Prove if a Square Matrix A is Diangonalizable

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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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