Why Does a Singular Matrix Imply Infinite or Zero Solutions?

In summary, a singular matrix implies that the matrix does not have an inverse and is degenerate. This means that the linear transformation represented by the matrix maps the vector space into a subspace of the output space. This results in either infinite or zero solutions, depending on whether the desired output is in the subspace or not. This can be seen through the dimension theorem and the concept of nullity and rank for linear transformations.
  • #1
member 428835
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!
 
Physics news on Phys.org
  • #2
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.)
 
  • #3
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.
 
  • #4
thanks this makes tons of sense!
 
  • #5


Singular matrix theory deals with matrices that have a determinant of 0. This means that the matrix is not invertible, and as a result, there are either infinite solutions or zero solutions to the equation Ax=B. The reason for this is that the determinant of a matrix is a measure of its "scale factor" or how much it "stretches" or "squishes" space. A determinant of 0 indicates that the matrix has no scale factor, and therefore, it cannot uniquely determine a solution to the equation.

To understand why this leads to either infinite or zero solutions, we can look at the geometric interpretation of matrices. When we multiply a vector by a matrix, it transforms the vector into a new vector. This transformation can be thought of as stretching, rotating, or reflecting the vector. When a matrix has a determinant of 0, it means that it is not able to stretch space in any direction. This can result in either all vectors being collapsed onto a single point (zero solutions) or all vectors being stretched onto a line or plane (infinite solutions).

To prove this mathematically, we can use the fact that a matrix is invertible if and only if its determinant is non-zero. Therefore, if a matrix has a determinant of 0, it is not invertible, and there exist either infinite or zero solutions to the equation Ax=B.

In summary, the singular matrix theory is important in understanding the solutions to linear equations and the properties of matrices. A determinant of 0 indicates a lack of scale factor, leading to either infinite or zero solutions to the equation Ax=B.
 

FAQ: Why Does a Singular Matrix Imply Infinite or Zero Solutions?

What is a singular matrix?

A singular matrix is a square matrix that does not have an inverse. This means that it cannot be multiplied by another matrix to get the identity matrix (a matrix with 1s along the diagonal and 0s everywhere else).

What is the significance of singular matrices?

Singular matrices are important in linear algebra and other areas of mathematics because they represent systems of equations that do not have a unique solution. This can occur when there are more equations than variables, or when the equations are not independent.

How can you determine if a matrix is singular?

A matrix is singular if its determinant is equal to 0. The determinant is a mathematical operation that can be performed on a square matrix to determine whether or not it has an inverse. If the determinant is 0, the matrix is singular; if it is non-zero, the matrix is non-singular.

What are some applications of singular matrix theory?

Singular matrix theory has applications in various fields such as physics, engineering, and economics. It is used to solve systems of linear equations, analyze matrices in quantum mechanics, and model economic systems.

Can a singular matrix be useful?

Yes, singular matrices can be useful in certain situations. For example, they can be used to model non-invertible systems or to find critical points in optimization problems. However, in most cases, singular matrices are considered problematic and steps are taken to avoid them in mathematical calculations.

Similar threads

Replies
15
Views
1K
Replies
4
Views
6K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
7
Views
2K
Back
Top