ie this.

try expaninding this (for arbitrary vectors)

x^T dot ( a.y_r + b.ys )?

so i assume x^T is the column for the X matrix. and r and s are the dependent vectors and the those dots between a.y and b.y i assume then are the dot products but where are the a's coming from? the entries of the x matrix?
 
  • #10
lanedance
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x_i^T is a row vector of X

say a column vector y_j can be written as a linear combination of other coulmn vectors y_r, y_s, for a,v constants

[tex] y_j = ay_r + a y_s [/tex]

[tex] x_i^T \bullet y_j = x_i^T \bullet (a y_r + b y_s) [/tex]
 
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  • #11
lanedance
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so consider the jth column of XY
[tex]
XY_j = \begin{pmatrix} x_1^T \bullet y_j\\ .. \\ x_i^T \bullet y_j\\.. \end{pmatrix}

[/tex]
 
  • #12
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OH i think i get it. because you assume all teh columns of Y are linearly dependent you can show (which is what a and b represent) are t9eh coefficients when you have a linearly combination between the sets. Now when you multiply X^T with it, that is simply a scalar multiple of the one before so it IS STILL a linearly dependent vector.

Correct me if im wrong. thanks!
 
  • #13
lanedance
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pretty much, however i would say all you need to assume is there is a linearly dependent column in Y, so you may want to generalise a bit
 
  • #14
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Yea but the question assumes every column in Y is linearly dependent so im set. thanks again !
 
  • #15
lanedance
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Yea but the question assumes every column in Y is linearly dependent so im set. thanks again !
no it doesn't that statement doesn't even make sense

it says "the columns of Y are linearly dependent" this means at least one cloumn can be written as a linear combination of the others
 

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