Quick diagonalize matrix (row reduce)

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The discussion revolves around diagonalizing a matrix and whether reducing it to reduced row-echelon form (RREF) affects the eigenvectors obtained. It clarifies that the original matrix and its RREF yield the same eigenvectors, specifically confirming that the eigenvector for the eigenvalue λ = 3 is (0,1,1). There is confusion regarding the eigenvector (0,1,-1/2) compared to the one listed on a referenced website, which states it should be (1/2,1,0). Participants suggest verifying the eigenvectors by substituting them into the equation Ax = λx to confirm their validity. The conversation emphasizes the importance of correctly identifying eigenvectors and understanding the implications of matrix transformations.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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