Proof that Non-Singularity of I-AB Implies Non-Singularity of I-BA

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In summary, the conversation is about proving that if the matrix I-AB is non-singular, then so is I-BA. The person asking the question lost 3 points on their test and is seeking clarification on what they did wrong. The expert summarizes the proof provided by the person and suggests possible reasons for the points lost.
  • #1
eyehategod
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Prove that if the matrix I-AB is non-singular, then so i I-BA.
This was one of my test questions and got 3 points off. Can anyone tell me what I did wrong.

my proof:
assume I-AB is nonsingular
then (I-AB)[tex]^{-1}[/tex] exists

Let C=(I-AB)[tex]^{-1}[/tex]
Consider (I+BCA)(I-BA)=I-BA+BCA-BCABA
=I-BA+BC(I-AB)A
=I-BA+B(I-AB)[tex]^{-1}[/tex] (I-AB)A
=I-BA+BIA
=I-BA+BA
=I
 
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  • #2
3 points over how many?

Maybe you lost some points because

i) You did not prove that if C is the left inverse of A, then it is also a right inverse. So you might have lost some points because you only showed that (I+BCA) is the left inverse of (I-BA).

ii) The definition of nonsingular in your book is that A is nonsingular if its determinant is nonzero. And so, you might have lost some points because you did not explain that since I-BA had an inverse, then its determinant was nonzero and hence I-BA is nonsingular.

But other than that, your proof looks very good! go see your instructor.
 
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  • #3


Therefore, (I-BA)^{-1} exists and I-BA is nonsingular.


One possible mistake in your proof is that you assumed that (I-AB)^{-1} = C, without explicitly proving it. In order to show that (I-AB)^{-1} exists, you need to show that it is invertible, i.e. that there exists a matrix D such that (I-AB)D = DI = I. This can be done by using the fact that (I-AB)(I+BCA) = I, and then solving for D.

Another issue is that in the step where you have I-BA+BC(I-AB)A, you are assuming that (I-AB)^{-1} exists, which is what you are trying to prove. Instead, you should use the fact that (I-AB)^{-1} exists and then show that (I-BA)^{-1} also exists.

A possible correct proof could be:

Assume (I-AB) is non-singular. This means that (I-AB)^{-1} exists.

Let C = (I-AB)^{-1}. Then (I-AB)C = CI = I.

Consider (I-BA)(I+ABC) = I-BA+ABC-BA(ABC)
=I-BA+AB(I-AB)C
=I-BA+ABCI
=I-BA+AB
=I

Therefore, (I-BA)^{-1} exists and (I-BA) is non-singular.
 

1. What is the significance of proving that non-singularity of I-AB implies non-singularity of I-BA?

The proof of this statement is important because it establishes a relationship between the non-singularity of two matrices, I-AB and I-BA. This relationship can be used in various applications, such as solving linear equations, determining invertibility of matrices, and understanding the properties of matrix multiplication.

2. Can you explain the concept of singularity in matrices?

A matrix is considered singular if its determinant is equal to zero. This means that the matrix is not invertible and does not have a unique solution for a system of linear equations. In other words, a singular matrix does not have an inverse.

3. How does the proof of this statement relate to matrix multiplication?

The proof shows that when two matrices, I-AB and I-BA, are non-singular, their product will also be non-singular. This is because the non-singularity of I-AB implies that the columns of AB are linearly independent, which in turn implies that the rows of BA are also linearly independent, making I-BA non-singular.

4. Is this statement true for all types of matrices?

Yes, this statement holds true for all square matrices, regardless of their size or type. It is a fundamental property of matrix multiplication and is not limited to any specific type of matrices.

5. What are some practical applications of this proof?

This proof has various applications in fields such as engineering, physics, and computer science. It can be used to solve systems of linear equations, determine the invertibility of matrices, and analyze the stability of dynamic systems. It also helps in understanding the properties of matrix multiplication and can be used in algorithms for efficient matrix operations.

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