Very Simple Eigenvalue Calculation Need explanation please

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The discussion revolves around calculating the eigenvalues of the matrix [[5, 2], [-3, 0]]. The correct method involves finding the determinant of (A - λI) and setting it to zero, leading to the equation (5 - λ)(-λ) + 6 = 0. Upon solving, the eigenvalues are determined to be 2 and 3, not 5 and 0 as initially thought. The confusion arose from a misunderstanding of the algebra involved in the determinant calculation. Ultimately, the correct eigenvalues are confirmed to be 2 and 3.
bmb2009
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Homework Statement



Calculate the eigenvalues of the matrix
5 2
-3 0



Homework Equations





The Attempt at a Solution



Ok we were taught that eigenvalues were calculated by taking the determinant( A - λI) = 0. So just subtract a "λ" value from the diagnol entries of the given matrix... so why aren't the eigenvalues 5 and 0? Please help me understand why the e values are 2 and 3?
 
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bmb2009 said:

Homework Statement



Calculate the eigenvalues of the matrix
5 2
-3 0



Homework Equations





The Attempt at a Solution



Ok we were taught that eigenvalues were calculated by taking the determinant( A - λI) = 0. So just subtract a "λ" value from the diagnol entries of the given matrix... so why aren't the eigenvalues 5 and 0? Please help me understand why the e values are 2 and 3?

What is the determinant of the matrix A - λI? What is the equation you get when you set this determinant to zero?

Besides: see what you get when you try to solve the equation Ax = 5x; that will tell you why 5 is not an eigenvalue.
 
bmb2009 said:
Ok we were taught that eigenvalues were calculated by taking the determinant( A - λI) = 0. So just subtract a "λ" value from the diagnol entries of the given matrix... so why aren't the eigenvalues 5 and 0? Please help me understand why the e values are 2 and 3?
I have to ask: Why do you think the eigenvalues would be 5 and 0?
 
I just realized that I butchered my algebra and now see that det(A- xI) = 0

(5-x)(-x) - (-6) = x^2 -5x + 6 = 0
(x-2) (x-3) = 0
x=2,3

Thanks for reminding me though appreciate it!
 
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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