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

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## Comments

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TopNewestDid you mean \(\displaystyle \sum_{i=i}^n \lambda_i = tr(A)\) ?

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Fixed it

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