Linear Algebra (2ed) Hoffman&Kunze 2.1

christangdt留下评论

本节的开头，Hoffman对线性代数这一概念有一个挺深刻的诠释：Linear algebra is that branch of mathematics which treats the common properties of algebraic systems which consists of a set, together with a reasonable notion of a ‘linear combination’ of elements in the set. 这个定义更抽象且更具有普遍性。严格的定义是：vector space含有F（scalars）、V（objects/vectors），以及加法、数乘两种运算，两种运算各需符合四种性质。
加法的四种性质：
\cdot 3(a) \alpha +\beta = \beta +\alpha
\cdot 3(b) \alpha +(\beta +\gamma) =(\alpha + \beta )+\gamma
\cdot 3(c) \exists 0\in V,\text{ s.t. }\alpha+0=\alpha,\forall\alpha \in V
\cdot 3(d) \forall\alpha \in V,\exists!-\alpha\in V,\text{ s.t. }\alpha+(-\alpha)=0
数乘的四种性质：
\cdot 4(a) 1\alpha=\alpha,\forall\alpha \in V
\cdot 4(b) (c_1c_2)\alpha=c_1(c_2\alpha)
\cdot 4(c) c(\alpha +\beta)=c\alpha +c\beta
\cdot 4(d) (c_1+c_2)\alpha=c_1\alpha +c_2\alpha
接下来，对vector有一段解释，目的是让读者消去对“向量”这一词汇的既有印象，或者说vector也是抽象的，不仅是几何空间里的向量，也是algebraic的对象的概念。后面举了五个例子，很精彩。第一个是常见的n维向量；第二个是m\times n矩阵，并提到第一个例子是第二个的特例（当m=1时）；第三个是泛函空间，并指出第一、第二个是第三个的特例；第四个例子是多项式函数，这实际上也是第三个例子的一个子集；第五个例子是说明：复数集可以看做实数集上的一个vector space。
从定义可以得到一些很简单但是有用的结论，比如c0=0,0\alpha =0，以及c\alpha =0\Rightarrow(c=0)\vee(\alpha =0)，还有(-1)\alpha=-\alpha。并且加法的结合律与交换律性质使得向量在相加时不需要考虑次序。
最后就可以定义linear combination的概念。

Exercises

1. If F is a field, verify that F^n (as defined in Example 1) is a vector space over the field F.

Solution: Since F is a field, any x_i,y_i,z_i\in F satisfies commutativity and associativity, thus

\alpha +\beta =(x_1+y_1,\dots,x_n+y_n )=(y_1+x_1,\dots,y_n+x_n )=\beta +\alpha

and similarly \alpha +(\beta +\gamma )=(\alpha +\beta )+\gamma .
The zero vector in F^n is (0,\dots,0), it’s easy to see \alpha +0=\alpha .
For any \alpha \in F^n, the vector -\alpha =(-x_1,\dots,-x_n ).
The properties 4(a)-4(d) can also be easily verified. Thus F^n is a vector space over F.

2. If V is a vector space over the field F, verify that

(\alpha_1+\alpha_2)+(\alpha_3+\alpha_4)=[\alpha_2+(\alpha_3+\alpha_1)]+\alpha_4

for all vectors \alpha_1,\alpha_2,\alpha_3 and \alpha_4 in V.

Solution: By 3(b), 3(a), 3(b) and 3(a) we have

\begin{aligned}(\alpha _1+\alpha _2 )+(\alpha _3+\alpha _4 )&=\big((\alpha _1+\alpha _2 )+\alpha _3 \big)+\alpha _4=\big((\alpha _2+\alpha _1 )+\alpha _3 \big)+\alpha _4 \\&=\big(\alpha _2+(\alpha _1+\alpha _3 )\big)+\alpha _4=\big(\alpha _2+(\alpha _3+\alpha _1 )\big)+\alpha _4\end{aligned}

3. If C is the field of complex numbers, which vectors in C^3 are linear combinations of (1,0,-1),(0,1,1), and (1,1,1)?

Solution: Since

\begin{bmatrix}1&0&-1\\0&1&1\\1&1&1\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-1\\0&1&1\\0&1&2\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-1\\0&1&1\\0&0&1\end{bmatrix}\rightarrow\begin{bmatrix}1&&\\&1&\\&&1\end{bmatrix}

we can see that all vectors in R^3 are linear combinations of the three vectors. In particular, (1,0,0),(0,1,0),(0,0,1) are linear combinations of the three vectors, thus (i,0,0),(0,i,0),(0,0,i) are linear combinations of the three vectors, and it follows that all vectors in C^3 are linear combinations of the three vectors.

4. Lef V be the set of all pairs (x,y) of real numbers, and let F be the field of real numbers. Define

(x,y)+(x_1,y_1)=(x+x_1,y+y_1)\ c(x,y)=(cx,y)

Is V, with these operations, a vector space over the field of real numbers?

Solution: Properties 3(a)-3(d) is easy to verify, since the addition definition coincide with that of F^n, to verify 4(a)-4(d), let \alpha =(x_1,y_1 ),\beta =(x_2,y_2 ), then

1\alpha =1(x_1,y_1 )=(1x_1,y_1 )=(x_1,y_1 )=\alpha

(c_1 c_2 )\alpha =(c_1 c_2 x_1,y_1 )=c_1 (c_2 x_1,y_1 )=c_1 (c_2 \alpha)

c(\alpha +\beta )=c(x_1+x_2,y_1+y_2 )=(cx_1+cx_2,y_1+y_2 )=(cx_1,y_1 )+cx_2,y_2)=c\alpha +c\beta

(c_1+c_2 )\alpha =((c_1+c_2 ) x_1,y_1 )=(c_1 x_1,y_1 )+(c_2 x_1,y_1 )=c_1 \alpha +c_2 \alpha

5. On R^n, define two operations

\alpha\oplus\beta=\alpha-\beta \\c \cdot\alpha=-c\alpha

The operations on the right are the usual ones. Which of the axioms for a vector spaces are satisfied by (R^n,\oplus,\cdot)?

Solution: Axiom 3(c) is satisfied, define 0 to be the ordinary zero vector, then \alpha \oplus0=\alpha -0=\alpha ,\forall\alpha \in R^n.
Axiom 3(d) is satisfied, for \forall \alpha \in R^n, the unique vector -\alpha =\alpha , since \alpha \oplus(-\alpha )=\alpha -\alpha =0.
Axiom 4(a) is satisfied, since 1\alpha =-\alpha =\alpha by Axiom 3(d).
Axiom 4( c ) is satisfied, since c\cdot (\alpha \oplus \beta )=c\cdot(\alpha -\beta )=-c\alpha +c\beta =c\cdot \alpha -c\cdot \beta =c\cdot \alpha \oplus c\cdot \beta .
The rest axioms are not satisfied.

6. Let V be the set of all complex-valued functions f on the real line such that (for all t in R)

f(-t)=\overline{f(t)}

The bar denotes complex conjugation. Show that V, with the operations

(f+g)(t)=f(t)+g(t)\\(cf)(t)=cf(t)

is a vector space over the field of real numbers. Give an example of a function in V which is not real-valued.

Solution: We first verify using the definition of addition and scalar multiplication, the results are still in V.
Let f,g\in V, then

(f+g)(-t)=f(-t)+g(-t)=\overline{f(t)} +\overline{g(t)}=\overline{f(t)+g(t)}=\overline{(f+g)(t)} \\(cf)(-t)=cf(-t)=c\overline{f(t)}=\overline{cf(t)}=\overline{(cf)(t)},\quad c\in R

Next, the axioms 3(a),3(b) are satisfied since C is also a field. The zero function is in V since

0(-t)=0=\overline{0}=\overline{0(t)}

so axiom 3( c ) is satisfied. Define -f by (-f)(t)=-f(t) for f\in V, then

(-f)(-t)=-\big(f(-t)\big)=-\overline{f(t)}=\overline{-f(t)}

thus -f\in V, and f+(-f)=0, so axiom 3(d) is satisfied.
It’s obvious that axiom 4(a)-4(d) is true.

7. Let V be the se of pairs (x,y) of real numbers and let F be the field of real numebrs. Define

(x,y)+(x_1,y_1)=(x+x_1,0)\ c(x,y)=(cx,0)

Is V, with these operations, a vector space?

Solution: No.
For axiom 3(c), there’s no zero vector such that (x,1)+0=(x,1).
For axiom 4(a), if y\neq 0, we have 1(x,y)=(1x,0)=(x,0)\neq (x,y).

Linear Algebra (2ed) Hoffman&Kunze 1.6

christangdt留下评论

可逆矩阵的引入和介绍就很有意思，如果P是一系列初等矩阵的乘积，那么B=PA和A是row-equivalent的，所以A和B也是row-equivalent的，故有一个一系列初等矩阵的乘积Q使得A=QB，如果A=I_m，那么就会有B=P,I_m=QB=QP，所以通过P得到了一个Q，因而有inverse matrix的概念，inverse可以有左有右，不必同时存在，但如果同时存在，必相等。Theorem 10说明：矩阵的逆矩阵也可逆且逆逆得自身，两个（可推广到有限个）逆矩阵乘积也可逆。Theorem 11则证明初等矩阵是可逆的，其逆矩阵就是单位矩阵在该初等矩阵对应初等行变换的逆变换下的矩阵，计算也很简单。
Theorem 12得出了重要的三个等价关系：可逆、与单位矩阵row-equivalent、是一系列初等矩阵的乘积，其两个重要推论，一是把A变成单位阵的行变换可以同时把I变成A^{-1}，二是两矩阵row-equivalent当且仅当一个矩阵等于某可逆阵左乘另一个矩阵（因为可逆阵就是一系列初等矩阵的乘积）。Theorem 13是另一组重要等价关系：A可逆、AX=0只有零解、AX=Y对每一个Y都有解。这两个定理说明了可逆阵的威力。推论一是如果是方阵，那么光有左逆或右逆就代表可逆。推论二是方阵的乘积可逆代表每个方阵均可逆，即Theorem 10在方阵约束下增强了。
这一节最后要说明的两个事，一个是如何用初等行变换或者初等矩阵乘积求逆，另一个是行的事情都可以复制到列上，只是初等矩阵会发生变化，且从左乘变成右乘。

Exercises

1. Let

A=\begin{bmatrix}1&2&1&0\\-1&0&3&5\\1&-2&1&1\end{bmatrix}

Find a row-reduced echelon matrix R which is row-equivalent to A and an invertible 3\times 3 matrix P such that R=PA.

Solution: We perform row operations on A'=\begin{bmatrix}A&Y\end{bmatrix} with Y=[y_1,y_2,y_3 ]^T, thus

\begin{aligned}A'&=\begin{bmatrix}1&2&1&0&y_1\\-1&0&3&5&y_2\\1&-2&1&1&y_3 \end{bmatrix}\rightarrow\begin{bmatrix}1&2&1&0&y_1\\0&2&4&5&y_2+y_1\\0&-4&0&1&y_3-y_1 \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&2&1&0&y_1\\0&2&4&5&y_2+y_1\\0&0&8&11&2y_2+y_3+y_1 \end{bmatrix}\rightarrow\begin{bmatrix}1&2&1&0&y_1\\0&1&2&\frac{5}{2}&\frac{1}{2} (y_2+y_1 )\\0&0&1&\frac{11}{8}&\frac{1}{8} (2y_2+y_3+y_1 ) \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&2&0&-\frac{11}{8}&\frac{7}{8} y_1-\frac{1}{4} y_2-\frac{1}{8} y_3\\0&1&0&-\frac{1}{4}&\frac{1}{4} (y_1-y_3 )\\0&0&1&\frac{11}{8}&\frac{1}{8} (2y_2+y_3+y_1 ) \end{bmatrix}\rightarrow\begin{bmatrix}1&0&0&-\frac{7}{8}&\frac{3}{8} y_1-\frac{1}{4} y_2+\frac{3}{8} y_3\\0&1&0&-\frac{1}{4}&\frac{1}{4} (y_1-y_3 )\\0&0&1&\frac{11}{8}&\frac{1}{8}(2y_2+y_3+y_1 )\end{bmatrix}\end{aligned}

thus R=\begin{bmatrix}1&0&0&-\frac{7}{8}\\0&1&0&-\frac{1}{4}\\0&0&1&\frac{11}{8}\end{bmatrix}, and the matrix P s.t. R=PA is P=\begin{bmatrix}3/8&-1/4&3/8\\1/4&0&-1/4\\1/8&1/4&1/8\end{bmatrix}.

2. Do Exercise 1, but with

A=\begin{bmatrix}2&0&i\\1&-3&-i\\i&1&1\end{bmatrix}

Solution: We perform row operations on A'=\begin{bmatrix}A&Y\end{bmatrix} with Y=[y_1,y_2,y_3 ]^T, thus

\begin{aligned}A'&=\begin{bmatrix}2&0&i&y_1\\1&-3&-i&y_2\\i&1&1&y_3 \end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{i}{2}&\frac{y_1}{2}\\0&-3&-\frac{3i}{2}&y_2-\frac{y_1}{2}\\0&1&\frac{3}{2}&y_3-\frac{i}{2} y_1 \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&\frac{i}{2}&\frac{y_1}{2}\\0&1&\frac{3}{2}&y_3-\frac{i}{2} y_1\\0&0&\frac{9-3i}{2}&3y_3-\frac{3i+1}{2} y_1+y_2 \end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{i}{2}&\frac{y_1}{2}\\0&1&\frac{3}{2}&y_3-\frac{i}{2} y_1\\0&0&1&\frac{3+i}{15} \left(3y_3-\frac{3i+1}{2} y_1+y_2 \right)\end{bmatrix}\end{aligned}

It’s easy to see R=I and to calculate P, we see A'\rightarrow\begin{bmatrix}I&Y'\end{bmatrix}, in which

\begin{aligned}Y'&=\begin{bmatrix}\dfrac{y_1}{2}-\dfrac{2(3+i)}{15i} \left(3y_3-\dfrac{3i+1}{2} y_1+y_2 \right)\\y_3-\dfrac{i}{2} y_1-\dfrac{2(3+i)}{45} \left(3y_3-\dfrac{3i+1}{2} y_1+y_2 \right)\\ \dfrac{3+i}{15} \left(3y_3-\dfrac{3i+1}{2} y_1+y_2\right)\end{bmatrix}\\&=\begin{bmatrix}\dfrac{1}{2}+\dfrac{2(3+i)(3i+1)}{30i}&-\dfrac{2(3+i)}{15i}&-\dfrac{2(3+i)}{5i}\\-\dfrac{i}{2}+\dfrac{(3+i)(3i+1)}{45}&-\dfrac{2(3+i)}{45}&1-\dfrac{2(3+i)}{15}\\-\dfrac{(3+i)(3i+1)}{30}&\dfrac{3+i}{15}&\dfrac{3+i}{5}\end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}\end{aligned}

thus P is the matrix on the left.

3. For each of the two matrices

\begin{bmatrix}2&5&-1\\4&-1&2\\6&4&1\end{bmatrix},\quad\begin{bmatrix}1&-1&2\\3&2&4\\0&1&-2\end{bmatrix}

use elementary row operations to discover whether it is invertible, and to find the inverse in case it is.

Solution: As

\begin{aligned}\begin{bmatrix}2&5&-1&1&0&0\\4&-1&2&0&1&0\\6&4&1&0&0&1\end{bmatrix}&\rightarrow\begin{bmatrix}2&5&-1&1&0&0\\0&-11&4&-2&1&0\\0&-11&4&-3&0&1\end{bmatrix}\\&\rightarrow\begin{bmatrix}2&5&-1&1&0&0\\0&-11&4&-2&1&0\\0&0&0&-1&-1&1\end{bmatrix}\end{aligned}

the first matrix is not invertible.
As

\begin{aligned}&\begin{bmatrix}1&-1&2&1&0&0\\3&2&4&0&1&0\\0&1&-2&0&0&1\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&2&1&0&0\\0&5&-2&-3&1&0\\0&1&-2&0&0&1\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&-1&2&1&0&0\\0&1&-2&0&0&1\\0&0&8&-3&1&-5\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&0&7/4&-1/4&5/4\\0&1&0&-3/4&1/4&-1/4\\0&0&1&-3/8&1/8&-5/8\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&0&1&0&1\\0&1&0&-3/4&1/4&-1/4\\0&0&1&-3/8&1/8&-5/8\end{bmatrix}\end{aligned}

the second matrix is invertible and its inverse is \begin{bmatrix}1&0&1\\-3/4&1/4&-1/4\\-3/8&1/8&-5/8\end{bmatrix}.

4. Let

A=\begin{bmatrix}5&0&0\\1&5&0\\0&1&5\end{bmatrix}

For which X does there exist a scalar c such that AX=cX?

Solution: If X=0, then AX=cX for any scalar c. If X\neq 0, then the system (A-cI)X=0 has non trivial solutions, if c\neq 5 we have

\begin{bmatrix}5-c&0&0\\1&5-c&0\\0&1&5-c \end{bmatrix} \rightarrow \begin{bmatrix}1&0&0\\0&5-c&0\\0&1&5-c \end{bmatrix}\rightarrow \begin{bmatrix}1&&\\&1&\\&&1 \end{bmatrix}

thus the system (A-cI)X=0 has only trivial solutions, a contradiction. If c=5, then as long as X=(0,0,k),k\neq 0, we have AX=5X. In conclusion we can say for X=(0,0,k),k\in R does there exists a scalar c s.t. AX=cX.

5. Discover whether

A=\begin{bmatrix}1&2&3&4\\0&2&3&4\\0&0&3&4\\0&0&0&4\end{bmatrix}

is invertible, and find A^{-1} if it exists.

Solution: A is invertible. we have

\begin{aligned}\begin{bmatrix}1&2&3&4&1&0&0&0\\0&2&3&4&0&1&0&0\\0&0&3&4&0&0&1&0\\0&0&0&4&0&0&0&1\end{bmatrix}&\rightarrow \begin{bmatrix}1&2&3&0&1&0&0&-1\\0&2&3&0&0&1&0&-1\\0&0&3&0&0&0&1&-1\\0&0&0&4&0&0&0&1\end{bmatrix}\\&\rightarrow \begin{bmatrix}1&2&0&0&1&0&-1&-1\\0&2&0&0&0&1&-1&-1\\0&0&3&0&0&0&1&-1\\0&0&0&4&0&0&0&1\end{bmatrix}\\&\rightarrow \begin{bmatrix}1&0&0&0&1&-1&-1&-1\\0&2&0&0&0&1&-1&-1\\0&0&3&0&0&0&1&-1\\0&0&0&4&0&0&0&1\end{bmatrix}\\&\rightarrow \begin{bmatrix}1&0&0&0&1&-1&-1&-1\\0&1&0&0&0&1/2&-1/2&-1/2\\0&0&1&0&0&0&1/3&-1/3\\0&0&0&1&0&0&0&1/4\end{bmatrix}\end{aligned}

thus A^{-1}=\begin{bmatrix}1&-1&-1&-1\\0&1/2&-1/2&-1/2\\0&0&1/3&-1/3\\0&0&0&1/4\end{bmatrix}.

6. Suppose A is a 2\times 1 matrix and that B is a 1\times 2 matrix. Prove that C=AB is not invertible.

Solution: Suppose A=\begin{bmatrix}a\\b\end{bmatrix},B=\begin{bmatrix}c&d\end{bmatrix}, then C=AB=\begin{bmatrix}ac&ad\\bc&bd\end{bmatrix}, use elementary row operation one can eliminate one of C’s rows if a\neq 0, and the first row of C vanishes if a=0, thus C can’t be row equivalent to the identity matrix, and C is not invertible.

7. Let A be an n\times n (square) matrix. Prove the following two statements:
(a) If A is invertible and AB=0 for some n\times n matrix B, then B=0.
(b) If A is not invertible, then there exists an n\times n matrix B such that AB=0 but B\neq 0.

Solution: (a) We have PA=I for some P, thus B=IB=PAB=P0=0.
(b) The system AX=0 has non trivial solutions, let x\neq 0 be a solution and let B=\begin{bmatrix}x&0&\dots&0\end{bmatrix}, then AB=0.

8. Let

A=\begin{bmatrix}a&b\\c&d\end{bmatrix}

Prove, using elementary row operations, that A is invertible if and only if (ad-bc)\neq 0.

Solution: If c=0, then to make A invertible if and only if (a\neq 0)\wedge (d\neq 0), which is equivalent to ad=ad-bc\neq 0.
If c\neq 0, then use elementary row operations, we have

A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\rightarrow\begin{bmatrix}ac&bc\\ac&ad\end{bmatrix}\rightarrow\begin{bmatrix}ac&bc\\0&ad-bc\end{bmatrix}\rightarrow\begin{bmatrix}a&b\\0&ad-bc\end{bmatrix},\quad a\neq 0 \\ A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\rightarrow\begin{bmatrix}c&d\\a&b\end{bmatrix}=\begin{bmatrix}c&d\\0&b\end{bmatrix},\quad a=0

thus if A is invertible, we shall have ad-bc\neq 0. Conversely if ad-bc\neq 0, then in the case a\neq 0 we directly have A is invertible since \begin{bmatrix}a&b\\0&ad-bc\end{bmatrix} is row equivalent to I_2, in the case a=0, notice then bc\neq 0 and thus b\neq 0, so \begin{bmatrix}c&d\\0&b\end{bmatrix} is row equivalent to I_2, in either case, A is invertible.

9. An n\times n matrix A is called upper-triangular if A_{ij}=0,i>j, that is, if every entry below the main diagonal is 0. Prove that an upper-triangular (square) matrix is invertible if and only if every entry on its main diagonal is different from 0.

Solution: If every entry on its main diagonal is not 0, then we use row n to eliminate the last column to e_n, and next use row n-1 to eliminate column n-1 to e_{n-1}, continuing we make A to I_n, thus A is invertible.
Conversely, suppose A is invertible and assume some entry on its main diagonal is 0, let k be the largest integer s.t. A_{kk}=0, then if k=n then row n is all 0, if not then repeating the same steps in the previous paragragh n-k times, we can get row k to be all 0, then A is not invertible, as the matrix can’t be row-equivalent to I_n.

10. Prove the following generalization of Exercise 6. If A is an m\times n matrix, B is an n\times m matrix and n<m, then AB is not invertible.

Solution: Since n<m, if row-reduce A to an row-deduced echelon form, the last row must be all 0, i.e. there’s an m\times m invertible matrix P s.t. PA=C, where the last row of C is all 0, now assume AB is invertible, as it’s square we can have D be it’s right inverse, so ABD=I, thus PABD=CBD=P, now the last row of C is 0, means the last row of CBD ,or P, is 0, this contradicts P being invertible.

11. Let A be an m\times n matrix. Show that by means of a finite number of elementary row and/or column operations one can pass from A to a matrix R which is both ‘row-reduced echelon’ and ‘column-reduced echelon’, i.e. R_{ij}=0 if i\neq j, R_{ii}=1,1\leq i\leq r, R_{ii}=0 if i>r. Show that R=PAQ, where P is an invertible m\times m matrix and Q is an invertible n\times n matrix.

Solution: We’ve proved use elementary row operations we can get a row-reduced echelon matrix R' such that R'=PA, where P is an invertible m\times m matrix. Starting from R' we can use elementary column operations to get a column-reduced echelon matrix R, and an invertible n\times n matrix Q s.t. R=R'Q, thus R=PAQ.

12. The result of Example 16 suggests that perhaps the matrix

\begin{bmatrix}1&\dfrac{1}{2}&\cdots&\dfrac{1}{n}\\ \dfrac{1}{2}&\dfrac{1}{3}&\cdots&\dfrac{1}{n+1}\\ \vdots&\vdots&&\vdots\\ \dfrac{1}{n}&\dfrac{1}{n+1}&\cdots&\dfrac{1}{2n-1}\end{bmatrix}

is invertible and A^{-1} has integer entries. Can you prove that?

Solution: This matrix is called the Hilbert matrix, it’s invertible and its inverse is given by B=(B_{ij}), where

B_ij=(-1)^{i+j} (i+j+1)\dbinom{i+j}{i}\dbinom{i+j}{j}\dbinom{n+i+1}{n-j}\dbinom{n+j+1}{n-i}

One can direct compute the product AB and verify that AB=I.

Linear Algebra (2ed) Hoffman&Kunze 1.5

christangdt2条评论

矩阵可以理解为线性变换在一组基下的表示，故矩阵乘法就是两个线性变换结合后的表示。Hoffman把矩阵更一般地作为函数，故其给矩阵乘法的引入是：矩阵乘法是对矩阵的行进行线性组合时产生的东西。在定义和后面的例子（Example 10）中，都体现了矩阵乘法的这一个思想：A某一行的每一个数与B的行向量进行线性组合，得到AB的这一行的结果。特别提到了矩阵乘法是not commutative的。
列矩阵（column matrix）引致一个有用的notation（可以从定义证得）：如果有矩阵B_{n\times p}，其中列向量为 B_1,\dots,B_p，则 B=\begin{bmatrix}B_1,\dots,B_p\end{bmatrix}并且 AB=\begin{bmatrix}AB_1,\dots,AB_p\end{bmatrix}。
Theorem 8 给出了矩阵乘法满足结合律的证明。其后说明了矩阵的幂是什么，以及 A(BC)=(AB)C说明linear combinations of linear combinations of the rows of C are again linear combinations of the rows of C，我认为这个诠释很有意思，Hoffman时刻在提示运算背后的思想。
最后就是要引入elementary matrix，与1.3节介绍的elementary row operations密切相关，elementary matrix的定义是从 I进行一次elementary row operations得到的矩阵，故其必是和 I同阶的方阵。Theorem 9 说明，如果用 e表示一个elementary row operation，则 e(A)=e(I)A，即行变换等于左乘对应初等矩阵。其推论显示： A,B两矩阵row equivalent的充要条件是 B=PA，其中 P是一堆初等矩阵乘积。

Exercises

1. Let

A=\begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix},\quad B=\begin{bmatrix}3\\1\\-1\end{bmatrix},\quad C=\begin{bmatrix}1&-1\end{bmatrix}

Compute ABC and CAB.

Solution:

ABC=\begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix} \begin{bmatrix}3\\1\\-1\end{bmatrix} \begin{bmatrix}1&-1\end{bmatrix} =\begin{bmatrix}4\\4\end{bmatrix} \begin{bmatrix}1&-1\end{bmatrix}=\begin{bmatrix}4&-4\\4&-4\end{bmatrix}\\CAB=\begin{bmatrix}1&-1\end{bmatrix}\begin{bmatrix}2&-1&1\\1&2&1\end{bmatrix}\begin{bmatrix}3\\1\\-1\end{bmatrix}=\begin{bmatrix}1&-3&0\end{bmatrix}\begin{bmatrix}3\\1\\-1\end{bmatrix}=0

2. Let

A=\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix},\quad B=\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}

Verify directly that A(AB)=A^2B

Solution:

\begin{aligned}A(AB)&=\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\left(\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}\right)\\&=\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\begin{bmatrix}5&-1\\8&0\\10&-2\end{bmatrix}=\begin{bmatrix}7&-3\\20&-4\\25&-5\end{bmatrix} \end{aligned}\\ \begin{aligned}A^2B&=\left(\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\right)\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}\\&=\begin{bmatrix}2&-1&1\\5&-2&3\\6&-3&4\end{bmatrix}\begin{bmatrix}2&-2\\1&3\\4&4\end{bmatrix}=\begin{bmatrix}7&-3\\20&-4\\25&-5\end{bmatrix}\end{aligned}

thus A(AB)=A^2 B

3. Find two different 2\times 2 matrices A such that A^2=0 but A\neq 0.

Solution: A_1=\begin{bmatrix}0&1\\0&0\end{bmatrix}\quad,A_2=\begin{bmatrix}0&0\\1&0\end{bmatrix}.

4. For the matrix A of Exercise 2, find elementary matrices E_1,E_2,\dots,E_k such that

E_k\dots E_2E_1A=I

Solution:

\begin{aligned}\begin{bmatrix}1&-1&1\\2&0&1\\3&0&1\end{bmatrix}\xrightarrow{\text{add} -2\times \text{(1) to (2), add }-3\times \text{ (1) to (3)}}&\begin{bmatrix}1&-1&1\\0&2&-1\\0&3&-2\end{bmatrix}\\ \xrightarrow{(2)\times(1/2)} &\begin{bmatrix}1&-1&1\\0&1&-1/2\\0&3&-2\end{bmatrix}\\ \xrightarrow{\text{add } -3\times(2) \text{ to }(3)}&\begin{bmatrix}1&-1&1\\0&1&-1/2\\0&0&-1/2\end{bmatrix}\\ \xrightarrow{(3)\times-2} &\begin{bmatrix}1&-1&1\\0&1&-1/2\\0&0&1\end{bmatrix}\\ \xrightarrow{\text{add } 1/2\times(3) \text{ to }(2),\text{ add } -1\times(3) \text{ to }(1)}&\begin{bmatrix}1&-1&0\\0&1&0\\0&0&1\end{bmatrix}\\ \xrightarrow{\text{add } (2) \text{ to }(1)}&\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\end{aligned}

so the corresponding elementary matrices are

E_1=\begin{bmatrix}1&0&0\\-2&1&0\\0&0&1\end{bmatrix},E_2=\begin{bmatrix}1&0&0\\0&1&0\\-3&0&1\end{bmatrix},E_3=\begin{bmatrix}1&0&0\\0&1/2&0\\0&0&1\end{bmatrix},E_4=\begin{bmatrix}1&0&0\\0&1&0\\0&-3&1\end{bmatrix}\\E_5=\begin{bmatrix}1&0&0\\0&1&0\\0&0&-2\end{bmatrix},E_6=\begin{bmatrix}1&0&0\\0&1&1/2\\0&0&1\end{bmatrix},E_7=\begin{bmatrix}1&0&-1\\0&1&0\\0&0&1\end{bmatrix},E_8=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}

5. Let

A=\begin{bmatrix}1&-1\\2&2\\1&0\end{bmatrix},\quad B=\begin{bmatrix}3&1\\-4&4\end{bmatrix}

Is there a matrix C such that CA=B?

Solution: One possible choice of C is C=\begin{bmatrix}1&1&0\\2&3&-12\end{bmatrix}.

6. Let A be an m\times n matrix and B and n\times k matrix. Show that the columns of C=AB are linear combinations of the columns of A. If \alpha_1,\dots,\alpha_n are the columns of A and \gamma_1,\dots,\gamma_k are the columns of C, then

\gamma_j=\sum_{r=1}^nB_{rj}\alpha_r

Solution: We write A=[\alpha_1,\dots,\alpha_n ], and C=[\gamma_1,\dots,\gamma_k ], then as C=AB, for 1\leq i\leq m, we have

C_{ij}=\sum\limits_{r=1}^nA_{ir} B_{rj}=\sum\limits_{r=1}^nB_{rj} A_{ir},\quad 1\leq i\leq m

thus

\gamma_j=\begin{bmatrix}C_{1j}\\ \vdots\\C_{mj}\end{bmatrix}=\begin{bmatrix}\sum_{r=1}^nB_{rj} A_{1r}\\ \vdots\\ \sum_{r=1}^nB_{rj} A_{mr}\end{bmatrix}=B_1j \begin{bmatrix}A_{11}\\ \vdots\\A_{m1}\end{bmatrix}+\cdots+B_nj \begin{bmatrix}A_{1n}\\ \vdots\\A_{mn}\end{bmatrix}= \sum_{r=1}^nB_{rj}\alpha_r

7. Let A and B be 2\times 2 matrices such that AB=I. Prove that BA=I.

Solution: We have ABA=IA=A, thus A(BA-I)=0, assume BA\neq I, then the system AX=0 has non-trivial solutions, thus A is row-equivalent to a row-reduced echelon matrix which is not I, so there’s P which is product of elementary matrices such that

PA=\begin{bmatrix}1&0\\0&0\end{bmatrix}

this means the second row of P=PAB is 0, a contradiction.

8. Let

C=\begin{bmatrix}C_{11}&C_{12}\\C_{21}&C_{22}\end{bmatrix}

be a 2\times 2 matrix. We inquire when it is possible to find 2\times 2 matrices A and B such that C=AB-BA. Prove that such matrices can be found if and only if C_{11}+C_{22}=0.

Solution: Let A=\begin{bmatrix}a&b\\c&d\end{bmatrix},B=\begin{bmatrix}e&f\\g&h\end{bmatrix}, then AB=\begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix},BA=\begin{bmatrix}ea+fc&eb+df\\ag+hc&gb+dh\end{bmatrix}, thus

AB-BA=\begin{bmatrix}bg-cf&af+bh-eb-df\\ce+dg-ag-hc&cf-bg\end{bmatrix}

if C=AB-BA, then C_{11}+C_{22}=bg-cf+cf-bg=0.
Conversely, if C_{11}+C_{22}=0, then C_{11}=-C_{22}, thus solve the equation

\begin{aligned}bg-cf&=C_{11}\\ af+bh-eb-df=(a-d)f+b(h-e)&=C_{12}\\ ce+dg-ag-hc=(d-a)g+c(e-h)&=C_{21}\end{aligned}

It’s a system of 3 equations with 8 unknowns, fix a=1,d=0,h=0,e=1, the system becomes

\begin{aligned}bg-cf&=C_{11}\\ f-b&=C_{12}\\ g-c&=C_{21}\end{aligned}

thus b(C_{21}+c)-c(C_{12}+b)=C_{11}, or bC_{21}=C_{11}+cC_{12}, if C_21\neq 0, let c=1; if C_{21}=0, let c=-C_{11}/C_{12} and b=1, all other unkowns can be solved.