Show that V is an internal direct sum of the eigenspaces

In summary, a method of representing matrices in a vector form using a specific operation was discussed, but it did not work well for general ##n\times n## matrices. To solve this, one could consider diagonalizability and use the equation ##\dfrac{1}{2}\left(A\pm A^\tau\right)## to separate the matrices into symmetric and antisymmetric parts. This approach ultimately helped in finding a solution to the problem.
  • #1
Karl Karlsson
104
12
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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  • #2
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?
 
  • #3
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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  • #4
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
 

1. What is an internal direct sum?

An internal direct sum is a decomposition of a vector space V into subspaces such that every vector in V can be uniquely expressed as a sum of vectors from those subspaces.

2. What are eigenspaces?

Eigenspaces are subspaces of a vector space V that correspond to eigenvalues of a linear transformation on V. They consist of all the vectors that are mapped to scalar multiples of themselves by the linear transformation.

3. How do you show that V is an internal direct sum of eigenspaces?

To show that V is an internal direct sum of eigenspaces, we need to prove two things: 1) that the eigenspaces are subspaces of V, and 2) that the sum of the eigenspaces is equal to V. This can be done by showing that every vector in V can be expressed as a unique sum of vectors from the eigenspaces.

4. Why is it important to show that V is an internal direct sum of eigenspaces?

Showing that V is an internal direct sum of eigenspaces allows us to understand the structure of V and the linear transformation on V. It also helps us to easily find the eigenvalues and eigenvectors of the linear transformation, which are important in many applications in mathematics and science.

5. Can V be an internal direct sum of eigenspaces for any type of linear transformation?

No, V can only be an internal direct sum of eigenspaces if the linear transformation on V is diagonalizable. This means that there exists a basis for V consisting of eigenvectors of the linear transformation. If this condition is not met, V cannot be an internal direct sum of eigenspaces.

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