Product and Sum of Eigenvalues

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

Solution

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

A=[a11a1nan1ann]A = \begin{bmatrix} {a}_{11} & \cdots &{a}_{1n} \\ \vdots &\ddots &\vdots \\ {a}_{n1} &\cdots & {a}_{nn}\\ \end{bmatrix}

AInλ=[a11λa1nan1annλ]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(AλIn)=det(A)+...+tr(A)(λ)n1+(λ)ndet(A - \lambda{I}_{n} ) = det(A) + ... + tr(A){(-\lambda)}^{n-1} + {(-\lambda)}^{n}.

Let r1,r2,...,rn{r}_{1}, {r}_{2}, ...,{r}_{n} be the roots of an n-order polynomial.

P(λ)=(r1λ)(r2λ)...(rnλ)P(\lambda) = ({r}_{1} - \lambda)({r}_{2} - \lambda)...({r}_{n} - \lambda) P(λ)=i=inri+...+i=inri(λ)n1+(λ)nP(\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)P(x) to det(AλIn)det(A - \lambda{I}_{n}). Therefore it is clear that i=inλi=det(A)\prod _{ i=i }^{ n }{ { \lambda }_{ i } = det(A)}

and

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

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

Note by Steven Zheng
5 years, 1 month ago

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

A Former Brilliant Member - 5 years, 1 month ago

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Fixed it

Steven Zheng - 5 years, 1 month ago

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what to do for a3*3 matrix

Harish Kumar Kumar - 1 month, 3 weeks ago

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Thanks

alka talwar - 3 weeks, 4 days ago

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