Show that V is an internal direct sum of the eigenspaces

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Homework Help Overview

The discussion revolves around demonstrating that a vector space V is an internal direct sum of its eigenspaces, particularly in the context of matrices. The original poster references a previous problem involving 2x2 matrices and expresses difficulty in generalizing their approach to n x n matrices.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the diagonalizability of the linear operator L and the conditions that eigenvalues must satisfy. There are suggestions to explore the relationship between eigenvalues and eigenvectors, as well as methods to separate matrices into symmetric and antisymmetric components.

Discussion Status

The discussion is ongoing, with various ideas being explored. Some participants have offered guidance on potential approaches, such as examining the structure of eigenvalues and eigenvectors, while others have reiterated methods for separating matrix components. There is no explicit consensus yet.

Contextual Notes

Participants are working under the constraints of linear algebra principles, specifically regarding eigenvalues and eigenspaces, and are considering the implications of matrix properties on their problem-solving approaches.

Karl Karlsson
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Homework Statement
If V = ##M_{n,n}(\mathbb R)## is the vectorspace of nxn matrices for some n and ##L(A) = A^T##, show that V is an internal direct sum
of the eigenspaces and show that L is diagonalizable.
Relevant Equations
xxx
I was in an earlier problem tasked to do the same but when V = ##M_{2,2}(\mathbb R)##. Then i represented each matrix in V as a vector ##(a_{11}, a_{12}, a_{21}, a_{22})## and the operation ##L(A)## could be represented as ##L(A) = (a_{11}, a_{21}, a_{12}, a_{22})##. This method doesn't really work well when we talk about general ##n\times n## matrices. How would one go about to solve this?

Thanks in advance!
 
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A bunch of ideas:

Have you been able to show that ##L## is diagonalisable? Maybe can you find some conditions the eigenvalues must satisfy? For example, you have that ##\lambda## is an eigenvalue with eigenvector ##A\in M_n(\mathbb{R})## if ##\lambda A =A^T##. Taking determinants, what conditions do you obtain? Does this help to find some eigenvectors? Can you find a basis of eigenvectors?
 
The standard equation to think about in such cases is to consider ##\dfrac{1}{2}\left(A\pm A^\tau\right)##, i.e. to separate the matrices in a symmetric and an antisymmetric part.
 
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fresh_42 said:
The standard equation to think about in such cases is to consider ##\dfrac{1}{2}\left(A\pm A^\tau\right)##, i.e. to separate the matrices in a symmetric and an antisymmetric part.
Thanks, that helped me solve the problem
 

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