Complex Eigenvalues and Eigenvectors of a 2x2 Matrix: Diagonalizable or Not?

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SUMMARY

The discussion centers on finding the eigenvalues and eigenvectors of the 2x2 matrix A = [0 -i | i 0]. The characteristic polynomial is derived as λ² - 1 = 0, leading to eigenvalues λ₁ = 1 and λ₂ = -1. The matrix A is confirmed to be diagonalizable, as it has two distinct eigenvalues. The diagonal matrix D can be constructed, and the corresponding matrix P can be determined to satisfy the equation A = PDP⁻¹.

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Homework Statement


Find the eigenvalues and eigenvectors of A. (Both eigenvalues and eigenvectors are now allowed
to be complex.) Is it diagonalizable? Explain why or why not. If it is diagonalizable, explicitly
find matrices P and D such that
A = PDP−1
where D is a diagonal 2 × 2 matrix.

A = [ 0 -i | i 0 ]

The Attempt at a Solution



I determined that A cannot be diagonalized because, by the characteristic polynomial equation we get [tex]\lambda[/tex]2 + 1 = 0

Therefore [tex]\lambda[/tex]1 = -i [tex]\lambda[/tex]2 = i

plugging [tex]\lambda[/tex]2 into my matrix A I get:

ix + iy = 0
-ix + iy = 0

but the only solution to this is x=y=0, I get the same result for [tex]\lambda[/tex]1

Is this correct? I have a feeling this trivial solution is wrong
I tried row reduction, but I still get the same result.
 
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so many minuses!

Hi Iconate! :smile:

(have a lambda: λ :wink:)
Iconate said:
A = [ 0 -i | i 0 ]

… by the characteristic polynomial equation we get [tex]\lambda[/tex]2 + 1 = 0

Noooo! :redface:
 


Ahhh I see
my determinant should be
λ2 - (-i)(i) = 0
λ2 + (i2) = 0
λ2 - 1 = 0

thus λ1 = 1 λ1 = -1

Thanks >.<
 

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