Quick diagonalize matrix (row reduce)

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The discussion centers on the diagonalization of matrices and the process of reducing them to reduced row-echelon form (RREF). Participants clarify that the matrices discussed are not eigenvectors, and they confirm that transforming a matrix to RREF does not alter the eigenvectors derived from it. The eigenvalues λ = 3 and λ = 6 are identified, with corresponding eigenvectors provided. A discrepancy in eigenvector values from different sources is also highlighted, emphasizing the importance of verification through substitution into the eigenvalue equation.

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  • Understanding of eigenvalues and eigenvectors
  • Knowledge of matrix transformations and RREF
  • Familiarity with linear algebra concepts
  • Ability to use online matrix calculators
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  • Learn how to compute eigenvalues and eigenvectors using characteristic polynomials
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Students and educators in linear algebra, mathematicians, and anyone involved in matrix theory or eigenvalue problems will benefit from this discussion.

Nope
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quick please..diagonalize matrix (row reduce)

Homework Statement



http://www.math4all.in/public_html/linear algebra/chapter10.1.html
10.1.5 Examples: (ii)
this part:

-----------------------------------------------------------------------------------------
april95.gif

are both eigenvector for the eigenvalue λ = 3. Similarly, for λ = 6, since
-----------------------------------------------------------------------------------------

Does it matter if I reduce it to rref?

Homework Equations


The Attempt at a Solution

 
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Nope said:

Homework Statement



http://www.math4all.in/public_html/linear algebra/chapter10.1.html
10.1.5 Examples: (ii)
this part:

-----------------------------------------------------------------------------------------
april95.gif

are both eigenvector for the eigenvalue λ = 3. Similarly, for λ = 6, since
-----------------------------------------------------------------------------------------

Does it matter if I reduce it to rref?
Your question doesn't make much sense. The matrices you show are NOT eigenvectors. In the link you gave, it shows two vectors that are the eigenvectors for the eigenvalue 3.

What do you mean by "reduce it to rref"?
 


oh i I figure it out...
I was just asking
Do I get the same eigenvector (which is 0,1,1) if I
transform
-3,0,0
0,1,-1
0,0,0
to rref

In reduced row-echelon form (RREF), the matrix above is
1 0 0
0 1 -1
0 0 0

The only thing that changed is row 1.
You'll get exactly the same eigenvector from this matrix as from the unreduced one.
 
Last edited by a moderator:


And using this calculator
http://wims.unice.fr/wims/en_tool~linear~matrix.html
I found that
Value Multiplicity Vector
6 1 (0,1,1)
3 2 (1,0,1), (0,1,-1/2)

(1,0,1) and (0,1,1) is right, but (0,1,-1/2) is different to the website , which is (1/2,1,0)
I m confused..
 


Nope said:
And using this calculator
http://wims.unice.fr/wims/en_tool~linear~matrix.html
I found that
Value Multiplicity Vector
6 1 (0,1,1)
3 2 (1,0,1), (0,1,-1/2)

(1,0,1) and (0,1,1) is right, but (0,1,-1/2) is different to the website , which is (1/2,1,0)
I m confused..

It's easy to check. Substitute each of these two vectors (i.e., <0, 1, -1/2> and <1/2, 1, 0> in the equation Ax = 3x. If they both make a true statement, they're both eigenvectors.
 

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