Solved: Both formulas follow from the product

Chapter 20, Problem 20.1.102

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The eigenvalues of A are the solutions \(\lambda\) of the characteristic equation

\(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\left|\begin{array}{cccc}a_{11}-\lambda & a_{12} & \cdots & a_{1 n} \\a_{21} & a_{22}-\lambda & \cdots & a_{2 n} \\\cdot & & \cdots & \cdot \\a_{n 1} & a_{n 2} & \cdots & a_{n n}-\lambda\end{array}\right|=0\).

Developing the characteristic determinant, we obtain the characteristic polynomial of A, which is of degree \(\lambda\) in \(\lambda\). Hence A has at least one and at most n numerically different eigenvalues. If A is real, so are the coefficients of the characteristic polynomial. By familiar algebra it follows that then the roots (the eigenvalues of A) are real or complex conjugates in pairs.

We shall usually denote the eigenvalues of A by

\(\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}\)

with the understanding that some (or all) of them may be equal.

The sum of these n eigenvalues equals the sum of the entries on the main diagonal of A, called the trace of A: thus

Text Transcription:

lambda

det(A-lambda I)=|a_11-lambda a_12 . . . a_1n

a_21 a_22-lambda . . . a_2n

. … .

a_n1 a_n2 …a_nn-lambda|=0

lambda_1, lambda_2 . . . lambda_n