Relations between trace, determinant and eigenvalues

Determinant of A equals to the product of its eigenvalues

Given a

$n\times n$ matrix

$A$, define

$$\begin{array}{}\text{(1)}& P(\lambda )=det(\lambda I-A),\end{array}$$
then we have

$$\begin{array}{}\text{(2)}& P(\lambda )={\lambda }^n+c_{n-1}{\lambda }^{n-1}\cdots +c_1\lambda +c_0.\end{array}$$
Since

$P$ is a nth-degree polynomial that has

$n$ roots

${\lambda }_1,\cdots ,{\lambda }_n$, we can rewrite

$P$ as

$$\begin{array}{}\text{(3)}& P(\lambda )=(\lambda -{\lambda }_1)\cdots (\lambda -{\lambda }_n).\end{array}$$

Finally, let us calculate

$P(0)$.

• Using (1)
  
  $$\begin{array}{}\text{(4)}& P(0)=det(-A)=(-1{)}^{n}det(A).\end{array}$$
• Using (3)
  
  $$\begin{array}{}\text{(5)}& P(0)=(-{\lambda }_1)\cdots (-{\lambda }_n)=(-1{)}^n{\lambda }_1\cdots {\lambda }_n.\end{array}$$

Therefore, using (4) and (5),

$$det(A)={\lambda }_1\cdots {\lambda }_n.$$

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Trace of A equals to the sum of its eigenvalues

Let

$A$ be a

$n\times n$ matrix written as

$$\begin{array}{}\text{(6)}& A=\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \cdots & a_{nn}\end{array}\right],\end{array}$$
so that

$$\begin{array}{}\text{(7)}& \lambda I-A=\left[\begin{array}{cccc}\lambda -a_{11}& -a_{12}& \cdots & -a_{1n}\\ -a_{21}& \lambda -a_{22}& \cdots & -a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{n1}& -a_{n2}& \cdots & \lambda -a_{nn}\end{array}\right].\end{array}$$

To calaulate

$det(\lambda I-A)$, we use cofactor expansion along the first column:

$$\begin{array}{}\text{(8)}& det(\lambda I-A)=(\lambda -a_{11})C_{11}+(-a_{21})C_{21}+\cdots ,\end{array}$$

where

$$C_{11}={\left[\begin{array}{ccc}\lambda -a_{22}& \cdots & -a_{2n}\\ ⋮& \ddots & ⋮\\ -a_{n2}& \cdots & \lambda -a_{nn}\end{array}\right]}_{(n-1)\times (n-1)},$$

and

$$C_{21}={\left[\begin{array}{cccc}-a_{12}& -a_{13}& \cdots & -a_{1n}\\ -a_{32}& \lambda -a_{33}& \cdots & -a_{3n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{n2}& \cdots & \cdots & \lambda -a_{nn}\end{array}\right]}_{(n-1)\times (n-1)}.$$

The important observation is that

$C_{21}$ is a polynomial of degree at most

$n-2$. Also, all the

$C_{k1}$,

$k>1$, are polynomials of degree at most

$n-2$. As a result, we have

$$\begin{array}{}\text{(9)}& det(\lambda I-A)=(\lambda -a_{11})C_{11}+{\hat{Q}}_{n-2}(\lambda ),\end{array}$$

where

${\hat{Q}}_{n-2}(\lambda )$ is a polynomial of degree at most

$n-2$.

Such an procedure can be performed iteratively, and eventually we have

$$\begin{array}{}\text{(10)}& det(\lambda I-A)=(\lambda -a_{11})(\lambda -a_{22})\cdots (\lambda -a_{nn})+{\tilde{Q}}_{n-2}(\lambda ),\end{array}$$

where

${\tilde{Q}}_{n-2}(\lambda )$ is a polynomial of degree at most

$n-2$.

Equation~(10) can then be rewritten as

$$\begin{array}{}\text{(11)}& det(\lambda I-A)={\lambda }^n-(a_{11}+a_{22}+\cdots +a_{nn}){\lambda }^{n-1}+Q_{n-2}(\lambda ),\end{array}$$

where

$Q_{n-2}(\lambda )$ is a polynomial of degree at most

$n-2$.

On the other hand, from (3) we have

$$\begin{array}{}\text{(12)}& P(\lambda )={\lambda }^n-({\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n){\lambda }^{n-1}+\cdots .\end{array}$$

Therefore, using (11) and (12),

$$a_{11}+a_{22}+\cdots +a_{nn}={\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n.$$