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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
5 years, 10 months ago

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

- 5 years, 10 months ago

Fixed it

- 5 years, 10 months ago

what to do for a3*3 matrix

- 10 months, 1 week ago

Thanks

- 9 months, 2 weeks ago