Problem 9

Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be 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 of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.

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

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We give two 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 property of determinants and traces.

Proof.

(1) Recall that eigenvalues are roots of the characteristic polynomial $p(\lambda)=\det(A-\lambda I_n)$.It follows that we have\begin{align*}&\det(A-\lambda I_n) \\&=\begin{vmatrix}a_{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\end{vmatrix} =\prod_{i=1}^n (\lambda_i-\lambda). \tag{*}\end{align*}

Letting $\lambda=0$, we see that $\det(A)=\prod_{i=1}^n \lambda_i$ and this completes the proof of part (a).

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

$$(-1)^{n-1}(a_{11}+a_{22}+\cdots a_{n n})=(-1)^{n-1}\tr(A).$$The coefficient of $\lambda^{n-1}$ of the determinant on the right side of (*) 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 exists an $n \times n$ invertible matrix $P$ such that\This is an upper triangular matrix and diagonal entries are eigenvalues.(If this is not familiar 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 have\begin{align*}\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),\end{align*}where we used the multiplicative property of the determinant.

(2) We take the trace of both sides of (**) and get\begin{align*}\sum_{i=1}^n \lambda_i =\tr(P^{-1}AP) =\tr(A).\end{align*}(Here for the last equality we used the 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 proof of (1) in the first method is simple, but that of (2) requires a bit observation, especially when we find the coefficient of the left-hand side.

The proof of (2) in the second method is simpler although you need to know about the Jordan normal/canonical form.

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These two formulas relate the determinant and the trace, and the eigenvalue of a matrix in a very simple way.