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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
3 years, 8 months ago

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

- 3 years, 8 months ago

Fixed it

- 3 years, 8 months ago