Can a 3x3 matrix have 4 eigenvalues?

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A 3x3 matrix cannot have four eigenvalues because the characteristic polynomial derived from the matrix is a cubic equation, which can have at most three roots. The process to find eigenvalues involves calculating the determinant of the matrix A minus lambda times the identity matrix, leading to a degree 3 equation. This confirms that a 3x3 matrix can only yield three eigenvalues. Therefore, it is impossible for a 3x3 matrix to have four eigenvalues. The discussion emphasizes the fundamental relationship between the degree of the characteristic polynomial and the maximum number of eigenvalues.
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Homework Statement



Prove or disprove the title of this thread.

Homework Equations


AX=(lamda)X


The Attempt at a Solution


I don't know where to start
 
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nicknaq said:

Homework Statement



Prove or disprove the title of this thread.

Homework Equations


AX=(lamda)X


The Attempt at a Solution


I don't know where to start

Start with the process you use to find the eigenvalues of a 3 x 3 matrix, which involves a determinant to get the characteristic equation for the matrix. What degree equation would you expect to get?
 
Mark44 said:
Start with the process you use to find the eigenvalues of a 3 x 3 matrix, which involves a determinant to get the characteristic equation for the matrix. What degree equation would you expect to get?

an equation of degree 3
 
So it's not possible for a 3 x 3 matrix to have four eigenvalues, right?
 
Mark44 said:
So it's not possible for a 3 x 3 matrix to have four eigenvalues, right?

right. Is there any proof that I can say for why an equation of degree 3 cannot have 4 solutions?

I guess it's obvious though.
 
no its not possible. I had completed this topic only today in my class and here is one interesting question.

how to solve for eigen values. I think we need to take determinant. of A -\lambda I

so we will get \lambda cube in the equation which obviously will give three values of \lambda
 

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