Proof of linear independence and dependence

[andorrak]

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?

 

[lanedance]

for the first part start with
$$Y = \begin{pmatrix} y_1 & .. & y_i .. & y_n \end{pmatrix}$$

$$X = \begin{pmatrix} x_1^T \\ .. \\ x_j^T \\.. \\ x_m^T \end{pmatrix}$$

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 )?

 

[andorrak]

Um could you elaborate? I have no idea what you are saying. sorry

 

[lanedance]

i re-wrote above for clarity & expanded a little

 

[andorrak]

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

 

[lanedance]

try expaninding this (for arbitrary vectors)

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

 

[andorrak]

what is the dot product? if you could answer it systematically it would be very helpful

 

[lanedance]

you don't know what a dot product is between 2 vectors?
http://en.wikipedia.org/wiki/Dot_product

 

[andorrak]

O that's 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?

 

[lanedance]

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

$$y_j = ay_r + a y_s$$

$$x_i^T \bullet y_j = x_i^T \bullet (a y_r + b y_s)$$

 

[lanedance]

so consider the jth column of XY
$$XY_j = \begin{pmatrix} x_1^T \bullet y_j\\ .. \\ x_i^T \bullet y_j\\.. \end{pmatrix} <br />$$

 

[andorrak]

OH i think i get it. because you assume all the 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 I am wrong. thanks!

 

[lanedance]

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

 

[andorrak]

Yea but the question assumes every column in Y is linearly dependent so I am set. thanks again !

 

[lanedance]

Yea but the question assumes every column in Y is linearly dependent so I am 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