Matrices

$\spadesuit$ ${\bf m \times n matrix }$ $\spadesuit$

As an example of vector space, we considered a set of geometric vectors in 1, a set of space vectors, a set of continuous functions, and a set of piecewise continuous functions.

In chapter 2, we treat a rectangular array of real numbers. A horizontal $n$-tuple is called row and a vertical $m$-tuples is called column. For example, $3 \times 4$ matrix is given in the following form.

$$\left( \begin{array}{llll} a_{11} & a_{12} & a_{13} & a_{14} ... ..._{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{array}\right)$$

Here $ij$th entry $a_{ij}$ is called a component, $i$ is a row number and $j$is a column number. Especially, $a_{ii}$ are called diagonal components. A matrix with $m$rows and $n$ columns is called an $m \times n$ matrix; the pair of numbers $(m,n)$ is called its size or shape. Also, a mtrix with$a_{ij}$ components is denoted by $A$ or $(a_{ij})$.

Matrices are often used in a natural science or a social science. Matrices are need to be considered as a mathematical objects. To do so, we define an addition and a scalar multiplication.

Definition 2..1  

For matrices with the same size $A = (a_{ij})$ and $B = (b_{ij})$, if corresponding components are the same; for every $i,j$, $a_{ij} = b_{ij}$. Then $A = B$.

$\spadesuit$Matrix Additon $\spadesuit$

A matrix addition is defined by additing corresponding entries.

Definition 2..2  

Let $A = (a_{ij})$ and $B = (b_{ij})$ be two matrices with the same size. Then the sum of $A$ and $B$, written $A + B$ is defined $A + B = (a_{ij} + b_{ij})$.

Example 2..1  

For matrices $A = \left ( \begin{array}{rrr} 7&3&4\\ -1&2&0\\ 2&-4&4\\ 3&0&-7 \end{array}\... ...t ( \begin{array}{ccc} 4&2&-1\\ 3&-1&7\\ 2&0&8\\ 6&3&-4 \end{array}\right )$, find $A + B$.

Answer $A + B = \left ( \begin{array}{ccc} 7+4&3+2&4-1\\ -1+3&2-1&0+7\\ 2+2&-4+0&4+8\... ...\begin{array}{rrr} 11&5&3\\ 2&1&7\\ 4&-4&12\\ 9&3&-11 \end{array}\right ) .$ $\blacksquare$

Note that for matrices with $m \times n$ satisfy the properties 1 thru 5 of a vector space. We recommend everyone to check it.

$\spadesuit$Scalar Multiplication $\spadesuit$

A scalar multiplication is defined by multiplying each entry by the scalar.

Definition 2..3  

A scalar multiplication $A = (a_{ij})$ by a scalar $\alpha$, written $A\alpha = \alpha A$ is defined

$$\alpha A = (\alpha a_{ij})$$

Example 2..2  

Let $A = \left ( \begin{array}{ccc} 2&5&3\\ -1&4&2 \end{array}\right )$ . Find $3A$.

Answer $3A = \left ( \begin{array}{rrr} 3 \cdot 2 & 3 \cdot 5 & 3 \cdot 3\\ 3 \cdot (-... ...{array}\right ) = \left(\begin{array}{rrr} 6&15&9\\ -3&12&6 \end{array}\right)$. $\blacksquare$

A scalar multiplication of matrices satisfies the properies 6 thru 9 of the vector space. From this fact, we can think of a set of $m \times n$ matrices as a vector space. When the entries are all real numbers, it is called real vector space. But we never say an $m \times n$ matrix a vector. As an exception, $m\times 1$ matrix or $1\times n$ matrix are called m-component row vector or n-component column vector.

Matrix Multiplication

Besides an addition and a scalar multiplication, it is possible to define a multiplicaiton of matrices. In the chapter 3, we study that a multiplication of two matrices represents a composition of transformations.

Definition 2..4  

Let $A = (a_{ik}), B = (b_{kj})$ be $m \times n$ matrix and $n\times r$ matrix. Then the matrix multiplication $AB$ is $m\times r$ matrix $C$ such that

$$C = (c_{ij}) = (\sum_{k=1}^{n} a_{ik}b_{kj}).$$

In other words, an ${ij}$ components of $C$ is given by taking inner product of $i$th row of the matrix $A$ and a $j$th column of the matrix $B$. From this, when you take a matrix multiplication of $A$ and $B$, the size of row of $A$ and the size of column of $B$ must be same.

Example 2..3  

Find the product of $A = \left(\begin{array}{ccc} 1&2&-3\\ 2&7&1 \end{array}\right)$ and $B = \left(\begin{array}{cc} 5&2\\ 6&3\\ -2&0 \end{array}\right)$.

Answer The product of two matrices $A$ and $B$ is $2\times 2$ matrix with the components,

$$c_{11} = (1)(5) + (2)(6) + (-3)(-2) = 23,$$

$$c_{12} = (1)(2) + (2)(3) + (-3)(0) = 8,$$

$$c_{21} = (2)(5) + (7)(6) + (1)(-2) = 50,$$

$$c_{22} = (2)(2) + (7)(3) + (1)(0) = 25.$$

Thus, we have

$$AB = \left(\begin{array}{rr} 23&8\\ 50&25 \end{array}\right) .$$

Furthermore, if we calculate $BA$, then

$$BA = \left(\begin{array}{rrr} 9&24&-13\\ 12&33&-15\\ -2&-4&6 \end{array}\right). \ensuremath{ \blacksquare}$$

As you can see, the operation of product of matrices is not commutative. In other words, $AB \neq BA$.

Example 2..4  

For $A \neq 0, B \neq 0$, find matrices $A$ and $B$ so that $AB = 0$.

Answer $A = \left(\begin{array}{cc} 1 & 2\\ 2 & 4 \end{array} \right)$ $B = \left(\begin{array}{cc} -2 & -6\\ 1 & 3 \end{array}\right)$とおくと

$$AB = \left(\begin{array}{cc} 1 & 2\\ 2 & 4 \end{array}\right)... ...n{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right) \ensuremath{ \blacksquare}$$

$\spadesuit$Block Matrices $\spadesuit$

Using a system of horizontal and vertical lines, we can partition a matrix $A$ into smaller matrices called blocks of $A$, The matrix $A$ is then called a block matirx.

Consider matrices $A$ and $B$.

$$A = \left(\begin{array}{ccc\vert c} 1 & 1 & 5 & -3\\ 4 & -3 & ... ... 3 & -3 & 8 & 2\\ -1 & 4 & 2 & 0\ \hline 2 & -1 & -3 & 2 \end{array}\right)$$

Each block divided by the horizontal and vertical lines is called sub-matrix, Here we let

$$A_{11} = \left(\begin{array}{ccc} 1 & 1 & 5 \\ 4 & -3 & 8 \\ ... ...-3 \end{array}\right), A_{22} = \left(\begin{array}{c} 4 \end{array}\right)$$

Then the matrix $A$ can be written

$$A = \left(\begin{array}{cc} A_{11} & A_{12}\\ A_{21} & A_{22} \end{array}\right)$$

Next we divide $B$ same as $A$. The the product of matrices $A$ and $B$ is given by the followings:

$$AB = \left(\begin{array}{cc} A_{11} & A_{12}\\ A_{21} & A_{22} \end{a... ...t)\left(\begin{array}{cc} B_{11} & B_{12}\\ B_{21} & B_{22} \end{array}\right)$$

$$= \left(\begin{array}{cc} A_{11}B_{11} + A_{12}B_{21} & A_{11}B_{12... ...\ A_{21}B_{11} + A_{22}B_{21} & A_{21}B_{12} + A_{22}B_{22} \end{array}\right)$$

$\spadesuit$Square Matrices $\spadesuit$

A matrix with the same number of rows as columns is called a square matrix. the number of rows and columns of the square matrix is called order. In other words, An $n \times n$ is a square matrix with the order $n$.

Now we introduce four different kind of square matrices. 1. A square matrix with the diagonal entries are all 0. i.e., For $a_{ij} = 0, i \neq j$, $A$ is called diagonal matrix.

2. A square matrix with $a_{ii} = 1$ for all $i$ and $a_{ij} = 0$ for all $i \neq j$ is called identity matrix and denoted by $I$.

$$A = \left (\begin{array}{ccc} 1&0&0\\ 0&-2&0\\ 0&0&3 \end{array}\right ), \ B = \left ( \begin{array}{cc} 1&0\\ 0&1 \end{array}\right )$$

As you can see $A$ is an diagonal matrix with $3 \times 3$ and $B$ is a identity matrix of $2\times 2$

Theorem 2..1  

Let $A,B$ be diagonal matrices of order $n$. Then show the followings:

1. $A + B$ is a diagonal matrix.
2. $AB, BA$ are diagonal matrices and $AB = BA$.

Proof 1. Let $A = (a_{ii}), B = (b_{ii})$. Then $A + B = (a_{ii}) + (b_{jj}) = (a_{ii} + b_{ii})$. Thus $A + B$ is a diagonal matrix。

2. Let $AB = C$. Then

$$C = (c_{ij}) = (\sum_{k=1}^{n} a_{ik}b_{kj}) = \left\{\begin{array}{ll} a_{ii}b_{ii} & i = j\\ 0 & i \neq j \end{array}\right.$$

Thus $AB, BA$ are diagonal matrices and $AB = BA$.

$\spadesuit$Transposed Matrix $\spadesuit$

The sum of a diagonal entries is called trace and denoted by

$$tr A = \sum_{i=1}^{n} a_{ii}$$

The transpose of a matrix $A$, written $A^{t}$, is the matrix obtained by writing the rows of $A$ , in order, as columns.

Example 2..5  

Let $A = \left(\begin{array}{ccc} 2&1&3\\ 3&0&1 \end{array}\right )$. Find $A^{t}$.

Answer $A^{t} = \left( \begin{array}{cc} 2&3\\ 1&0\\ 3&1 \end{array}\right ) .$ $\blacksquare$

The transpose operation on matrices satifies the following properties:

Theorem 2..2  

For matrices $A,B$, the followings hold.

1. $(A + B)^{t} = A^{t} + B^{t}$
2. $(A^{t})^{t} = A$
3. $(AB)^{t} = B^{t}A^{t}$

Proof 1. Let $A = (a_{ij})_{n \times k}, B = (b_{ij})_{n \times k}$. Then $A + B = (a_{ij} + b_{ij})_{n \times k}$. Thus the transposed matrix is given by $(a_{ji} + b_{ji})_{k \times n}$. On the other hand, $A^{t} = (a_{ji})_{k \times n}, B^{t} = (b_{ji})_{k \times n}$. Thus we have $(A + B)^{t} = A^{t} + B^{t}$.

2. Let $A = (a_{ij})_{n \times k}$. Then $A^{t} = (a_{ji})_{k \times n}$. Thus, $(A^{t})^{t} = (a_{ij})_{n \times k} = A$.

3. Let $A,B$ be $n \times k, k \times m$ matrices. Then the matrix $AB$ is the size of $n \times m$. Also, $(AB)^{t}$ is the size of $m \times n$. By the definition of transposed matrix, $(i,j)$ element of $(AB)^{t}$ is the $(j, i)$ element of $AB$. So, $AB$ can be expressed as $a_{j1}b_{1i} + a_{j2}b_{2i} + \cdots + a_{jk}b_{ki}$. Now $B^{t}$ is the order fo $m \times k$and $A^{t}$ is the order of $k \times n$. Thus, $B^{t}A^{t}$ is the order of $m \times n$. $(i,j)$ element of $B^{t}A^{t}$ is the inner product of $i$th row of $B^{t}$ and $j$th column of $A^{t}$ . Thus, $i$th row and $j$th column of the matrix $B^{t}A^{t}$ is $a_{j1}b_{1i} + a_{j2}b_{2i} + \cdots + a_{jk}b_{ki}$. Therefore, $(AB)^{t} = B^{t}A^{t}$. $\blacksquare$

$\spadesuit$Symmetric Matrices $\spadesuit$

When the square matrix $A$ and its transpose matrix $A^{t}$ are the same, we say $A$ is symmetric matrix. When the matrix $A$ satisfies $A^{t} = -A$, we say $A$ is skew symmetric matrix.

Example 2..6  

Determine whether $A$ is a symmetric matrix.

$A = \left(\begin{array}{rrr} 1&-1&1\\ -1&2&0\\ 1&0&3 \end{array}\right )$

Answer $A^{t} = \left(\begin{array}{rrr} 1&-1&1\\ -1&2&0\\ 1&0&3 \end{array}\right ) = A$. Thus $A$ is a symmetric matrix. $\blacksquare$

For the square matrix $A$ with the order of $n \times n$. We define the power as follows:

$$A^{2} = AA, A^{3} = A^{2}A, \ldots , A^{n} = A^{n-1}A$$

.

When the matrix $A$ satifies $A^{k} = 0$ for some natural number $k$, we say the matrix $A$ is nilpotent.

Exercise2-2

1. For matrices $A = \left(\begin{array}{rr} 2&-3\\ 4&2 \end{array}\right), B = \left(\begin{array}{rr} -1&2\\ 3&0 \end{array}\right )$, evaluate the followings:

(a) $A + B$ (b) $2A - 3B$ (c) $AB, BA$

2. For matrices $A = \left(\begin{array}{rrr} 3&1&7\\ 5&2&-4 \end{array}\right), B = \left(\begin{array}{rr} 2&-3\\ 3&6\\ 4&1 \end{array}\right )$, find $AB, BA$.

3. For the matrix $A = \left(\begin{array}{rrr} 2&3&0\\ 1&4&1\\ 2&0&1 \end{array}\right)$, calculate $A^{2} - 5A + 6I$.

4. Let $A$ and $B$ be symmetric matrices of the order $n$. Show that $A + B$ is a symmetric matrix.

5. Let $A$ and $B$ be symmetric matrices of the order $n$. Find the necessary and sufficient conditions so that $AB$ is a symmetric matrix.

6. Suppose that $A$ is a skew symmetric matrix. Then show that $A^{2}$ is a symmetric marix.

7. Find matrices so that the product of $\left(\begin{array}{ccc} a_{1}&0&0\\ 0&a_{2}&0\\ 0&0&a_{3} \end{array}\right)$ is interchangeable. Here, $a_{1},a_{2},a_{3}$ are different real numbers.

8. Show that any square matrix $A$ can be expressed by the sum of a symmetric matrix and a skew symmetric matrix.

9. Find the product of $A$ and $B$, where $A = \left(\begin{array}{cc\vert c} 1 & 1 & 1\\ 2 & -1 & 0\ \hline -1 & 0 & ... ... 1 & 2 & 3 & -1\\ 3 & -1 & 1 & 0\ \hline 0 & 0 & -3 & 1 \end{array}\right)$.