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Product and Sum of Eigenvalues

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Given a square matrix \(A\), prove that the sum of its eigenvalues is equal to the trace of \(A\), and the product of its eigenvalues is equal to the determinant of \(A\).

Solution

This proof requires the investigation of the characteristic polynomial of \(A\), which is found by taking the determinant of \((A - \lambda{I}_{n})\).

\[A = \begin{bmatrix} {a}_{11} & \cdots &{a}_{1n} \\ \vdots &\ddots &\vdots \\ {a}_{n1} &\cdots & {a}_{nn}\\ \end{bmatrix}\]

\[A - {I}_{n}\lambda = \begin{bmatrix} {a}_{11} - \lambda & \cdots &{a}_{1n} \\ \vdots &\ddots &\vdots \\ {a}_{n1} &\cdots & {a}_{nn} - \lambda\\ \end{bmatrix}\]

Observe that \(det(A - \lambda{I}_{n} ) = det(A) + ... + tr(A){(-\lambda)}^{n-1} + {(-\lambda)}^{n}\).

Let \({r}_{1}, {r}_{2}, ...,{r}_{n}\) be the roots of an n-order polynomial.

\[P(\lambda) = ({r}_{1} - \lambda)({r}_{2} - \lambda)...({r}_{n} - \lambda)\] \[P(\lambda) = \prod _{ i=i }^{ n }{ { r }_{ i } } +...+\sum _{ i=i }^{ n }{ { r }_{ i }{(-\lambda)}^{n-1} + {(-\lambda)}^{n} } \]

Since the eigenvalues are the roots of a matrix polynomial, we can match \(P(x)\) to \(det(A - \lambda{I}_{n})\). Therefore it is clear that \[\prod _{ i=i }^{ n }{ { \lambda }_{ i } = det(A)}\]

and

\[\sum _{ i=i }^{ n }{ { \lambda }_{ i } = tr(A)}.\]

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
2 years, 9 months ago

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Did you mean \(\displaystyle \sum_{i=i}^n \lambda_i = tr(A)\) ? Ethan Robinett · 2 years, 9 months ago

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@Ethan Robinett Fixed it Steven Zheng · 2 years, 9 months ago

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