1.4 Row-Reduced Echelon Matrices

我已经忘了这个概念中文如何翻译，但是其本质就是在Row-reduced的基础上，作行的对调，使得所有非零行在上面且每一行开头的1以从左至右自上而下的方式排列、所有零行在下面。这也是Theorem 5的结论，即任何 m × n m\times n m×n矩阵都和一个row-reduced echelon matrix是行等价的。
前两节讨论了如此多的row-reduced问题，目的是为了解方程组，因为row-reduced的形式使得解方程非常方便，并且如果非零行数比未知数的个数少，那么一定有非退化解（non-trivial solution），即不是0向量的解。Theorem 6是一个更强的假设和更弱一些的结论：如果 A A A是 m × n m\times n m×n矩阵且 m < n m<n m<n，那么 A X = 0 AX=0 AX=0一定有non-trivial solution。Theorem 7是一个特殊一点的结论： n × n n\times n n×n方阵行等价于单位矩阵的充要条件是 A X = 0 AX=0 AX=0只有0解。
以上已经讨论了足够多的关于homogeneous system的结论。如果将条件放松，即 A X = 0 AX=0 AX=0变为 A X = Y AX=Y AX=Y，那么需要考虑augmented matrix [ A Y ] \begin{bmatrix}A&Y\end{bmatrix} [AY], 在inhomogeneous system中，首先要考虑的是有没有解（不像homogeneous system里一定有0解），按照Theorem 4和Theorem 5可以将 A A A变为一个row-reduced echelon matrix A ′ A' A′，对应的会把 Y Y Y变成一个新的列向量 Z Z Z, 那么这个system有解的充要条件是 A ′ A' A′里全为零的最后几行对应的 Z Z Z的最后几个分量也都是0。
最后有一个看似很奇怪但确实正确的结论： A X = Y AX=Y AX=Y如果 A , Y A,Y A,Y都是在 F F F的一个subfield F 1 F_1 F1上，那么其在 F F F中有解可以推出其在 F 1 F_1 F1中有解。

Exercises

1.Find all solutions to the following systems of equations by row-reducing the coefficient matrix:

1 3 x 1 + 2 x 2 − 6 x 3 = 0 − 4 x 1 + 5 x 3 = 0 − 3 x 1 + 6 x 2 − 13 x 3 = 0 − 7 3 x 1 + 2 x 2 − 8 3 x 3 = 0 \begin{aligned}\frac{1}{3}&x_1+2x_2&-6x_3=0\\-4&x_1&+5x_3=0\\-3&x_1+6x_2&-13x_3=0\\-\frac{7}{3}&x_1+2x_2&-\frac{8}{3}x_3=0\end{aligned} 31−4−3−37x1+2x2x1x1+6x2x1+2x2−6x3=0+5x3=0−13x3=0−38x3=0
Solution: We have
[ 1 / 3 2 − 6 − 4 0 5 − 3 6 − 13 − 7 / 3 2 − 8 / 3 ] → [ 1 6 − 18 0 24 − 67 0 24 − 67 − 7 6 − 8 ] → [ 1 6 − 18 0 24 − 67 0 0 0 0 48 − 134 ] → [ 1 6 − 18 0 24 − 67 0 0 0 0 0 0 ] \begin{bmatrix}1/3&2&-6\\-4&0&5\\-3&6&-13\\-7/3&2&-8/3\end{bmatrix}\rightarrow\begin{bmatrix}1&6&-18\\0&24&-67\\0&24&-67\\-7&6&-8\end{bmatrix}\rightarrow\begin{bmatrix}1&6&-18\\0&24&-67\\0&0&0\\0&48&-134\end{bmatrix}\rightarrow\begin{bmatrix}1&6&-18\\0&24&-67\\0&0&0\\0&0&0\end{bmatrix} ⎣⎢⎢⎡1/3−4−3−7/32062−65−13−8/3⎦⎥⎥⎤→⎣⎢⎢⎡100−7624246−18−67−67−8⎦⎥⎥⎤→⎣⎢⎢⎡1000624048−18−670−134⎦⎥⎥⎤→⎣⎢⎢⎡100062400−18−6700⎦⎥⎥⎤
let x 3 = c x_3=c x3=c, then all solutions is of the form:
x 1 = 1062 67 c , x 2 = 24 67 c , x 3 = c , c ∈ R x_1=\dfrac{1062}{67} c,\quad x_2=\dfrac{24}{67} c,\quad x_3=c,\quad c\in R x1=671062c,x2=6724c,x3=c,c∈R

2. Find a row-reduced echelon matrix which is row-equivalent to

A = [ 1 − i 2 2 i 1 + i ] A=\begin{bmatrix}1&-i\\2&2\\i&1+i\end{bmatrix} A=⎣⎡12i−i21+i⎦⎤

What are the solutions of A X = 0 AX=0 AX=0?

Solution: We have
A = [ 1 − i 2 2 i 1 + i ] → [ 1 1 0 − i − 1 i 1 + i ] → [ 1 1 0 1 0 − i ] → [ 1 0 0 1 0 0 ] A=\begin{bmatrix}1&-i\\2&2\\i&1+i\end{bmatrix}\rightarrow\begin{bmatrix}1&1\\0&-i-1\\i&1+i\end{bmatrix}\rightarrow\begin{bmatrix}1&1\\0&1\\0&-i\end{bmatrix}\rightarrow\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix} A=⎣⎡12i−i21+i⎦⎤→⎣⎡10i1−i−11+i⎦⎤→⎣⎡10011−i⎦⎤→⎣⎡100010⎦⎤
the solutions of A X = 0 AX=0 AX=0 are trivial solutions.

3.Describe explicitly all 2 × 2 2\times 2 2×2 row-reduced echelon matrices.

Solution: They are (due to the nonzero entries of the matrix)
[ 0 0 0 0 ] , [ 1 0 0 0 ] , [ 1 b 0 0 ] , [ 1 0 0 1 ] \begin{bmatrix}0&0\\0&0\end{bmatrix},\quad\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad\begin{bmatrix}1&b\\0&0\end{bmatrix},\quad\begin{bmatrix}1&0\\0&1\end{bmatrix} [0000],[1000],[10b0],[1001]

4.Consider the system of equations

x 1 − x 2 + 2 x 3 = 1 2 x 1 + 2 x 3 = 1 x 1 − 3 x 2 + 4 x 3 = 2 \begin{aligned}&x_1-x_2+&2x_3=1\\2&x_1&+2x_3=1\\&x_1-3x_2&+4x_3=2\end{aligned} 2x1−x2+x1x1−3x22x3=1+2x3=1+4x3=2

Does this system have a solution? If so, describe explicitly all solutions.

Solution: We have
[ 1 − 1 2 1 2 0 2 1 1 − 3 4 2 ] → [ 1 − 1 2 1 0 2 − 2 − 1 0 − 2 2 1 ] → [ 1 − 1 2 1 0 1 − 1 − 1 2 0 0 0 0 ] → [ 1 0 1 1 2 0 1 − 1 − 1 2 0 0 0 0 ] \begin{bmatrix}1&-1&2&1\\2&0&2&1\\1&-3&4&2\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&2&1\\0&2&-2&-1\\0&-2&2&1\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&2&1\\0&1&-1&-\frac{1}{2}\\0&0&0&0\end{bmatrix}\rightarrow\begin{bmatrix}1&0&1&\frac{1}{2}\\0&1&-1&-\frac{1}{2}\\0&0&0&0\end{bmatrix} ⎣⎡121−10−3224112⎦⎤→⎣⎡100−12−22−221−11⎦⎤→⎣⎡100−1102−101−210⎦⎤→⎣⎡1000101−1021−210⎦⎤
this system has a solution, let x 3 = c x_3=c x3=c, then all solutions of the system is
( 1 2 − c , c − 1 2 , c ) \left(\dfrac{1}{2}-c,c-\dfrac{1}{2},c\right) (21−c,c−21,c)

5. Give an example of a system of two linear equations in two unknowns which has no solution.

Solution: An example may be
{ x 1 + x 2 = 1 2 x 1 + 2 x 2 = 3 \begin{cases}x_1+x_2&=1\\2x_1+2x_2&=3\end{cases} {x1+x22x1+2x2=1=3

6.Show that the system

x 1 − 2 x 2 + x 3 + 2 x 4 = 1 x 1 + x 2 − x 3 + x 4 = 2 x 1 + 7 x 2 − 5 x 3 − x 4 = 3 \begin{array}{r}x_1-2x_2+x_3+2x_4=1\\x_1+x_2-x_3+x_4=2\\x_1+7x_2-5x_3-x_4=3\end{array} x1−2x2+x3+2x4=1x1+x2−x3+x4=2x1+7x2−5x3−x4=3

has no solution.

Solution: Use row operation on the augmented matrix of the system we get
[ 1 − 2 1 2 1 1 1 − 1 1 2 1 7 − 5 − 1 3 ] → [ 1 − 2 1 2 1 0 3 − 2 − 1 1 0 9 − 6 − 3 2 ] → [ 1 − 2 1 2 1 0 3 − 2 − 1 1 0 0 0 0 − 1 ] \begin{bmatrix}1&-2&1&2&1\\1&1&-1&1&2\\1&7&-5&-1&3\end{bmatrix}\rightarrow\begin{bmatrix}1&-2&1&2&1\\0&3&-2&-1&1\\0&9&-6&-3&2\end{bmatrix}\rightarrow\begin{bmatrix}1&-2&1&2&1\\0&3&-2&-1&1\\0&0&0&0&-1 \end{bmatrix} ⎣⎡111−2171−1−521−1123⎦⎤→⎣⎡100−2391−2−62−1−3112⎦⎤→⎣⎡100−2301−202−1011−1⎦⎤
thus the system has no solution

7.Find all solutions of

2 x 1 − 3 x 2 − 7 x 3 + 5 x 4 + 2 x 5 = − 2 x 1 − 2 x 2 − 4 x 3 + 3 x 4 + x 5 = − 2 2 x 1 − 4 x 3 + 2 x 4 + x 5 = 3 x 1 − 5 x 2 − 7 x 3 + 6 x 4 + 2 x 5 = − 7 \begin{aligned}2x_1-3x_2-7x_3+5x_4+2x_5&=-2\\x_1-2x_2-4x_3+3x_4+x_5&=-2\\2x_1-4x_3+2x_4+x_5&=3\\x_1-5x_2-7x_3+6x_4+2x_5&=-7\end{aligned} 2x1−3x2−7x3+5x4+2x5x1−2x2−4x3+3x4+x52x1−4x3+2x4+x5x1−5x2−7x3+6x4+2x5=−2=−2=3=−7
Solution: We have
[ 2 − 3 − 7 5 2 − 2 1 − 2 − 4 3 1 − 2 2 0 − 4 2 1 3 1 − 5 − 7 6 2 − 7 ] → [ 1 − 2 − 4 3 1 − 2 0 1 1 − 1 0 2 0 4 4 − 4 − 1 7 0 − 3 − 3 3 1 − 5 ] → [ 1 − 2 − 4 3 1 − 2 0 1 1 − 1 0 2 0 0 0 0 − 1 − 1 0 0 0 0 1 1 ] → [ 1 0 − 2 1 1 2 0 1 1 − 1 0 2 0 0 0 0 1 1 0 0 0 0 0 0 ] \begin{bmatrix}2&-3&-7&5&2&-2\\1&-2&-4&3&1&-2\\2&0&-4&2&1&3\\1&-5&-7&6&2&-7\end{bmatrix}\rightarrow\begin{bmatrix}1&-2&-4&3&1&-2\\0&1&1&-1&0&2\\0&4&4&-4&-1&7\\0&-3&-3&3&1&-5\end{bmatrix}\\\rightarrow\begin{bmatrix}1&-2&-4&3&1&-2\\0&1&1&-1&0&2\\0&0&0&0&-1&-1\\0&0&0&0&1&1\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-2&1&1&2\\0&1&1&-1&0&2\\0&0&0&0&1&1\\0&0&0&0&0&0\end{bmatrix} ⎣⎢⎢⎡2121−3−20−5−7−4−4−753262112−2−23−7⎦⎥⎥⎤→⎣⎢⎢⎡1000−214−3−414−33−1−4310−11−227−5⎦⎥⎥⎤→⎣⎢⎢⎡1000−2100−41003−10010−11−22−11⎦⎥⎥⎤→⎣⎢⎢⎡10000100−21001−10010102210⎦⎥⎥⎤
thus x 5 = 1 x_5=1 x5=1, if we let x 3 = c 1 , x 4 = c 2 x_3=c_1,x_4=c_2 x3=c1,x4=c2, then all solutions of the system is
x 1 = 1 + 2 c 1 − c 2 , x 2 = 2 − c 1 + c 2 , x 3 = c 1 , x 4 = c 2 , x 5 = 1 x_1=1+2c_1-c_2,x_2=2-c_1+c_2,x_3=c_1,x_4=c_2,x_5=1 x1=1+2c1−c2,x2=2−c1+c2,x3=c1,x4=c2,x5=1

8.Let

A = [ 3 − 1 2 2 1 1 1 − 3 0 ] A=\begin{bmatrix}3&-1&2\\2&1&1\\1&-3&0\end{bmatrix} A=⎣⎡321−11−3210⎦⎤

For which triples ( y 1 , y 2 , y 3 ) (y_1,y_2,y_3) (y1,y2,y3) does the system A X = Y AX=Y AX=Y have a solution?

Solution:
[ 3 − 1 2 y 1 2 1 1 y 2 1 − 3 0 y 3 ] → [ 1 − 3 0 y 3 0 7 1 y 2 − 2 y 3 0 8 2 y 1 − 3 y 3 ] → [ 1 − 3 0 y 3 0 7 1 y 2 − 2 y 3 0 8 2 y 1 − 3 y 3 ] → [ 1 − 3 0 y 3 0 7 1 y 2 − 2 y 3 0 1 1 y 1 − y 2 − y 3 ] \begin{bmatrix}3&-1&2&y_1\\2&1&1&y_2\\1&-3&0&y_3 \end{bmatrix}→\begin{bmatrix}1&-3&0&y_3\\0&7&1&y_2-2y_3\\0&8&2&y_1-3y_3 \end{bmatrix}→\begin{bmatrix}1&-3&0&y_3\\0&7&1&y_2-2y_3\\0&8&2&y_1-3y_3 \end{bmatrix}→\begin{bmatrix}1&-3&0&y_3\\0&7&1&y_2-2y_3\\0&1&1&y_1-y_2-y_3\end{bmatrix} ⎣⎡321−11−3210y1y2y3⎦⎤→⎣⎡100−378012y3y2−2y3y1−3y3⎦⎤→⎣⎡100−378012y3y2−2y3y1−3y3⎦⎤→⎣⎡100−371011y3y2−2y3y1−y2−y3⎦⎤
Since A A A is row equivalent to the identity matrix, A X = Y AX=Y AX=Y has a solution for all ( y 1 , y 2 , y 3 ) (y_1,y_2,y_3 ) (y1,y2,y3).

9.Let

A = [ 3 − 6 2 − 1 − 2 4 1 3 0 0 1 1 1 − 2 1 0 ] A=\begin{bmatrix}3&-6&2&-1\\-2&4&1&3\\0&0&1&1\\1&-2&1&0\end{bmatrix} A=⎣⎢⎢⎡3−201−640−22111−1310⎦⎥⎥⎤

For which ( y 1 , y 2 , y 3 , y 4 ) (y_1,y_2,y_3,y_4) (y1,y2,y3,y4) does the system A X = Y AX=Y AX=Y have a solution?

Solution:
[ 3 − 6 2 − 1 y 1 − 2 4 1 3 y 2 0 0 1 1 y 3 1 − 2 1 0 y 4 ] → [ 1 − 2 1 0 y 4 0 0 1 1 y 3 0 0 − 1 3 y 2 − 2 y 4 0 0 − 1 − 1 y 1 − 3 y 4 ] → [ 1 − 2 1 0 y 4 0 0 1 1 y 3 0 0 0 4 ∗ 0 0 0 0 y 1 − 3 y 4 + y 3 ] \begin{bmatrix}3&-6&2&-1&y_1\\-2&4&1&3&y_2\\0&0&1&1&y_3\\1&-2&1&0&y_4\end{bmatrix}→\begin{bmatrix}1&-2&1&0&y_4\\0&0&1&1&y_3\\0&0&-1&3&y_2-2y_4\\0&0&-1&-1&y_1-3y_4 \end{bmatrix}→\begin{bmatrix}1&-2&1&0&y_4\\0&0&1&1&y_3\\0&0&0&4&*\\0&0&0&0&y_1-3y_4+y_3\end{bmatrix} ⎣⎢⎢⎡3−201−640−22111−1310y1y2y3y4⎦⎥⎥⎤→⎣⎢⎢⎡1000−200011−1−1013−1y4y3y2−2y4y1−3y4⎦⎥⎥⎤→⎣⎢⎢⎡1000−200011000140y4y3∗y1−3y4+y3⎦⎥⎥⎤
so for triples ( y 1 , y 2 , y 3 , y 4 ) (y_1,y_2,y_3,y_4 ) (y1,y2,y3,y4) which satisfies y 1 = 3 y 4 − y 3 y_1=3y_4-y_3 y1=3y4−y3 does the system have a solution

10. Suppose R R R and R ′ R' R′ are 2 × 3 2\times 3 2×3 row-reduced echelon matrices and that the systems R X = 0 RX=0 RX=0 and R ′ X = 0 R'X=0 R′X=0 have exactly the same solutions. Prove that R = R ′ R=R' R=R′.

Solution: If R R R and R ′ R' R′ are zero matrix then there’s nothing to prove.
If both are nonzero and assume they have different number of rows, let
R = [ 1 0 a 0 1 b ] , R ′ = [ 1 c d 0 0 0 ] R=\begin{bmatrix}1&0&a\\0&1&b\end{bmatrix},\quad R'=\begin{bmatrix}1&c&d\\0&0&0\end{bmatrix} R=[1001ab],R′=[10c0d0]
then ( c , − 1 , 0 ) (c,-1,0) (c,−1,0) is a solution for the system of R ′ R' R′, but not a solution for the system of R R R, a contradiction, thus R R R and R ′ R' R′ must have the same number of rows.
If both R R R and R R R have one row, let
R = [ 1 a b 0 0 0 ] , R ′ = [ 1 c d 0 0 0 ] R=\begin{bmatrix}1&a&b\\0&0&0\end{bmatrix},\quad R'=\begin{bmatrix}1&c&d\\0&0&0\end{bmatrix} R=[10a0b0],R′=[10c0d0]
then ( c , − 1 , 0 ) (c,-1,0) (c,−1,0) is a solution for the system of R ′ R' R′, thus a solution for the system of R R R, so
c − a + 0 ⋅ b = 0 ⇒ c = a c-a+0⋅b=0 \Rightarrow c=a c−a+0⋅b=0⇒c=a
similarly use the solution ( d , 0 , − 1 ) (d,0,-1) (d,0,−1) we get b = d b=d b=d, thus R = R ′ R=R' R=R′ in this case.
If both R R R and R R R have two rows, let
R = [ 1 0 a 0 1 b ] , R ′ = [ 1 0 c 0 1 d ] R=\begin{bmatrix}1&0&a\\0&1&b\end{bmatrix},\quad R'=\begin{bmatrix}1&0&c\\0&1&d\end{bmatrix} R=[1001ab],R′=[1001cd]
then it’s easy to see b = d b=d b=d, otherwise ( 0 , b , − 1 ) (0,b,-1) (0,b,−1) can’t be a solution for the system of R ′ R' R′, also a = c a=c a=c, since ( a , 0 , − 1 ) (a,0,-1) (a,0,−1) is a solution for the system of R ′ R' R′ and we have a − c = 0 a-c=0 a−c=0. The proof is complete.