 # 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 $\left[\begin{array}{cc}a& c+id\\ c-id& b\end{array}\right]$ 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 $|A-\lambda I|=det\left(A-\lambda I\right)=0$ 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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The characteristic polynomial is $\left(\lambda -a\right)\left(\lambda -b\right)-\left({c}^{2}+{d}^{2}\right)$. 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.

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