problem 9

Let $A$ it is in an $n\times n$ matrix and also let $\lambda_1, \dots, \lambda_n$ it is in its eigenvalues.Show that

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

(2) $$\tr(A)=\sum_i=1^n \lambda_i$$

Here $\det(A)$ is the determinant that the procession $A$ and $\tr(A)$ is the trace of the matrix $A$.

You are watching: Why is determinant the product of eigenvalues

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

*
Add to solve laterSponsored Links


We give two different proofs.

setup 1.

Use the an interpretation of eigenvalues (the properties polynomial).Compare coefficients.

arrangement 2.

Make $A$ top triangular matrix or in the Jordan normal/canonical form.Use the home of determinants and also traces.

Proof.

(1) Recall that eigenvalues room roots that the characteristic polynomial $p(\lambda)=\det(A-\lambda I_n)$.It complies with that us have\beginalign*&\det(A-\lambda I_n) \\&=\beginvmatrixa_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-\lambda\endvmatrix =\prod_i=1^n (\lambda_i-\lambda). \tag*\endalign*

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

(2) to compare the coefficients of $\lambda^n-1$ of the both sides of (*).The coefficient of $\lambda^n-1$ the the determinant top top the left side of (*) is

$$(-1)^n-1(a_11+a_22+\cdots a_n n)=(-1)^n-1\tr(A).$$The coefficient that $\lambda^n-1$ the the determinant ~ above the best side that (*) is$$(-1)^n-1\sum_i=1^n \lambda_i.$$Thus we have $\tr(A)=\sum_i=1^n \lambda_i$.

Proof.

Observe that there exist an $n \times n$ invertible matrix $P$ together that\This is an upper triangular matrix and diagonal entries space eigenvalues.(If this is not acquainted to you, then study a “triangularizable matrix” or “Jordan normal/canonical form”.)

(1) because the determinant that an top triangular procession is the product of diagonal line entries, us have\beginalign*\prod_i=1^n \lambda_i & =\det(P^-1 A P)=\det(P^-1) \det(A) \det(P) \\&= \det(P)^-1\det(A) \det(P)=\det(A),\endalign*where we used the multiplicative building of the determinant.

(2) we take the trace of both sides of (**) and get\beginalign*\sum_i=1^n \lambda_i =\tr(P^-1AP) =\tr(A).\endalign*(Here because that the critical equality we used the residential or commercial property of the trace that $\tr(AB)=\tr(BA)$ for any $n\times n$ matrices $A$ and $B$.)Thus we obtained the result $\tr(A)=\sum_i=1^n \lambda_i$.

Comment.

The evidence of (1) in the first method is simple, but that of (2) needs a little bit observation, especially when we discover the coefficient of the left-hand side.

The proof of (2) in the second technique is easier although you should know around the Jordan normal/canonical form.

See more: Tales From The Borderlands Episode 2 Choices, Atlas Mugged

These 2 formulas relate the determinant and also the trace, and also the eigenvalue that a matrix in a very basic way.