Proving Singular Matrix A Has Nonzero Matrix B: Linear Algebra Problem

Click For Summary

Discussion Overview

The discussion revolves around proving that for every square singular matrix A, there exists a nonzero matrix B such that the product AB equals the zero matrix. The scope includes theoretical aspects of linear algebra and properties of singular matrices.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states they derived AB to equal the identity matrix but struggles to show it equals the zero matrix.
  • Another participant suggests viewing A as a linear transformation, noting that its singularity implies it maps a subspace to the zero subspace, and proposes finding the kernel of A to define matrix B.
  • A third participant questions the first post's claim of finding an inverse, pointing out that a singular matrix does not have an inverse.
  • A fourth participant mentions the determinant properties, stating that since A is singular, its determinant is zero, and discusses the implications for the determinant of the product AB.

Areas of Agreement / Disagreement

Participants express differing views on the approach to proving the existence of matrix B, with no consensus reached on the correctness of the initial claims or the methods proposed.

Contextual Notes

There are unresolved assumptions regarding the properties of singular matrices and the implications of determinant calculations. The discussion does not clarify the steps needed to transition from the identity matrix to the zero matrix.

Braka
Messages
5
Reaction score
0
The problem is prove that for every square singular matrix A there is a nonzero matrix B, such that AB equals the zero matrix.

I got AB to equal the idenity matrix, but have no clue how to get it to the zero matrix.
 
Physics news on Phys.org
If A is viewed as the matrix of a linear transformation, then it being singular is the same as saying that it maps a subspace in the domain to the 0 subspace of the range. Find the kernel of A and let B be a matrix of vectors in that space.
 
you said:
I got AB to equal the idenity matrix, but have no clue how to get it to the zero matrix.

how did you do that? If A is singular then it doesn't have an inverse, but you found one namely B.
 
If A is singular, det(A)= 0. Also det(AB)= det(A)det(B). You, apparently, have proved that det(AB)= 0(det(B)= 0= det(I)= 1. A miracle indeed!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
5K
  • · Replies 34 ·
2
Replies
34
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K