Proving Diagonalizability of an nxn Matrix

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Homework Help Overview

The discussion revolves around proving the diagonalizability of an nxn matrix A with m distinct eigenvalues L1,...,Lm. Participants are exploring the implications of the matrix expression (A-L1*I)(A-L2*I)...(A-Lm*I) equating to zero.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the conceptual understanding of the expression (A-LI) and its relation to eigenvalues. There is exploration of how to demonstrate that a matrix can be the zero matrix and the implications of eigenvectors forming a basis.

Discussion Status

The discussion is active, with participants offering insights into the nature of eigenvectors and their linear combinations. Some guidance has been provided regarding the conceptual approach to the problem, but multiple interpretations and lines of reasoning are still being explored.

Contextual Notes

There is an emphasis on understanding the properties of diagonalizable matrices and the definitions of zero matrices, along with the constraints of the problem setup. Participants are also considering the implications of linear combinations of eigenvectors in their reasoning.

buzzmath
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can anyone help with this problem?

Consider a diagonalizable nxn matrix A with m distinct eigenvalues L1,...,Lm show that (A-L1*I)(A-L2*I)****(A-Lm*I)=0
 
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Do you recognize the form (A-LI)? Have you ever solved something like (A-LI)x=0?
 
isn't that just used to find the eigenvalues? how would you get rid of x?
 
Don't think in terms of "getting rid of x." Think about the problem conceptually. What ways do you know to show that some matrix is the zero matrix?
 
what do you mean by showing that a matrix is the zero matrix? A matrix is a zero matrix if anything multiplied by it is zero? I know that A is similar do a diagonal matrix D with the eigienvalues down the diagonal. I'm not sure if that would help though?
 
Right, a matrix is a zero matrix if any vector multiplied by it is zero. Write the vector as a linear combination of eigenvectors.
 
So since A is diagonizable the eigenvectors form an eigenbasis for A. Thus any vector is a linear combination of these eigenvectors. Does that mean that is x is any linear combination of these eigenvectors that (A-L1*I)***(A-Lm*I)*x=0? and if so does it mean since any vector mulitplied by this matrix is zero that this matrix is zero then?
 
buzzmath said:
Does that mean that is x is any linear combination of these eigenvectors that (A-L1*I)***(A-Lm*I)*x=0?
Can you think of a reason? Try actually writing x as a linear combination rather than just saying that it is.
 

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