“矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授。
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Summary
- Matrix
A matrix is a rectangular array of elements. Matrix $A$ is denoted by $$A = egin{bmatrix}a_{11} & cdots & a_{1n}\ vdots&vdots&vdots \ a_{m1}& cdots & a_{mn} end{bmatrix}$$ - Vector
A vector is a matrix that has only one row or one column. For example, $[1, 2, 3]$ is a row vector of dimension 3, and $egin{bmatrix}1 \ 2 \ 3 end{bmatrix}$ is a column vector of dimension 3. - Equal matrices
Two matrices $[A]$ and $[B]$ are equal if the size of $[A]$ and $[B]$ is the same, that is, the number of rows and columns of $[A]$ are same as that of $[B]$. And $a_{ij}=b_{ij}$ for all $i$ and $j$. - Zero matrix
A matrix whose all entries are zero is called a zero matrix, that is, $a_{ij}=0$ for all $i$ and $j$. For example, $$A = egin{bmatrix}0 & 0 & 0\ 0 & 0 & 0end{bmatrix}$$ - Submatrix
If some rows or/and columns of a matrix $[A]$ are deleted (no rows or columns may be deleted), the remaining matrix is called a submatrix of $[A]$. For example, some of the submatrix of $egin{bmatrix}1 & 2 \ 3 & 4\ 5 & 6 end{bmatrix}$ are $$[1], [1, 2], egin{bmatrix}1\3\5end{bmatrix}, egin{bmatrix} 1 & 2\3 & 4 end{bmatrix}, egin{bmatrix} 1 & 2\5 & 6 end{bmatrix}, egin{bmatrix}1 & 2 \ 3 & 4\ 5 & 6 end{bmatrix}.$$ - Square matrix
If the number of rows of a matrix is equal to the number of columns of a matrix, then the matrix is called a square matrix. For example, $$A = egin{bmatrix}1 & 2 & 3\ 4 & 5 & 6\ 7 & 8 & 9 end{bmatrix}$$ - Diagonal matrix
A square matrix with all non-diagonal elements equal to zero is called a diagonal matrix, that is, only the diagonal entries of the square matrix can be non-zero, $a_{ij} = 0$ for $i eq j$. For example, $$A=egin{bmatrix}1 & 0 & 0\ 0 & 3 & 0\ 0 & 0 & 5 end{bmatrix}$$ - Identity matrix
A diagonal matrix with all diagonal elements equal to 1 is called an identity matrix, that is, $a_{ij}=0$, $i eq j$ for all $i$, $j$ and $a_{ii}=1$ for all $i$. For example, $$A = egin{bmatrix}1 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 1 end{bmatrix}$$ - Upper triangular matrix
A $n imes n$ matrix for which $a_{ij} = 0$, $i > j$ for all $i$, $j$ is called an upper triangular matrix. That is, all the elements below the diagonal entries are zeros. For example, $$A = egin{bmatrix}1 & 0 & 3\ 0 & 5 & 6\ 0 & 0 & 9 end{bmatrix}$$ - Lower triangular matrix
A $n imes n$ matrix for which $a_{ij} = 0$, $j > i$ for all $i$, $j$ is called a lower triangular matrix. That is, all the elements above the diagonal entries are zeros. For example, $$A = egin{bmatrix}1 & 0 & 0\ 4 & 5 & 0\ 0 & 8 & 9 end{bmatrix}$$ - Tridiagonal matrix
A tridiagonal matrix is a square matrix in which all elements not on the following are zero: the major diagonal, the diagonal above the major diagonal, and the diagonal below the major diagonal. For example, $$A = egin{bmatrix}1 & 2 & 0 & 0\ 4 & 5 & 6 & 0\ 0 & 0 & 7 & 8\ 0& 0& -1& 2 end{bmatrix}$$ Note that a non-square matrix also has diagonal entries. For an $m imes n$ matrix, the diagonal entries are $a_{11}$, $cdots$, $a_{kk}$ where $k=min{m, n}$. For example, $$A = egin{bmatrix}1& 2\ 3& 4 \ 5& 6end{bmatrix}$$ the diagonal elements are $a_{11}=1$ and $a_{22}=4$. - Diagonally dominant matrix
An $n imes n$ square matrix $[A]$ is a diagonal dominant matrix of $$|a_{ii}|geq sum_{j=1, i eq j}^{n}|a_{ij}|$$ for $i = 1, cdots, n$ and $$|a_{ii}| > sum_{j=1, i eq j}^{n}|a_{ij}|$$ for at least one $i$. That is, for each row, the absolute value of the diagonal element is greater than or equal to the sum of the absolute values of the rest of the elements of that row, and that the inequality is strictly greater than for at least one row. For example, $$A = egin{bmatrix}15& 6& 7\ 2& -4& -2\ 3& 2& 6 end{bmatrix}$$ is a diagonal dominant matrix since $$egin{cases}|a_{11}| = 15 geq |a_{12}| + |a_{13}| =13\ |a_{22}|= 4 geq |a_{21}| + |a_{23}| = 4\ |a_{33}| = 6 geq |a_{31}| + |a_{32}| = 5 end{cases}$$ and for at least one row, that is row 1 and row 3 in this case, the inequality is a strictly greater than inequality.
Selected Problems
1. Given $$A=egin{bmatrix}6& 2& 3& 9\ 0& 1& 2& 3\ 0& 0& 4& 5\ 0& 0& 0& 6 end{bmatrix}$$ then $[A]$ is a ( ) matrix.
Solution:
This is an upper triangular matrix.
2. A square matrix $[A]$ is lower triangular if ( ).
Solution:
Lower triangular matrix: $a_{ij} = 0$ for $j > i$.
3. Given $$A = egin{bmatrix} 12.3& -12.3& 20.3\ 11.3& -10.3& -11.3\ 10.3& -11.3& -12.3end{bmatrix}, B = egin{bmatrix} 2& 4\ -5& 6\ 11& -20end{bmatrix}$$ then if $[C] = [A]cdot[B]$, then $c_{31}= ( )$.
Solution:
$$c_{31} = egin{bmatrix}10.3 & -11.3 &-12.3end{bmatrix}cdot egin{bmatrix}2\ -5\ 11 end{bmatrix}$$ $$= 10.3 imes2 + (-11.3) imes(-5) + (-12.3) imes11= -58.2$$
4. The following system of equations has ( ) solutions. $$egin{cases}x + y =2\ 6x + 6y=12 end{cases}$$
Solution:
$x=2-y$ where $y$ is arbitrary. Thus it has infinite solutions.
5. Consider there are only two computer companies in a country. The companies are named Dude and Imac. Each year, Dude keeps ${1/5}^{th}$ of its customers, while the rest switch to Imac. Each year, Imac keeps ${1/3}^{rd}$ of its customers, while the rest switch to Dude. If in 2003, Dude had ${1/6}^{th}$ of the market and Imac had ${5/6}^{th}$ of the market, what will be the share of Dude computers when the market becomes stable?
Solution:
Since we want when the market is stable, the market share should not change from year to year. Let $D$ and $M$ denote the market of Dude and Imac, respectively. Thus we have $$egin{cases} D_n = {1over5}D + {2over3}M\ M_n= {4over5}D + {1over3}Mend{cases}Rightarrow egin{bmatrix}D_n\ M_n end{bmatrix} = egin{bmatrix}{1over5} & {2over3}\ {4over5}& {1over3} end{bmatrix}cdot egin{bmatrix}D\ Mend{bmatrix}$$ $D_n = D$ and $M_n=M$ eventually. That is $$egin{cases}{4over5}D - {2over3} M=0 \ D+M=1end{cases}Rightarrowegin{cases}D = {5over11}\ M= {6over11} end{cases}$$ Hence the final market share of Dude will be $displaystyle{5over11}$.
6. Three kids - Jim, Corey and David receive an inheritance of 2,253,453. The money is put in three trusts but is not divided equally to begin with. Corey's trust is three times that of David's because Corey made an A in Dr. Kaw's class. Each trust is put in an interest generating investment. The three trusts of Jim, Corey and David pays an interest of 6%, 8%, 11%, respectively. The total interest of all the three trusts combined at the end of the first year is 190,740.57. The equations to find the trust money of Jim (J), Corey (C) and David (D) in a matrix form is ( ).
Solution:
From the given conditions, we have $$egin{cases}J + C +D =2253453\ C=3D\ 0.06J + 0.08C + 0.11D = 190740.57end{cases}$$ $$Rightarrow egin{cases}J + C +D =2253453\ C-3D = 0\ 0.06J + 0.08C + 0.11D = 190740.57end{cases}$$ $$Rightarrow egin{bmatrix}1& 1& 1\ 0& 1& -3\ 0.06& 0.08& 0.11 end{bmatrix}cdot egin{bmatrix}J\ C\ D end{bmatrix} = egin{bmatrix}2253453\ 0\ 190740.57 end{bmatrix}$$
7. Which of the following matrices are strictly diagonally dominant? $$A = egin{bmatrix}15 &6 &7\ 2 &-4 &2\ 3& 2 &6 end{bmatrix}, B = egin{bmatrix}5 &6 &7\ 2 &-4 &2\ 3& 2 &-5 end{bmatrix}, C = egin{bmatrix}5&3 &2\ 6 &-8 &2\ 7& -5 &12 end{bmatrix}. $$
Solution:
For $A$, $$egin{cases}|a_{11}|=15 > |a_{12}| + |a_{13}| = 13\ |a_{22}| = 4 = |a_{21}| + |a_{23}| = 4\ |a_{33}| = 6 > |a_{31}| + |a_{32}| = 5 end{cases}$$ So it is strictly diagonal dominant. For $B$, $$|b_{11}| = 5 < |b_{12}| + |b_{13}| = 13$$ So it is not strictly diagonal dominant. For $C$, $$egin{cases}|c_{11}|=5 = |c_{12}| + |c_{13}| = 5\ |c_{22}| = 8 = |c_{21}| + |c_{23}| = 8\ |c_{33}| = 12 = |c_{31}| + |c_{32}| = 12 end{cases}$$ So it is not strictly diagonal dominant.