rjawbreakerca

2022-07-14

A (2 x 2) matrix $\left[\begin{array}{cc}a& c+id\\ c-id& b\end{array}\right]$ 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?

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?

pampatsha

Beginner2022-07-15Added 15 answers

The characteristic polynomial is $(\lambda -a)(\lambda -b)-({c}^{2}+{d}^{2})$. Note that the first term already has roots a,b, so the discriminant is already positive; ${c}^{2}+{d}^{2}\ge 0$ 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.

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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