Is Every Skew Symmetric Matrix Invertible When Added to the Identity Matrix?

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Homework Help Overview

The discussion revolves around the properties of skew symmetric matrices, specifically focusing on the invertibility of the matrix formed by adding a skew symmetric matrix \( A \) to the identity matrix \( I \). The original poster presents a problem involving proving that \( I + A \) is invertible, following an initial proof related to the quadratic form \( u^T A u = 0 \) for any vector \( u \).

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of the quadratic form and explore the conditions under which \( (I + A)u = 0 \). There is an attempt to clarify the relationship between the kernel of \( I + A \) and its invertibility, with some participants expressing confusion about terminology such as "kernel" and "injective operator."

Discussion Status

There is an ongoing exploration of the properties of the matrix \( I + A \) and its kernel. Some participants have offered hints and guidance, while others express uncertainty about the concepts being discussed. The conversation reflects a mix of understanding and confusion, with no explicit consensus reached yet.

Contextual Notes

Participants note a lack of familiarity with certain linear algebra concepts, such as the kernel and injective operators, which may be affecting their ability to engage fully with the problem. There is also an acknowledgment of previous discussions on similar topics, indicating a shared learning environment.

weasel3000
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A is a skew symmetric nxn matrix with entries in R. AT=-A.
1). Prove uTAu=0 for every u in R. (I made it)
2). Prove In + A is a invertible matrix.
I have no idea how to prove the 2nd.
Thanks
 
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Hmm, I guess one of your classmates also asked this question recently, cause I can remember it.

Anyway, I'll give a hint. For which vectors u, does (I+A)u=0. Use (1) for this.
 
micromass said:
Hmm, I guess one of your classmates also asked this question recently, cause I can remember it.

Anyway, I'll give a hint. For which vectors u, does (I+A)u=0. Use (1) for this.

? not get it. 0*u=0; (I+A)*0=0; non-zero vector *non-zero vector=0 also possible.
 
You're right, if u=0, then (I+A)u=0. Are there other possibilities for u?
 
thank you for your hint. However I am so dumb ,and unable to get it. R u going to get det(I+A) not 0 or find (I+A)^-1
 
No, we are trying to establish that the kernel of I+A is trivial. This implies that I+A is an injective operator. And by some theorem of linear algebra, this implies that I+A is invertible...
 
micromass said:
No, we are trying to establish that the kernel of I+A is trivial. This implies that I+A is an injective operator. And by some theorem of linear algebra, this implies that I+A is invertible...

oh my god. i am going to find a place to cry. i just know the null space. never learn what kernel and injective operators are. Thank you again. I am going to learn that damn THINGS tonight.
 
Don't worry :smile: The kernel is exactly thesame as the null space. And injective (or one-to-one or monomorphism) just means that the null space is trivial.
 

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