A (2 xx 2) matrix [[a,c+id],[c-id,b]] where a, b, c, d are real constants will have two different eigenvalues unless it is a multiple of the identity matrix. i used this |A-lambda I|=det(A-lambda I)=0 to prove that it will have 2 eigenvalues. How do i prove that it will only have 1 when it is a multiple of the identity matrix?

rjawbreakerca 2022-07-14 Answered
A (2 x 2) matrix [ a c + i d c i d b ] where a, b, c, d are real constants will have two different eigenvalues unless it is a multiple of the identity matrix.
i used this | A λ I | = d e t ( A λ I ) = 0 to prove that it will have 2 eigenvalues. How do i prove that it will only have 1 when it is a multiple of the identity matrix?
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pampatsha
Answered 2022-07-15 Author has 15 answers
The characteristic polynomial is ( λ a ) ( λ b ) ( c 2 + d 2 ). Note that the first term already has roots a,b, so the discriminant is already positive; c 2 + d 2 0 and increasing c 2 + d 2 will only increase the discriminant further.
The only possibility for the matrix to have only one eigenvalue is therefore c=d=0 and a=b, i.e. the matrix is a multiple of the identity matrix.

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