Let A∈Rn× n be a non-symmetric matrix. Prove that |λ1| is real, provided that |λ1|>|λ2|≥|λ3|≥...≥|λn| where...

Let A∈R^{n× n} be a non-symmetric matrix.

Prove that |λ₁| is real, provided that |λ₁|>|λ₂|≥|λ₃|≥...≥|λ_n| where λ_i , i= 1,...,n are the eigenvalues of A, while others can be real or not real.

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Expert Solution

Consider the characteristic equation . The eigenvalues of the matrix are the roots of this characteristic equation.

Suppose that are the eigenvalues. Suppose that is not real; then for some reals with . Since is a polynomial equation with real coefficients, and since is a root of this polynomial, its conjugate is also a root of this polynomial. Thus, for some . Note that since ; however, .

Thus, whenever a complex eigenvalue occurs then its absolute value is attained by at least two distinct complex eigenvalues. In particular, if the eigenvalue is such that , then must be real.

To show that the other eigenvalues can be real or complex, we note that

and

are both non-symmetric; while the eigenvalues of are , those of are . Thus, the other eigenvalues can be real or complex.

Colby Messinger answered 1 year ago

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