Prove if a Square Matrix A is Diangonalizable

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Homework Help Overview

The discussion revolves around the diagonalizability of a non-diagonal n-by-n matrix A of rank m, particularly focusing on the implications of having a set of linearly independent vectors that are eigenvectors corresponding to the eigenvalue 10. Participants are tasked with proving the diagonalizability and providing an example for specific dimensions.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definition of diagonalizability and the conditions under which a matrix has a basis of eigenvectors. They discuss the known eigenvalue of 10 and question how to find other eigenvalues and corresponding eigenvectors. There is also exploration of the implications of the matrix's rank on its eigenvalues.

Discussion Status

The discussion is active, with participants questioning how to construct a basis of eigenvectors and considering the implications of the matrix's rank on its eigenvalues. Some guidance has been offered regarding the relationship between the rank and the eigenvalues, but no consensus has been reached on the methods to find the eigenvectors.

Contextual Notes

Participants are working under the constraints of the problem statement, particularly the conditions of rank and the nature of the eigenvalues. There is a focus on understanding the implications of these conditions without providing direct solutions.

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Homework Statement



Let A be a non-diagonal n-by-n matrix of rank m. Suppose that a set of vectors, v1 ... vm, are linearly independent vectors in Rn such that for i = 1,...,m,
Avi = 10vi (vi is a vector in the given set of linearly independent vectors).

(a) Prove that A is diagonalizable.
(b) Give an example of a matrix satisfying these conditions for n = 2, m = 2.

Homework Equations


D = S-1AS

The Attempt at a Solution


I am very confused on how to do part a. I understand that one of the eigenvalue for vi is 10 but I do not know what to do with that...
 
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A matrix is diagonalizable if and only if it has a basis of eigenvectors. Can you find such a basis?
 
Okay so we know that an eigenbasis consists of eigenvectors. We know one of the eigenvalues, 10. But, how do you find the other eigenvalues and corresponding eigenvectors?
 
raveenamc said:
Okay so we know that an eigenbasis consists of eigenvectors. We know one of the eigenvalues, 10. But, how do you find the other eigenvalues and corresponding eigenvectors?

You also know another eigenvalue (hint: the rank of the matrix is m).

What are some eigenvectors of the eigenvalue 10?? What dimension is the space spanned by those eigenvectors?? Can you find other eigenvectors that extend it to a basis?
 
Since the rank is m, the matrix is not full rank. Therefore, is the other eigenvalue 0?
 
raveenamc said:
Since the rank is m, the matrix is not full rank. Therefore, is the other eigenvalue 0?

Yes (except if n=m of course). Can you construct a basis of eigenvectors now?
 
So, for eigenvalue=0, you just find the kernel of matrix A and the span will be one of the eigenvectors? What about for eigenvalue=10?
 
raveenamc said:
So, for eigenvalue=0, you just find the kernel of matrix A and the span will be one of the eigenvectors? What about for eigenvalue=10?

Ok, that made no sense. How can "the span" be one of the eigenvectors?
 
The span of the kernel is an eigenvector for the corresponding eigenvalue, is it not?
 
  • #10
What do you mean with "the span of the kernel"??
 
  • #11
Okay, completely forgetting about the span, how else would you find the eigenvectors?
 

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