Register to reply

Singular matrix theory

by joshmccraney
Tags: matrix, singular, theory
Share this thread:
Jan15-13, 02:29 AM
P: 346
hey guys

given [itex]Ax=B[/itex] where A is a square matrix and x and B are vectors, can anyone tell me why a singular matrix (that is, the determinant = 0) implies one of two situations: infinite solutions or zero solutions? a proof would be nice. i read through pauls notes but there was no proof.

thanks all!
Phys.Org News Partner Science news on
New type of solar concentrator desn't block the view
Researchers demonstrate ultra low-field nuclear magnetic resonance using Earth's magnetic field
Asian inventions dominate energy storage systems
Simon Bridge
Jan15-13, 03:12 AM
Sci Advisor
HW Helper
Simon Bridge's Avatar
P: 12,739
If ##\mathbf{A}\vec{x}=\vec{y}## then ##\mathbf{A}^{-1}\vec{y}=\vec{x}## provided the inverse exists.

If the matrix ##\mathbf{A}## is singular, it does not have an inverse.
Another name for it is "degenerate".

What does that tell you about the solutions?
(Think about it in terms of solving simultaneous equations.)
Jan15-13, 07:09 AM
Sci Advisor
PF Gold
P: 39,488
An n by n square matrix represents a linear transformation, A, from Rn to Rn. If it is "non-singular", then it maps all of Rn to all of Rn. That is, it is a "one to one" mapping- given any y in Rn there exist a unique x in Rn such that Ax= y.

But we can show that, for any linear transformation, A, from one vector space, U, to another, V, the "image" of A, that is, the set of all vectors y, of the form y= Ax for some x, is a subspace of V and that the "null space" of A, the set of all vectors, x, in U such that Ax= 0, is a subspace of U. Further, we have the "dimension theorem". If "m" is dimension of the image of A (called the "rank" of A) and "n" is the dimension of the nullspace of A (called the "nullity" of A) then m+ n is equal to the dimension of V. In particuar, if U and V have the same dimension, n, and the rank of A is m with m< n, then the nullity of A= m-n> 0.

It is further true that if A(u)= v and u' is in the nullspace of A then A(u+ u')= A(u)+ A(u')= v+ 0= v.

The result of all of that is this: If A is a singular linear transformation from vector space U to vector space V, then it maps U into some subspace of V. If y is NOT in that subspace then there is NO x such that Ax= y. If y is in that subspace then there exist x such that Ax= y but also, for any v in the nullity of A (which has non-zero dimension and so contains an infinite number of vectors) A(x+ v)= y also so there exist an infinite number of such vectors.

Jan16-13, 08:38 PM
P: 346
Singular matrix theory

thanks this makes tons of sense!

Register to reply

Related Discussions
Singular matrix Calculus & Beyond Homework 2
Can we still diagnolize a matrix if its eigenvectors matrix is singular? Linear & Abstract Algebra 1
MatLab Warning: Matrix is singular, close to singular or badly scaled Engineering, Comp Sci, & Technology Homework 1
Singular matrix General Math 5
Singular values of a matrix times a diagonal matrix Linear & Abstract Algebra 1