Can You Find the Minimal and Characteristic Polynomials of This Operator?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on determining the minimal and characteristic polynomials of the operator $\bigwedge^2 T$ on a 4-dimensional real vector space $\mathbb{V}$, given that the characteristic polynomial of $T$ is $X^4 - 1$. The characteristic polynomial of $\bigwedge^2 T$ is established to have a degree of 6, derived from the dimension formula for wedge products. No solutions were provided during the discussion, and the original poster plans to update with a full solution within 24 hours.

PREREQUISITES
  • Understanding of linear operators on vector spaces
  • Familiarity with characteristic and minimal polynomials
  • Knowledge of wedge products in linear algebra
  • Comprehension of combinatorial dimensions, specifically ${n \choose k}$
NEXT STEPS
  • Research the properties of characteristic and minimal polynomials in linear algebra
  • Study the application of wedge products in vector spaces
  • Explore the implications of polynomial degrees in linear transformations
  • Learn about the structure of operators on exterior powers of vector spaces
USEFUL FOR

Mathematicians, students studying linear algebra, and anyone interested in the properties of linear operators and their polynomials.

Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Here's this week's problem.

-----

Problem: Let $T: \mathbb{V}\rightarrow \mathbb{V}$ be an operator on a 4-dimensional real vector space $\mathbb{V}$. Assume that the characteristic polynomial of $T$ is $X^4-1$. Determine the minimal and characteristic polynomials for the operator $\bigwedge^2 T:\bigwedge^2\mathbb{V}\rightarrow \bigwedge^2\mathbb{V}$.

-----

Remark: Note that if $\mathbb{V}$ is an $n$-dimensional vector space, then $\displaystyle\dim\bigwedge\!\!\,^k\mathbb{V} = {n\choose k}$. It then follows that in our problem, the characteristic polynomial of $\bigwedge^2 T$ has degree 6.

 
Physics news on Phys.org
No one answered this week's question.

As I have been busy as of late, I don't have a full solution right now to post -- I'll have one in the next 24 hours and will update this post, so stay tuned!
 

Similar threads

Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K