Linear Algebra (2ed) Hoffman & Kunze 4.3

christangdt留下评论

这一节其实是两部分内容：Lagrange interpolation formula和linear algebra的isomorphism。Lagrange interpolation 公式提供了一种用n+1个点的值去固定所有不高于n阶多项式的方法，其在标准基下的矩阵是Vandermonde matrix，这个矩阵的行列式相信折磨过不少刚学高等代数的学子。
第二部分先把polynomial（本质是一个很长的vector）和polynomial function（从t到f(t)）联系起来，并定义乘法(f^{\sim}g^{\sim})(t)=f^{\sim}(t)g^{\sim}(t)，通过上一节的Theorem 2可知，polynomial function的空间也是一个linear algebra with identity over F. 在给出两个algebra如何才是isomorphic的定义后，可以证明本节最重要的结论（即Theorem 3）：f\to f^{\sim}是一个algebra的isomorphism。当然，EXAMPLE 4也很精彩，给予线性算子的多项式和其矩阵以一种新的看法。

Exercises

1.Use the Lagrange interpolation formula to find a polynomial f with real coefficients such that f has degree \leq 3 and f(-1)=-6,f(0)=2,f(1)=-2,f(2)=6.

Solution: We have t_0=-1,t_1=0,t_2=1,t_3=2, and

P_0=\dfrac{x(x-1)(x-2)}{(-1-0)(-1-1)(-1-2)}=-\dfrac{1}{6}(x^3-3x^2+2x) \\ P_1=\dfrac{(x+1)(x-1)(x-2)}{(0+1)(0-1)(0-2)}=\dfrac{1}{2}(x^3-2x^2-x+2) \\ P_2=\dfrac{(x+1)x(x-2)}{(1+1)(1-0)(1-2)}=-\dfrac{1}{2}(x^3-x^2-2x)\\ P_3=\dfrac{(x+1)x(x-1)}{(2+1)(2-0)(2-1)}=\dfrac{1}{6}(x^3-x)

Using Lagrange interpolation formula we have

\displaystyle{\begin{aligned}f&=f(-1)P_0+f(0)P_1+f(1)P_2+f(2)P_3\\&=-6P_0+2P_1-2P_2+6P_3\\&=(x^3-3x^2+2x)+(x^3-2x^2-x+2)+(x^3-x^2-2x)+(x^3-x)\\&=4x^3-6x^2-2x+2\end{aligned}}

2.Let \alpha,\beta,\gamma,\delta be real numbers. We ask when it is possible to fine a polynomial f over R, of degree not more than 2, such that f(-1)=\alpha,f(1)=\beta,f(3)=\gamma and f(0)=\delta. Prove that this is possible if and only if

\displaystyle{3\alpha+6\beta-\gamma-8\delta=0.}

Solution: In the subspace of degree \leq 2, we let t_0=-1,t_1=1,t_2=3 and thus

P_0=\dfrac{(x-1)(x-3)}{(-1-1)(-1-3)}=\dfrac{1}{8}(x^2-4x+3) \\ P_1=\dfrac{(x+1)(x-3)}{(1+1)(1-3)}=-\dfrac{1}{4}(x^2-2x-3) \\ P_2=\dfrac{(x+1)(x-1)}{(3+1)(3-1)}=\dfrac{1}{8}(x^2-1)

So to satisfy f(-1)=\alpha,f(1)=\beta,f(3)=\gamma, f must be

\displaystyle{\begin{aligned}f&=\alpha P_0 +\beta P_1+\gamma P_2\\&=\frac{\alpha}{8}(x^2-4x+3)-\frac{\beta}{4}(x^2-2x-3)+\frac{\gamma}{8}(x^2-1)\end{aligned}}

As f(0)=\delta, we have

\displaystyle{f(0)=\frac{3\alpha}{8}+\frac{3\beta}{4}-\frac{\gamma}{8}=\delta \Leftrightarrow 3\alpha+6\beta-\gamma-8\delta=0}

3.Let F be the field of real numbers,

\displaystyle{A=\begin{bmatrix}2&0&0&0\\0&2&0&0\\0&0&3&0\\0&0&0&1\end{bmatrix}\qquad p=(x-2)(x-3)(x-1)}

( a ) Show that p(A)=0.
( b ) Let P_1,P_2,P_3 be the Lagrange polynomials for t_1=2,t_2=3,t_3=1. Compute E_i=P_i(A),i=1,2,3.
( c ) Show that E_1+E_2+E_3=I,E_iE_j=0 if i\neq j, E_i^2=E_i.
( d ) Show that A=2E_1+3E_2+E_3.

Solution:
( a ) A direct calculation shows

\displaystyle{\begin{aligned}p(A)&=(A-2I)(A-3I)(A-I)\\&=\begin{bmatrix}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&-1\end{bmatrix}\begin{bmatrix}-1&0&0&0\\0&-1&0&0\\0&0&0&0\\0&0&0&-4\end{bmatrix}\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&2&0\\0&0&0&0\end{bmatrix}\\&=0\end{aligned}}

( b ) We have

P_1=\dfrac{(x-3)(x-1)}{(2-3)(2-1)}=-(x^2-4x+3) \\ P_2=\dfrac{(x-2)(x-1)}{(3-2)(3-1)}=\dfrac{1}{2}(x^2-3x+2) \\ P_3=\dfrac{(x-2)(x-3)}{(1-2)(1-3)}=\dfrac{1}{2}(x^2-5x+6)

and we have A^2=\begin{bmatrix}4&&&\\&4&&\\&&9&\\&&&1\end{bmatrix}, thus

E_1=P_1(A)=-\left(\begin{bmatrix}4&&&\\&4&&\\&&9&\\&&&1\end{bmatrix}-4\begin{bmatrix}2&&&\\&2&&\\&&3&\\&&&1\end{bmatrix}+3\begin{bmatrix}1&&&\\&1&&\\&&1&\\&&&1\end{bmatrix}\right)=\begin{bmatrix}1&&&\\&1&&\\&&0&\\&&&0\end{bmatrix}

E_2=P_2(A)=\dfrac{1}{2}\left(\begin{bmatrix}4&&&\\&4&&\\&&9&\\&&&1\end{bmatrix}-3\begin{bmatrix}2&&&\\&2&&\\&&3&\\&&&1\end{bmatrix}+2\begin{bmatrix}1&&&\\&1&&\\&&1&\\&&&1\end{bmatrix}\right)=\begin{bmatrix}0&&&\\&0&&\\&&1&\\&&&0\end{bmatrix}

E_3=P_3(A)=\dfrac{1}{2}\left(\begin{bmatrix}4&&&\\&4&&\\&&9&\\&&&1\end{bmatrix}-5\begin{bmatrix}2&&&\\&2&&\\&&3&\\&&&1\end{bmatrix}+6\begin{bmatrix}1&&&\\&1&&\\&&1&\\&&&1\end{bmatrix}\right)=\begin{bmatrix}0&&&\\&0&&\\&&0&\\&&&1\end{bmatrix}

( c ) All are obvious and easy to verify.
( d ) Obvious.

4.Let p=(x-2)(x-3)(x-1) and let T be any linear operator on R^4 such that p(T)=0. Let P_1,P_2,P_3 be the Lagrange polynomials of Exercise 3, and let E_i=P_i(T),i=1,2,3. Prove that

\displaystyle{E_1+E_2+E_3=I,\qquad E_iE_j=0 \quad \text{if}\quad i\neq j, \\E_i^2=E_i,\quad \text{and}\quad T=2E_1+3E_2+E_3.}

Solution: From p(T)=0 we get T^3-6T^2+11T-6I=0, now

E_1=P_1(T)=-(T-3I)(T-I)=-T^2+4T-3I \\ E_2=P_2(T)=\dfrac{1}{2}(T-2I)(T-I)=\frac{1}{2}T^2-\dfrac{3}{2}T+I \\ E_3=P_3(T)=\dfrac{1}{2}(T-2I)(T-3I)=\dfrac{1}{2}T^2-\frac{5}{2}T+3I

A direct calculation shows E_1+E_2+E_3=I, and we have

E_1E_2=E_2E_1=-\dfrac{1}{2}(T-I)p(T)=0 \\ E_1E_3=E_3E_1=-\dfrac{1}{2}(T-3I)p(T)=0 \\ E_2E_3=E_3E_2=\dfrac{1}{4}(T-2I)p(T)=0

thus E_iE_j=0 if i\neq j. Notice p(T)=0 gives T^3=6T^2-11T+6I, so

\displaystyle{\begin{aligned}E_1^2&=T^4-8T^3+22T^2-24T+9I\\&=T(6T^2-11T+6I)-8T^3+22T^2-24T+9I\\&=-2(6T^2-11T+6I)+11T^2-18T+9I\\&=-T^2+4T-3I=E_1\end{aligned}}

Similarly we can show E_2^2=E_2 and E_3^2=E_3. Finally

\displaystyle{\begin{aligned}&\quad2E_1+3E_2+E_3\\&=2(-T^2+4T-3I)+\frac{3}{2}T^2-\frac{9}{2}T+3I+\frac{1}{2}T^2-\frac{5}{2}T+3I\\&=T\end{aligned}}

5.Let n be a positive integer and F a field. Suppose A is an n\times n matrix over F and P is an invertible n\times n matrix over F. If f is any polynomial over F, prove that f(P^{-1}AP)=P^{-1}f(A)P.

Solution: Let f=\sum_{i=0}^nf_ix^i, and thus f(A)=\sum_{i=0}^nf_iA^i. Notice that for any k\geq 0,

\displaystyle{(P^{-1}AP)^k=(P^{-1}AP)\cdots (P^{-1}AP)=P^{-1}A(PP^{-1})\cdots (PP^{-1})AP=P^{-1}A^kP}

thus

\displaystyle{P^{-1}f(A)P=P^{-1}\left(\sum_{i=0}^nf_iA^i\right)P=\sum_{i=0}^nf_i(P^{-1}A^iP)=\sum_{i=0}^nf_i(P^{-1}AP)^i=f(P^{-1}AP)}

6.Let F be a field. We have considered certain special linear functions on F[x] obtained via ‘evaluation at t‘: L(f)=f(t). Such functionals are not only linear but also have the property that L(fg)=L(f)L(g). Prove that if L is any linear functional on F[x] such that L(fg)=L(f)L(g) for all f and g, then either L=0 or there is a t in F such that L(f)=f(t) for all f.

Solution: If L=0 then the conclusion is obvious, now suppose L\neq 0, then \exists g\in F[x], such that L(g)\neq0. For the scalar polynomial cx^0 we have L(cx^0g)=L(cg)=L(c)L(g)=cL(g), thus c=L(c) for all c\in F. Next if we denote t:=L(x), then L(x^2)=L(xx)=[L(x)]^2=t^2 and by induction we can see L(x^n)=t^n for all n\geq 1. Thus if f=\sum_{i=0}^nf_ix^i, we shall have

\displaystyle{L(f)=L\left(\sum_{i=0}^nf_ix^i\right)=\sum_{i=0}^nf_iL(x^i)=f_0+\sum_{i=1}^nf_it^i=\sum_{i=0}^nf_it^i=f(t)}

Linear Algebra (2ed) Hoffman & Kunze 4.2

christangdt留下评论

这一章讲多项式是我读过的线性代数教材中讲多项式最为拔高的，在定义了什么是一个域上的linear algebra（实质是vector space加上向量乘法，不是数乘）之后，定义F[x]为无限维向量中所有有限维（即从某一分量之后全为0）的向量张成的子空间，其中的元素称为polynomial。这是一种全新的定义方式，而多项式中的x实际上是向量(0,1,0,\dots)，在这种定义下，一方面既往意义上多项式的很多性质是保留的（Theorem 1），另一方面又能将多项式和Linear algebra的本质更好的联系（Corollary 1），Corollary 2得到了一个类似于消去律（cancellation law）的结论。如此定义多项式后，对多项式在一个linear algebra with identity over F上的元素进行类似于“取值”操作也更加不违背常理，Theorem 2说明这种取值定义满足线性性质和乘积不变性质。

Exercises

1.Let F be a subfield of the complex numbers and let A=\begin{bmatrix}2&1\\-1&3\end{bmatrix} be a 2\times 2 matrix over F. For each of the following polynomials f over F, compute f(A).
( a ) f=x^2-x+2;
( b ) f=x^3-1;
( c ) f=x^2-5x+7.

Solution:
( a ) A^2=\begin{bmatrix}2&1\\-1&3\end{bmatrix}\begin{bmatrix}2&1\\-1&3\end{bmatrix}=\begin{bmatrix}3&5\\-5&8\end{bmatrix}, so

\displaystyle{f(A)=\begin{bmatrix}3&5\\-5&8\end{bmatrix}-\begin{bmatrix}2&1\\-1&3\end{bmatrix}+2\begin{bmatrix}1&\\&1\end{bmatrix}=\begin{bmatrix}3&4\\-4&7\end{bmatrix}}

( b ) A^3=\begin{bmatrix}3&5\\-5&8\end{bmatrix}\begin{bmatrix}2&1\\-1&3\end{bmatrix}=\begin{bmatrix}1&18\\-18&19\end{bmatrix}, so

\displaystyle{f(A)=\begin{bmatrix}1&18\\-18&19\end{bmatrix}-\begin{bmatrix}1&\\&1\end{bmatrix}=\begin{bmatrix}0&18\\-18&18\end{bmatrix}}

( c ) f(A)=\begin{bmatrix}3&5\\-5&8\end{bmatrix}-5\begin{bmatrix}2&1\\-1&3\end{bmatrix}+7\begin{bmatrix}1&\\&1\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}

2.Let T be the linear operator on R^3 defined by T(x_1,x_2,x_3)=(x_1,x_3,-2x_2-x_3). Let f be the polynomial over R defined by f=-x^3+2. Find f(T).

Solution: Let A be the matrix of T with the standard basis of R^3, then

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

It is easy to compute

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

thus -A^3+2I=A, which means f(T)=-T^3+2I=T.

3.Let A be an n\times n diagonal matrix over the field F, i.e., a matrix satisfying A_{ij}=0 for i\neq j. Let f be the polynomial over F defined by f=(x-A_{11})\cdots(x-A_{nn}). What is the matrix f(A)?

Solution: If A=(A_{ij}) and B=(B_{ij}) are two n\times n diagonal matrices, then use matrix multiplication we have

\displaystyle{AB=\begin{bmatrix}A_{11}&&\\&\ddots&\\&&A_{nn} \end{bmatrix}\begin{bmatrix}B_{11}&&\\&\ddots&\\&&B_{nn} \end{bmatrix}=\begin{bmatrix}A_{11}B_{11}&&\\&\ddots&\\&&A_{nn}B_{nn} \end{bmatrix}}

As f(A)=(A-A_{11}I)\cdots(A-A_{nn}I) is a product of n diagonal matrix, we have

\displaystyle{f(A)=\begin{bmatrix}\prod_{i=1}^n(A_{11}-A_{ii})&&\\&\ddots&\\&&\prod_{i=1}^n(A_{nn}-A_{ii})\end{bmatrix}}

Since \prod_{i=1}^n(A_{jj}-A_{ii})=0 for j=1,\dots,n, we have f(A)=0.

4.If f and g are independent polynomials over a field F and h is a non-zero polynomial over F, show that fh and gh are independent.

Solution: Let c,d\in F such that cfh+dgh=0, then (cf+dg)h=0 and h\neq 0 means cf+dg=0, since f and g are independent polynomials, c=d=0 and the conclusion follows.

5.If F is a field, show that the product of two non-zero elements of F^{\infty} is non-zero.

Solution: Let f=(f_0,f_1,f_2,\dots) and g=(g_0,g_1,g_2,\dots) be two non-zero elements of F^{\infty}, and let

\displaystyle{m:=\min\{k\in N:f_k\neq 0\},\quad n:=\min\{k\in N:g_k\neq 0\}}

Obviously m\geq 0,n\geq 0, thus

\displaystyle{\begin{aligned}(fg)_{m+n}&=\sum_{i=0}^{m+n}f_i g_{m+n-i}\\&=\sum_{i=0}^{m-1}f_i g_{m+n-i}+f_m g_n+\sum_{i=m+1}^{m+n}f_i g_{m+n-i}\\&=f_m g_n\neq 0\end{aligned}}

which means fg is non-zero.

6.Let S be a set of non-zero polynomials over a field F. If no two elements of S have the same degree, show that S is an independent set in F[x].

Solution: Assume S is not an independent set in F[x], then we shall find s_1,\dots,s_n in S such that there are scalars c_1,\dots,c_n\in F, not all zero which satisfies c_1 s_1+\cdots+c_n s_n=0, we can further suppose \deg s_i<\deg s_j if i<j, otherwise we just switch s_i,s_j and also switch the corresponding c_i,c_j. Now define k=\max\{1\leq i\leq n:c_i\neq 0\}, then we can rewrite c_k s_k=\sum_{i=1}^{k-1}c_i s_i, then

\deg s_k =\deg(c_k s_k)=\deg \left(\sum_{i=1}^{k-1}c_i s_i\right)\leq \max_{1\leq i\leq k-1}\deg s_i <\deg s_k

this is a contradiction.

7.If a and b are elements of a field F and a\neq 0, show that the polynomials 1,ax+b,(ax+b)^2,(ax+b)^3,\dots form a basis of F[x].

Solution: Let S=\{(ax+b)^i:i\in N\}, then no two elements of S have the same degree, thus S is an independent set in F[x] by Exercise 6. We only have to show S spans F[x].
Let f=f_0 x^0+f_1 x+\cdots+f_n x^n be any vector in F[x], then if we solve the system of equations f=\sum_{i=0}^nx_i (ax+b)^i about the unknowns x_0,\dots,x_n, it follows that the augmented matrix of this system is

\displaystyle{A'=\begin{bmatrix}1&b&b^2&\cdots&b^n&f_0\\ &a&2ab&\cdots&\binom{n}{1}ab^{n-1}&f_1\\ & &a^2&\cdots&\binom{n}{2}a^2 b^{n-2}&f_2\\ & & &\ddots&\vdots&\vdots\\ & & & &a^n&f_n \end{bmatrix}}

since a\neq 0, the row rank of A' is n+1 and must equal to the row rank of the coefficient matrix, which is also n+1. Thus the system of equations has a solution and so S spans F[x].

8.If F is a field and h is a polynomial over F of degree \geq 1, show that the mapping f\to f(h) is a one-one linear transformation of F[x] into F[x]. Show that this transformation is an isomorphism of F[x] onto F[x] if and only if \deg h=1.

Solution: First if there is f,g\in F[x] such that f(h)=g(h), then 0=f(h)-g(h)=(f-g)(h), by definition, denote f-g=\sum_{i=0}^nf_i x^i, then (f-g)(h)=\sum_{i=0}^nf_i h^i, since \deg h\geq 1, h^i\neq 0 for 0\leq i\leq n, it follows that f_i=0,0\leq i\leq n and so f-g=0, thus f=g and f\to f(h) is one-one.
Now if \deg h=1, we can write h=cx,c\neq 0, so for f=\sum_{i=0}^nf_i x^i \in F[x], there is g=\sum_{i=0}^nc^{-i}f_i x^i\in F[x] such that f=g(h), which means the mapping is onto. Conversely assume \deg h\geq 2, then no f\in F[x] could make f(h)=x, since f must have the form f=\sum_{i=0}^nf_i x^i, and thus f(h)=\sum_{i=0}^nf_i h^i, if f_i=0 for i\geq 1, then \deg f(h)=0, otherwise \deg f(h)\geq 2. Thus if we suppose this transformation is an isomorphism, then \deg h<2, and the conclusion follows.

9.Let F be a subfield of the complex numbers and let T,D be the transformation on F[x] defined by

\displaystyle{T\left(\sum_{i=0}^nc_ix^i\right)=\sum_{i=0}^n\frac{c_i}{1+i}x^{i+1}\quad \text{and}\quad D\left(\sum_{i=0}^nc_ix^i\right)=\sum_{i=1}^nic_ix^{i-1}}

( a ) Show that T is a non-singular linear operator on F[x]. Show also that T is not invertible.
( b ) Show that D is a linear operator on F[x] and find its null space.
( c ) Show that DT=I and TD\neq I.
( d ) Show that T[(Tf)g]=(Tf)(Tg)-T[f(Tg)] for all f,g in F[x].
( e ) State and prove a rule for D similar to the one given for T in (d).
( f ) Supopse V is a non-zero subspace of F[x] such that Tf belongs to V for each f in V. Show that V is not finite-dimensional.
( g ) Suppose V is a finite-dimensional subspace of F[x]. Prove there is an integer m\geq 0 such that D^mf=0 for each f in V.

Solution:
( a ) Let f=\sum_{i=0}^mf_ix^i and g=\sum_{i=0}^ng_ix^i be elements of F[x] and c a complex number, then for m\geq n we have

\displaystyle{\begin{aligned}T(cf+g)&=T\left(c\sum_{i=0}^mf_ix^i+\sum_{i=0}^ng_ix^i\right)\\&=T\left(\sum_{i=0}^n(cf_i+g_i)x^i+\sum_{i=n+1}^mcf_ix^i\right)\\&=\sum_{i=0}^n\frac{cf_i+g_i}{1+i}x^{i+1}+\sum_{i=n+1}^m\frac{cf_i}{1+i}x^{i+1}\\&=c\sum_{i=0}^m\frac{f_i}{1+i}x^{i+1}+\sum_{i=0}^n\frac{g_i}{1+i}x^{i+1}\\&=cT(f)+T(g)\end{aligned}}

for m<n we have

\displaystyle{\begin{aligned}T(cf+g)&=T\left(c\sum_{i=0}^mf_ix^i+\sum_{i=0}^ng_ix^i\right)\\&=T\left(\sum_{i=0}^m(cf_i+g_i)x^i+\sum_{i=m+1}^ng_ix^i\right)\\&=\sum_{i=0}^m\frac{cf_i+g_i}{1+i}x^{i+1}+\sum_{i=m+1}^n\frac{g_i}{1+i}x^{i+1}\\&=c\sum_{i=0}^m\frac{f_i}{1+i}x^{i+1}+\sum_{i=0}^n\frac{g_i}{1+i}x^{i+1}\\&=cT(f)+T(g)\end{aligned}}

Thus T is a linear operator on F[x].
Suppose f=\sum_{i=0}^mf_ix^i and Tf=0, then we have
\displaystyle{T\left(\sum_{i=1}^nf_ix^i\right)=\sum_{i=0}^n\frac{f_i}{1+i}x^{i+1}=0 \implies f_i=0,0\leq i\leq n\implies f=0}
To see T is not invertible, since no f\in F[x] can make T(f)=1, T is not onto, by Theorem 9 of Chapter 3 we see T is not invertible.

( b ) Let f=\sum_{i=0}^mf_ix^i and g=\sum_{i=0}^ng_ix^i be elements of F[x] and c a complex number, then for m\geq n we have

\displaystyle{\begin{aligned}D(cf+g)&=D\left(c\sum_{i=0}^mf_ix^i+\sum_{i=0}^ng_ix^i\right)\\&=D\left(\sum_{i=0}^n(cf_i+g_i)x^i+\sum_{i=n+1}^mcf_ix^i\right)\\&=\sum_{i=1}^ni(cf_i+g_i)x^{i-1}+\sum_{i=n+1}^mi(cf_i)x^{i-1}\\&=c\sum_{i=1}^mif_ix^{i-1}+\sum_{i=1}^nig_ix^{i-1}\\&=cD(f)+D(g)\end{aligned}}

for m<n we have

\displaystyle{\begin{aligned}D(cf+g)&=D\left(c\sum_{i=0}^mf_ix^i+\sum_{i=0}^ng_ix^i\right)\\&=D\left(\sum_{i=0}^m(cf_i+g_i)x^i+\sum_{i=m+1}^ng_ix^i\right)\\&=\sum_{i=1}^mi(cf_i+g_i)x^{i-1}+\sum_{i=m+1}^nig_ix^{i-1}\\&=c\sum_{i=1}^mif_ix^{i-1}+\sum_{i=1}^nig_ix^{i-1}\\&=cD(f)+D(g)\end{aligned}}

Thus D is a linear operator on F[x]. To find its null space, suppose f=\sum_{i=0}^nf_ix^i\in \text{null }(D), then D(f)=\sum_{i=1}^nif_ix^{i-1}=0, thus f_i=0,i\geq 1, so the null space of D is the subspace of scalar polynomials in F[x].

( c ) We have

\displaystyle{DT\left(\sum_{i=0}^nc_ix^i\right)=D\left(\sum_{i=0}^n\frac{c_i}{1+i}x^{i+1}\right)=D\left(\sum_{i=1}^n\frac{c_{i-1}}{i}x^{i}\right)=\sum_{i=1}^nc_{i-1}x^{i-1}=\sum_{i=0}^nc_ix^i}

and since TD(1)=T(0)=0\neq 1, we have TD\neq I.

( d ) If f=\sum_{i=0}^mf_ix^i and g=\sum_{i=0}^ng_ix^i, then the polynomials on both side of the equation has degree m+n+2, and for 0\leq k\leq m+n+1 we first have

\begin{aligned}((Tf)g+f(Tg))_k&=\sum_{i=0}^k(Tf)_ig_{k-i}+\sum_{i=0}^kf_i(Tg)_{k-i}\\&=\sum_{i=1}^k\dfrac{f_{i-1}g_{k-i}}{i}+\sum_{i=0}^{k-1}\dfrac{f_{i}g_{k-i-1}}{k-i}\\&=\sum_{i=1}^k\dfrac{f_{i-1}g_{k-i}}{i}+\sum_{i=1}^{k}\dfrac{f_{i-1}g_{k-i}}{k+1-i}\\&=\sum_{i=1}^kf_{i-1}g_{k-i}\left(\dfrac{1}{i}+\dfrac{1}{k+1-i}\right)\\&=\sum_{i=1}^k\dfrac{(k+1)f_{i-1}g_{k-i}}{(i)(k+1-i)}\end{aligned}

thus

\displaystyle{\begin{aligned}T[(Tf)g+f(Tg)]_{k+1}&=\frac{1}{k+1}\left(\sum_{i=1}^k\frac{(k+1)f_{i-1}g_{k-i}}{(i)(k+1-i)}\right)\\&=\sum_{i=1}^{k}\frac{f_{i-1}}{i}\frac{g_{k-i}}{k+1-i},\quad 1\leq k\leq m+n+1\end{aligned}}

While we also have

((Tf)(Tg))_{k+1}=\sum\limits_{i=0}^{k+1}(Tf)_i(Tg)_{k+1-i}=\sum\limits_{i=1}^{k}\dfrac{f_{i-1}}{i}\dfrac{g_{k-i}}{k+1-i},\quad 1\leq k\leq m+n+1

thus T[(Tf)g+f(Tg)]=(Tf)(Tg), and the statement holds.

( e ) We can prove one similar but slightly different result about D:

\displaystyle{D[(Df)g]=D^2(fg)-D[f(Dg)]}

( f ) Assume V is finite-dimensional, then there are f_1,\dots,f_n\in F[x] such that \{f_1,\dots,f_n\} spans V. We choose 1\leq j\leq n such that \deg f_j=\max_{1\leq i\leq n}\deg f_i, then any g\in V shall satisfy \deg g\leq \deg f_j. Since f_j\in V we have Tf_j\in V, but \deg Tf_j=\deg f_j+1>\deg f_j, this is a contradiction.

( g ) Since V is finite-dimensional, there are f_1,\dots,f_n\in F[x] such that \{f_1,\dots,f_n\} spans V. For 1\leq j\leq n, if f_j=\sum_{i=0}^{n_j}f_{ij}x^i, then it is easy to see that D^{n_j+1}f_j=0, thus let m=1+\max_{1\leq j\leq n}n_j, we have D^mf_i=0,1\leq i\leq n, thus for each f\in V, we express f=\sum_{i=1}^na_if_i, then

\displaystyle{D^mf=D^m\left(\sum_{i=1}^na_if_i\right)=\sum_{i=1}^na_i(D^mf_i)=0}

Linear Algebra (2ed) Hoffman & Kunze 3.7

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Theorem 21说明，从T\in L(V,W)可以自然的得到一个$latex L(W^,V^)$的元素，即对于g\in W^*，g(T\alpha),\alpha \in V是一个从V到F的Linear functional。如果记这个元素为T^t，那么有T^t(g)(\alpha)=g(T\alpha)。这个概念大部分书上叫adjoint，但本书称为transpose。Theorem 22非常有用，其说明：null(T^t)=(range(T))^0，如果在有限维空间中，则进一步有rank(T)=rank(T^t)以及range(T^t)=(null(T))^0。可以看到T和T^t也有一种对偶关系。Theorem 23可以看做是对transpose的解释，如果T在一组基下的矩阵是A，那么T^t在对应的那组dual basis下的矩阵就是A^t。由此可以得到矩阵行列秩相等的另一种证法（Theorem 24)。

Exercises

Exercise 3.7.1. Let F be a field and let f be the linear functional on F^w defined by f(x_1,x_2)=ax_1+bx_2. For each of the following linear operators T, let g=T^tf, and find g(x_1,x_2).
( a ) T(x_1,x_2)=(x_1,0);
( b ) T(x_1,x_2)=(-x_2,x_1);
( c ) T(x_1,x_2)=(x_1-x_2,x_1+x_2).

Solution:
( a ) g(x)=T^t f(x)=f(Tx), thus g(x_1,x_2 )=f(x_1,0)=ax_1.
( b ) g(x_1,x_2 )=f(-x_2,x_1 )=-ax_2+bx_1.
( c ) g(x_1,x_2 )=f(x_1-x_2,x_1+x_2 )=a(x_1-x_2 )+b(x_1+x_2 )=(a+b) x_1+(b-a) x_2.

Exercise 3.7.2. Let V be the vector space of all polynomial functions over the field of real numbers. Let a and b be fixed real numbers and let f be the linear functional on V defined by

\displaystyle{f(p)=\int_a^bp(x)dx.}

If D is the differentiation operator on V, what is D^tf?

Solution: Since D^t f(p)=f(D(p)), we can write

\displaystyle{D^t f(p)=\int_a^b {p'(x)dx}=\int_a^b{dp(x)}=p(b)-p(a)}

Exercise 3.7.3. Let V be the space of all n\times n matrices over the field F and let B be a fixed n\times n matrix. If T is the linear operator on V defined by T(A)=AB-BA, and if f is the trace function, what is T^tf?

Solution: For any A\in V we have

\displaystyle{T^t f(A)=f(T(A))=f(AB-BA)=0}

Thus T^t f is the zero functional on V.

Exercise 3.7.4. Let V be a finite-dimensional vector space over the field F and let T be a linear operator on V. Let c be a scalar and suppose there is a non-zero vector \alpha in V such that T\alpha=c\alpha. Prove that there is a non-zero linear functional f on V such that T^tf=cf.

Solution: Let {\dim V}={\dim V^*}=n, if c=0, the conclusion is trivial since we can define f=0 on V. Now suppose c\neq 0. As \alpha\neq 0 and \alpha\in null(T-cI), we know \dim null(T-cI)\geq 1, thus \dim range(T-cI)\leq n-1,notice for any f\in V^* and v\in V, we have

\displaystyle{\begin{aligned}((T-cI)^t f)(v)&=f((T-cI)v)=f(Tv-cv)=f(Tv)-cf(v)\&=(T^t f)(v)-(cf)(v)=(T^t f-cf)(v)=((T^t-cI)f)(v)\end{aligned}}

thus (T-cI)^t=T^t-cI and by Theorem 22 we have

\displaystyle{\dim range(T-cI)^t =\dim range(T-cI)\leq n-1}

which gives \dim null(T-cI)^t \geq 1, so there exists a nonzero f\in V^* such that

\displaystyle{(T-cI)^t f=(T^t-cI)f=0}

this means T^t f=cf.

Exercise 3.7.5. Let A be an m\times n matrix with real entries. Prove that A=0 if and only if \text{trace }(A^tA)=0.

Solution: Let A=(A_ij ),i=1,\dots ,m,j=1,\dots ,n, then A^t=(A_{ji} ) and

\displaystyle{(A^t A)_{ii}=\sum_{j=1}^m A_{ji}^2 ,\qquad i=1,\dots ,n}

so trace(A^t A)=\sum_{i=1}^n \sum_{j=1}^m A_{ji}^2 , thus A=0 if and only if A_{ij}=0 for i=1,\dots ,m,j=1,\dots ,n, if and only if trace(A^t A)=\sum_{i=1}^n \sum_{j=1}^m A_{ji}^2 =0.

Exercise 3.7.6. Let n be a positive integer and let V be the space of all polynomial functions over the field of real numbers which have degree at most n, i.e., functions of the form

\displaystyle{f(x)=c_0+c_1x+\cdots+c_nx^n.}

Let D be the differentiation operator on V. Find a basis for the null space of the transpose operator D^t.

Solution: Notice {\dim V}=n+1={\dim V^*}, and {\dim range(D)}=n={\dim range(D^t)}, thus {\dim null(D^t)}=1, and if g\in null(D^t), g must be an annihilator of range(D). One choice of g is

\displaystyle{g(f(x))=c_n,\quad f(x)=c_0+c_1 x+\cdots+c_n x^n\in V}

Exercise 3.7.7. Let V be a finite-dimensional vector space over the field F. Show that T\to T^t is an isomorphism of L(V,V) onto $latex L(V^,V^)$.

Solution: Let \mathfrak B be a basis for V, and \mathfrak B^* be its dual, then for any T\in L(V,V), let A=(A_{ij}) be the matrix of T with respect to the basis \mathfrak B, then (A_{ji}) is the matrix of T^t with respect to the basis \mathfrak B^*, since T\to A_{ij} is an isomorphism and A_{ji}\to T^t is an isomorphism, the conclusion follows since A_{ij}\to A_{ji} is an obvious isomorphism.

Exercise 3.7.8. Let V be the vector space of n\times n matrices over the field F.
( a ) If B is a fixed n\times n matrix, define a function f_B on V by f_B(A)=\text{trace }(B^tA). Show that f_B is a linear functional on V.
( b ) Show that every linear functional on V is of the above form, i.e., is f_B for some B.
( c ) Show that B\to f_B is an isomorphism of V onto V^*.

Solution:
( a ) Let A,D\in V, c\in F, then

\displaystyle{\begin{aligned}f_B (cA+D)&=trace(B^t (cA+D))=trace(cB^t A+B^t D)\\&=c\cdot trace(B^t A)+trace(B^t D)=cf_B (A)+f_B (D)\end{aligned}}

( b ) Let E_{ij} be the matrix with 1 in the position row i, column j and 0 otherwise, then {E_{ij}: i,j=1,\dots,n} is a basis for V, and let {f_{ij}: i,j=1,\dots,n} be its dual space on V^*, i.e.

\displaystyle{f_{kh} (E_{ij} )=\delta_{ki} \delta_{hj},\quad i,j,k,h=1,\dots,n}

if f\in V^*, then there is n^2 scalars x_{ij}:i,j=1,\dots,n in F such that

\displaystyle{f=\sum\limits_{i=1}^n \sum\limits_{j=1}^n x_{ij} f_{ij}}

then for A=(A_{ij})\in V, f(A)= \sum\limits_{i=1}^n \sum\limits_{j=1}^n x_{ij} A_{ij}, if we let

\displaystyle{B=\left[\begin{matrix} x_{11}&\cdots&x_{1n}\\ \vdots&\ddots&\vdots \\ x_{n1}&\cdots &x_{nn}\end{matrix}\right]}

then

\displaystyle{B^t A= \left[\begin{matrix} x_{11}&\cdots&x_{n1}\\ \vdots&\ddots&\vdots \\ x_{1n}&\cdots &x_{nn}\end{matrix}\right] \left[\begin{matrix} A_{11}&\cdots&A_{1n}\\ \vdots&\ddots&\vdots \\ A_{n1}&\cdots &A_{nn}\end{matrix}\right]}

which we can see trace(B^t A)=\sum_{j=1}^n \sum_{i=1}^n x_{ij}A_{ij}=f(A)

( c ) From (b) we can see B\to f_B is surjective. Now let f_B=0 on V^* with

\displaystyle{B=\left[\begin{matrix} x_{11}&\cdots&x_{1n}\\ \vdots&\ddots&\vdots \\ x_{n1}&\cdots &x_{nn}\end{matrix}\right]}

then

\displaystyle{f_B (E_{ij})=x_{ij}=0\quad ,i,j=1,\dots,n}

it follows that B=0 and so B\to f_B is injective, thus an isomorphism.