Problem 9

Let $A$ be an $n imes n$ matrix and let $lambda_1, dots, lambda_n$ be its eigenworths.Show that

(1) $$det(A)=prod_i=1^n lambda_i$$

(2) $$ r(A)=sum_i=1^n lambda_i$$

Here $det(A)$ is the determinant of the matrix $A$ and also $ r(A)$ is the map of the matrix $A$.

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Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and also (2) the trace of $A$ is the amount of the eigenworths.

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We provide 2 different proofs.

Plan 1.

Use the definition of eigenvalues (the characteristic polynomial).Compare coefficients.

Plan 2.

Make $A$ upper triangular matrix or in the Jordan normal/canonical form.Use the building of factors and traces.


(1) Recall that eigenvalues are roots of the characteristic polynomial $p(lambda)=det(A-lambda I_n)$.It adheres to that we haveeginalign*&det(A-lambda I_n) \&=eginvmatrixa_1 1- lambda & a_1 2 & cdots & a_1,n \a_2 1 & a_2 2 -lambda & cdots & a_2,n \vdots & vdots & ddots & vdots \a_n 1 & a_m 2 & cdots & a_n n-lambdaendvmatrix =prod_i=1^n (lambda_i-lambda). ag*endalign*

Letting $lambda=0$, we view that $det(A)=prod_i=1^n lambda_i$ and this completes the proof of component (a).

(2) Compare the coefficients of $lambda^n-1$ of the both sides of (*).The coeffective of $lambda^n-1$ of the determinant on the left side of (*) is

$$(-1)^n-1(a_11+a_22+cdots a_n n)=(-1)^n-1 r(A).$$The coeffective of $lambda^n-1$ of the determinant on the right side of (*) is$$(-1)^n-1sum_i=1^n lambda_i.$$Hence we have $ r(A)=sum_i=1^n lambda_i$.


Observe that there exists an $n imes n$ invertible matrix $P$ such thatThis is an upper triangular matrix and diagonal entries are eigenvalues.(If this is not acquainted to you, then study a “triangularizable matrix” or “Jordan normal/canonical form”.)

(1) Since the determinant of an upper triangular matrix is the product of diagonal entries, we haveeginalign*prod_i=1^n lambda_i & =det(P^-1 A P)=det(P^-1) det(A) det(P) \&= det(P)^-1det(A) det(P)=det(A),endalign*wright here we provided the multiplicative building of the determinant.

(2) We take the trace of both sides of (**) and geteginalign*sum_i=1^n lambda_i = r(P^-1AP) = r(A).endalign*(Here for the last etop quality we provided the residential or commercial property of the map that $ r(AB)= r(BA)$ for any kind of $n imes n$ matrices $A$ and $B$.)Hence we obtained the result $ r(A)=sum_i=1^n lambda_i$.


The proof of (1) in the initially approach is simple, yet that of (2) requires a bit observation, particularly once we uncover the coeffective of the left-hand side.

The proof of (2) in the second approach is easier although you need to recognize about the Jordan normal/canonical form.

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These 2 formulas relate the determinant and also the trace, and also the eigenworth of a matrix in an extremely simple way.