Let $\mathbf{A}$ be an $n \times n$ matrix. The determinant $|\mathbf{A}-\lambda \mathbf{I}|$ is a polynomial in the (complex) variable $\lambda$ of degree $n$ and is called the characteristic polynomial of $\mathbf{A}$. The equation
$$
|\mathbf{A}-\lambda \mathbf{I}|=0
$$
is called the characteristic equation of $\mathbf{A}$. By the fundamental theorem of algebra, the equation has $n$ roots, and these roots are called the eigenvalues of $\mathbf{A}$.

The eigenvalues may not all be distinct. The number of times an eigenvalue occurs as a root of the characteristic equation is called the algebraic multiplicity of the eigenvalue.
We factor the characteristic polynomial as
$$
|\mathbf{A}-\lambda \mathbf{I}|=\left(\lambda_{1}-\lambda\right) \cdots\left(\lambda_{n}-\lambda\right)
$$
Setting $\lambda=0$ in (4) we see that $|\mathbf{A}|$ is just the product of the eigenvalues of $\mathbf{A}$. Similarly by equating the coefficient of $\lambda^{n-1}$ on either side of (4) we see that the trace of $\mathbf{A}$ equals the sum of the eigenvalues.
A principal submatrix of a square matrix is a submatrix formed by a set of rows and the corresponding set of columns. A principal minor of $\mathbf{A}$ is the determinant of a principal submatrix.
A square matrix $\mathbf{A}$ is called symmetric if $\mathbf{A}=\mathbf{A}^{\prime} .$ An $n \times n$ matrix $\mathbf{A}$ is said to be positive definite if it is symmetric and if for any nonzero vector $\mathbf{x}, \mathbf{x}^{\prime} \mathbf{A} \mathbf{x}>0$
The identity matrix is clearly positive definite and so is a diagonal matrix with only positive entries along the diagonal.

If $\mathbf{A}$ is positive definite, then it is nonsingular.

PROOF. If $\mathbf{A} \mathbf{x}=\mathbf{0}$, then $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x}=0$, and since $\mathbf{A}$ is positive definite, $\mathbf{x}=\mathbf{0}$ Therefore, A must be nonsingular.

The next result is obvious from the definition.

For $0 \leq \alpha \leq 1$, define

For $0 \leq \alpha \leq 1$, define
$$
f(\alpha)=|\alpha \mathbf{A}+(1-\alpha) \mathbf{I}|
$$

$\alpha \mathbf{A}+(1-\alpha) \mathbf{I}$ is positive definite, and therefore by 8.1, $f(\alpha) \neq 0,0 \leq$ $\alpha \leq 1$. Clearly, $f(0)=1$, and since $f$ is continuous, $f(1)=|\mathbf{A}|>0$

If $\mathbf{A}$ is positive definite, then any principal submatrix of $\mathbf{A}$ is positive definite.

ProoF. Since $\mathbf{A}$ is positive definite, $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x}>0$ for all $\mathbf{x} \neq \mathbf{0}$. Apply this condition to the set of vectors that have zeros in coordinates $j_{1}, \ldots, j_{s} .$ For such a vector $\mathbf{x}, \mathbf{x}^{\prime} \mathbf{A} \mathbf{x}$ reduces to an expression of the type $\mathbf{y}^{\prime} \mathbf{B y}$ where $\mathbf{B}$ is the principal submatrix of A formed by deleting rows and columns $j_{1}, \ldots, j_{s}$ from $\mathbf{A}$. It follows that $\mathbf{B}$, and similarly any principal submatrix of $\mathbf{A}$, is positive definite.

A symmetric $n \times n$ matrix $\mathbf{A}$ is said to be positive semidefinite if $\mathbf{x}^{\prime} \mathbf{A} \mathbf{x} \geq 0$ for all $\mathbf{x} \in R^{n}$

If $\mathbf{A}$ is a symmetric matrix, then the eigenvalues of $\mathbf{A}$ are all real.

PROOF. Suppose $\mu$ is an eigenvalue of $\mathbf{A}$ and let $\mu=\alpha+i \beta$, where $\alpha, \beta$ are real and $i=\sqrt{-1}$. Since $|\mathbf{A}-\mu \mathbf{I}|=0$, we have
$$
|(\mathbf{A}-\alpha \mathbf{I})-i \beta \mathbf{I}|=0
$$
Taking the complex conjugate of the above determinant and multiplying the two, we get
$$
|(\mathbf{A}-\alpha \mathbf{I})-i \beta \mathbf{I} |(\mathbf{A}-\alpha \mathbf{I})+i \beta \mathbf{I}|=0
$$
Thus
$$
\left|(\mathbf{A}-\alpha \mathbf{I})^{2}+\beta^{2} \mathbf{I}\right|=0
$$
Since $\mathbf{A}$ is symmetric, it is true that $\mathbf{A}^{2}$ is positive semidefinite (it follows from the definition that $\mathbf{B B}^{\prime}$ is positive semidefinite for any matrix $\left.\mathbf{B}\right)$. Thus if $\beta \neq 0$, then $\left|(\mathbf{A}-\alpha \mathbf{I})^{2}+\beta^{2} \mathbf{I}\right|$ is positive definite, and then by $\mathbf{8 . 1},(5)$ cannot hold. Thus $\beta=0$, and $\mu$ must be real.