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−λIn).
A=⎣⎢⎡a11⋮an1⋯⋱⋯a1n⋮ann⎦⎥⎤
A−Inλ=⎣⎢⎡a11−λ⋮an1⋯⋱⋯a1n⋮ann−λ⎦⎥⎤
Observe that det(A−λIn)=det(A)+...+tr(A)(−λ)n−1+(−λ)n.
Let r1,r2,...,rn be the roots of an n-order polynomial.
P(λ)=(r1−λ)(r2−λ)...(rn−λ)
P(λ)=i=i∏nri+...+i=i∑nri(−λ)n−1+(−λ)n
Since the eigenvalues are the roots of a matrix polynomial, we can match P(x) to det(A−λIn). Therefore it is clear that
i=i∏nλi=det(A)
and
i=i∑nλi=tr(A).
Check out my other notes at Proof, Disproof, and Derivation
#Algebra
#Matrices
#Eigenvalues
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Did you mean i=i∑nλi=tr(A) ?
Log in to reply
Fixed it
what to do for a3*3 matrix
Thanks