[Linear Algebra] prove that A is singular (A is a square matrix)?

Click For Summary

Homework Help Overview

The discussion revolves around proving that a matrix A, defined as the product of two matrices B (8x3) and C (3x8), is singular. The participants explore the implications of matrix dimensions and properties related to determinants and ranks.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the dimensions of matrices B and C and the resulting matrix A. There is an attempt to understand how to demonstrate that A cannot be non-singular without numerical examples. Questions arise about the rank of A and the formation of its columns.

Discussion Status

Some participants have provided hints and guidance regarding the rank of A and its implications for singularity. There is an ongoing exploration of the concepts of linear transformations and subspaces, but no consensus has been reached on the proof itself.

Contextual Notes

Participants note the challenge of proving the singularity of A without resorting to numerical examples, and there is a focus on understanding the underlying linear algebra concepts involved.

nooonz
Messages
3
Reaction score
0

Homework Statement


Let A = BC
where B is 8*3 and C is 3*8
Prove that A is singular.


Homework Equations


A is singular when Det(A)=0



The Attempt at a Solution


When B and C are multiplied the result is A which is an 8*8 matrix. However, top prove it can't be non singular i chose a smaller matrix (2*3) and (3*2) and the result showed that if i chose the same numbers the answer is 0. But how do i prove it without numbers?
 
Physics news on Phys.org
I don't really understand the question, I think.
Yes, if det(A) = 0, then A is singular.
Yes, det(A) = det(B) det(C).

But for arbitrary 8x3 and 3x8 matrices B and C, respectively, it is not true (if det(B) and det(C) are non-zero, so is det(A)).

Have you quoted the complete question?
 
nooonz said:

Homework Statement


Let A = BC
where B is 8*3 and C is 3*8
Prove that A is singular.


Homework Equations


A is singular when Det(A)=0



The Attempt at a Solution


When B and C are multiplied the result is A which is an 8*8 matrix. However, top prove it can't be non singular i chose a smaller matrix (2*3) and (3*2) and the result showed that if i chose the same numbers the answer is 0. But how do i prove it without numbers?

You want to show that the rank of A is less than 8; in other words, you want to show that the number of linearly independent columns of A is less than 8. How is column j of A formed?

RGV
 
CompuChip said:
I don't really understand the question, I think.
Yes, if det(A) = 0, then A is singular.
Yes, det(A) = det(B) det(C).

But for arbitrary 8x3 and 3x8 matrices B and C, respectively, it is not true (if det(B) and det(C) are non-zero, so is det(A)).

Have you quoted the complete question?

yes, i quoted the entire question
 
Ray Vickson said:
You want to show that the rank of A is less than 8; in other words, you want to show that the number of linearly independent columns of A is less than 8. How is column j of A formed?

RGV

What do you mean how is column j of A formed? How do i show that the rank is less than 8 without using numbers?
 
If I told you the answers to these questions, I would be doing the whole question for you. I gave you hints, and will stop there.

RGV
 
nooonz said:
What do you mean how is column j of A formed? How do i show that the rank is less than 8 without using numbers?

OK, I will give a bit more of a hint. You have C = A*B, where C is 8x8, A is 8x3 and B is 3x8. Look at it as a linear transformation on E8 = E^8 (8-dimensional real space). It is composed of two transformations B:E8 --> E3, a linear transformation from an 8-dimensional space to a 3-dimensional one, followed by a transformation A:E3 --> E8, from a 3-dimensional space to an 8-dimensional one. Altogether, C:E8 --> E8 transforms E8 into a subspace C(E8) of E8, and you are asked to show that this transformation is not the identity. What is the dimension of the subspace C(E8)? Remember that we get to C(E8) from a subspace B(E8) of E3.

RGV
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
4K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
9
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K