Some Special Matrices and Their Properties

Symmetric matrix.

A square matrix is called a symmetric matrix when the pairs of elements in similar positions across the diagonal are equal.

$\begin{indisplay} \left[ \begin{array}{ccc} 1 & x & y \\ x & 2 & z \\ y & z & 3 \\ \end{array} \right] \end{indisplay}$

The transpose of a matrix is the matrix obtained by writing the rows as columns or by writing the columns as rows.

The symbol for the transpose of matrix is .

$\begin{indisplay} A= \left[ \begin{array}{rrr} 3 & -1& 4 \\ 0 & 2 & 3 \\ 1 & 1 & 2 \\ \end{array} \right], \end{indisplay}$

$\begin{indisplay} A^T= \left[ \begin{array}{rrr} 3 & 0& 1 \\ -1 & 2 & 1 \\ 4 & -3& 2 \\ \end{array} \right] \end{indisplay}$

If all the elements above/below the diagonal are zero, a matrix is called lower/upper-triangular (L/U);

$\begin{indisplay} L= \left[ \begin{array}{ccc} x & 0 & 0 \\ x & x & 0 \\ x & x & x \\ \end{array} \right],\end{indisplay}$

$\begin{indisplay} U= \left[ \begin{array}{ccc} x & x& x \\ 0 & x& x \\ 0 & 0 & x \\ \end{array} \right] \end{indisplay}$

We will deal with square matrices.

Sparse matrix. In some important applied problems, only a few of the elements are nonzero.

Such a matrix is termed a sparse matrix and procedures that take advantage of this sparseness are of value.

Division of matrices is not defined, but we will discuss the inverse of a matrix.

The determinant of a square matrix is a number.

The method of calculating determinants is a lot of work if the matrix is of large size.
Methods that triangularize a matrix, as described in next section, are much better ways to get the determinant.

If a matrix, $B$

, is triangular (either upper or lower), its determinant is just the product of the diagonal elements:

$\displaystyle det(B)=\Pi B_{ii},~~~~~~i = 1,\ldots, n$

$\begin{indisplay} det\left\vert \begin{array}{rrr} 4 & 0 & 0 \\ 6 &-2 & 0 \\ 1 &-3 & 5 \\ \end{array} \right\vert=-40 \end{indisplay}$

If we have a square matrix and the coefficients of the determinant are nonzero, there is a unique solution.

Determinants can be used to obtain the characteristic polynomial and the eigenvalues of a matrix, which are the roots of that polynomial.

If a matrix is triangular, its eigenvalues are equal to the diagonal elements.

This follows from the fact that

its determinant is just the product of the diagonal elements and
its characteristic polynomial is the product of the terms $(a_{ii}-\lambda)$ with going from to , the number of rows of the matrix.

$\begin{indisplay} A=\left[ \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{array} \right], \end{indisplay}$

$\begin{indisplay} det(A-\lambda I)=det\left\vert \begin{array}{ccc} 1-\lambda &... ...da \\ \end{array} \right\vert=(1-\lambda)(4-\lambda)(6-\lambda) \end{indisplay}$

whose roots are clearly 1, 4, and 6.

It does not matter if the matrix is upper- or lower-triangular.