Matrices problems: any help appreciated

dreadnaut · 13-04-2009: 13th April 2009 17:32

Hi all,

In my revision I've come across a few gaps in my notes and things that aren't so clear.

First off: Principal minors

I have the matrix (not sure about doing this in latex, apologies)
A=
a11 a12
a21 a22

Find the first and second principal minors of matrix A, wiki seems to have set me on the right tracks i.e det(A) has to be positive for the first P.M....is the second simply the element a11?

Next up is the business of linear independance/singular matrices....are these both revolving round whether det(A) = 0 or not?

det(A) = 0: linearly dependant, singular (for nxn matrix)

My course booklet doesnt really distinguish between linearly independant and non singular....are they the same/similar thing?

thats all for now, though more will probably leave me stuck sooner or later

all and any help/comments much appreciated

thanks

DoMakeSayThink · 13-04-2009: 13th April 2009 17:44

If we take an n by n matrix A, and assign to each of it's columns a vector c_i

, the matrix is linearly independent if $\lambda_1c_1 + \lambda_2c_2 + \dots + \lambda_n c_n = 0$ implies that $\lambda_1 = \lambda_2 = \dots = \lambda_n=0$ . If the converse is true, i.e. A is linearly dependent, we can write at least one of the columns of a as a linear combination of the others. You may be aware that the determinant of a matrix is unaffected by row and column operations (a proof of which is beyond the scope of this post), and consequently if some column $c_i = \lambda_{j_1} c_{j_1} + \dots + \lambda_{j_k} c_{j_k}$ one can make an entire column of values equal to zero by subtracting the $\lambda_j c_j$ 's. This would then give the matrix a determinant of zero.

In short, linear dependence implies a zero determinant, and hence a singular matrix.

dreadnaut · 13-04-2009: 13th April 2009 17:53

Originally Posted by DoMakeSayThink

If we take an n by n matrix A, and assign to each of it's columns a vector c_i

, the matrix is linearly independent if $\lambda_1c_1 + \lambda_2c_2 + \dots + \lambda_n c_n = 0$ implies that $\lambda_1 = \lambda_2 = \dots = \lambda_n=0$ . If the converse is true, i.e. A is linearly dependent, we can write at least one of the columns of a as a linear combination of the others. You may be aware that the determinant of a matrix is unaffected by row and column operations (a proof of which is beyond the scope of this post), and consequently if some column $c_i = \lambda_{j_1} c_{j_1} + \dots + \lambda_{j_k} c_{j_k}$ one can make an entire column of values equal to zero by subtracting the $\lambda_j c_j$ 's. This would then give the matrix a determinant of zero.

In short, linear dependence implies a zero determinant, and hence a singular matrix.

cool beans. thanks very much. I think thats all i need to quote in my exam....its only economics so they dont really want much from us in the LA department relative to other subjects

cheers

anyone got anything on principal minors?

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Matrices problems: any help appreciated

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