Quick Matrix Algrebra Question
 

When a Matrix is singular, how does one show that one row is a linear combination of the other rows?

Re: Quick Matrix Algrebra Question
 

solve A*x = 0

Re: Quick Matrix Algrebra Question
 

Thanks for the help attempt, but I'm too awful with mathematical language to understand how to do it from that answer.

Re: Quick Matrix Algrebra Question
 

[ QUOTE ]
solve A*x = 0

[/ QUOTE ]

This is true for all large...errr..real...ah.. them symbol things there that begin with a 0.

pokervintage

Re: Quick Matrix Algrebra Question
 

Oops, I misread row for column the first time. The first thing you wanted to do is look at the transpose of your matrix. In the transpose matrix, the ith column vector is the ith row vector of the original matrix.

Below is how to express a column vector as a linear combo of the others.

Let A be an m x n matrix with column vectors A1,...,An

Let x be a n x 1 vector with entries x1,...,xn.

If A*x = 0 ( m x 1 zero vector), then A1*x1 + A2*x2 + .... + An*xn = 0. Suppose you want to represent A1 as a linear combination of the others. Then A1 = -(A2*x2 + A3*x3 + .... + An*xn) / x1.

Re: Quick Matrix Algrebra Question
 

It's not clear from your question whether you want to prove that any singular matrix has linearly dependent rows (or colummns), or rather prove a particular matrix has linearly dependent rows.

If the latter, you may use the fact that the following statements are equivalent for a square matrix:

(i) The matrix is singular.
(ii) The matrix has linearly dependent rows (and columns).
(iii) The determinant of the matrix is zero.

Thus, it will suffice to compute the determinant of the matrix, and if it's zero, that immediately implies (ii).

matrix determinant

Re: Quick Matrix Algrebra Question
 

I am asking for the latter in general, but I suppose my question is not how to prove it, but rather, how to express a row as a linear combo of the others.

Re: Quick Matrix Algrebra Question
 

Think of A as the matrix and X as a single column vector.

ie , X={x1,x2,x3,...xn }T where T is the transpose .

And A can be written as {V1 V2 V3 ... Vn} and the Vi's are column vectors .

In other words , you wish to solve for x1*V1 + x2*V2 +...xn*Vn = ( 0,0,0)

The rhs is the 0 vector which may be written as (0,0,0) .

ie , V1=(0,0,1) V2= (1,0,1) V3= (0,1,1) and the xi's are scalars .

So we wish to solve for x1*(0,0,1) + x2*(1,0,1) + x3*(0,1,1) =(0,0,0)

Or x2=0 , x3=0 and x1+x2+x3=0 which implies x1=0 .

Since all three of the xi's are zero , this implies that the vectors V1,V2 and V3 are linearly independent .

Also it's important to know that a set of vectors is linear dependent if and only if some vi can be written as a linear combination of the others .If you can prove independence , then you can prove that a single vector cannot be written as a linear combination of the others .

Re: Quick Matrix Algrebra Question
 

I see, you want to explicitly compute the dependent relationship.

Let R1, ... ,Rn be the n linearly dependent rows of a singular (real) matrix. (Thus each row, Ri, is an n-dimensional vector.) By definition, there exists a set of n real numbers, (X1, ... ,Xn), not all zero, which satisfy the set of n linear equations:

X1*R1 + X2*R2 + ... + Xn*Rn = zero vector (size nx1)

The preceding is n linear equations for the n unknowns X1 through Xn. Just use basic algebra to solve these equations. (This is equavalent to the matrix equation a previous poster posted.)