Can a 3x3 matrix have 4 eigenvalues?

nicknaq
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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
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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