Definition and Theorems (Chapter 3)

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Definition. Let V and W be vector spaces over the field F. A linear tranformation from V into W is a function T from V into W such that

\displaystyle{T(c{\alpha}+{\beta})=c(T{\alpha})+T{\beta},\quad\forall {\alpha},{\beta}\in V,\forall c\in F}

Theorem 1. Let V be a finite-dimensional vector space over the field F and let \{{\alpha}_1,\dots,{\alpha}_n\} be an ordered basis for V. Let W be a vector space over the same field F and let \{{\beta}_1,\dots,{\beta}_n\} be any vectors in W. Then there is precisely one linear transformation T from V into W such that

\displaystyle{T{\alpha}_j={\beta}_j,\qquad j=1,\dots,n}

Definition. Let V and W be vector spaces over the field F and let T be a linear transformation from V into W. The null space of T is the set of all vectors \alpha in V such that T{\alpha}=0. If V is finite-dimensional, the rank of T is the dimension of the range of T and the nullity of T is the dimension of the null space of T.

Theorem 2. Let V and W be vector spaces over the field F and let T be a linear transformation from V into W. Suppose that V is finite-dimensional. Then

\displaystyle{\text{rank}(T)+\text{nullity}(T)=\dim(V)}

Theorem 3. If A is an m\times n matrix with entries in the field F, then \text{row rank}(A)=\text{column rank}(A).

Theorem 4. Let V and W be vector spaces over the field F. Let T and U be linear transformations from V into W. The function (T+U) defined by

\displaystyle{(T+U)(\alpha)=T{\alpha}+U{\alpha}}

is a linear transformation from V into W. If c is any element of F, the function (cT) defined by

\displaystyle{(cT)(\alpha)=c(T{\alpha})}

is a linear transformation from V into W. The set of all linear transformations from V into W, together with the addition and scalar multiplication defined above, is a vector space over the field F.

Theorem 5. Let V be an n-dimensional vector space over the field F, and let W be an m-dimensional vector space over F. Then the space L(V,W) is finite-dimensional and has dimension mn.

Theorem 6. Let V,W,Z be vector spaces over the field F. Let T be a linear transformation from V into W and U a linear transformation from W into Z. Then the composed function UT defined by (UT)({\alpha})=U(T(\alpha)) is a linear transformation from V into Z.

Definition. If V is a vector space over the field F, a linear operator on V is a linear transformation from V into V.

Lemma. Let V be a vector space over the field F; let U,T_1,T_2 be linear operators on V; let c be an element of F.
( a ) IU=UI=U;
( b ) U(T_1+T_2)=UT_1+UT_2;(T_1+T_2)U=T_1U+T_2U;
( c ) c(UT_1)=(cU)T_1=U(cT_1).

Theorem 7. Let V and W be vector spaces over the field F and let T be a linear transformation from V into W. If T is invertible, then the inverse function T^{-1} is a linear transformation from W onto V.

Theorem 8. Let T be a linear transformation from V into W. Then T is non-singular if and only if T carries each linearly independent subset of V onto a linearly independent subset of W.

Theorem 9. Let V and W be finite-dimensional vector spaces over the field F such that \dim V=\dim W. If T is a linear transformation from V into W, the following are equivalent:
(i) T is invertible.
(ii) T is non-singular.
(iii) T is onto, that is, the range of T is W.
(iv) If \{{\alpha}_1,\dots,{\alpha}_n\} is basis for V, then \{T{\alpha}_1,\dots,T{\alpha}_n\} is a basis for W.
(v) There is some basis \{{\alpha}_1,\dots,{\alpha}_n\} for V such that \{T{\alpha}_1,\dots,T{\alpha}_n\} is a basis for W.

Definition. A group consists of a set G and a rule (or operation) which associates with each pair of elements x,y\in G an element xy\in G in such a way that
( a ) x(yz)=(xy)z,\forall x,y,z\in G (associativity);
( b ) there is some e\in G such that ex=xe=x,\forall x\in G;
( c ) to each element x\in G there corresponds an element x^{-1}\in G such that xx^{-1}=x^{-1}x=e.

Theorem 10. Every n-dimensional vector space over the field F is isomorphic to the space F^n.

Theorem 11. Let V be an n-dimensional vector space over the field F and W an m-dimensional vector space over F. Let \mathfrak B be an ordered basis for V and \mathcal B' an ordered basis for W. For each linear transformation T from V into W, there is an m\times n matrix A with entries in F such that [T\alpha]_{\mathfrak B'}=A[\alpha]_{\mathfrak B} for every vector \alpha \in V. Furthermore, T\to A is a one-one correspondence between the set of all linear transformations from V into W and the set of all m\times n matrices over the field F.

Theorem 12. Let V be an n-dimensional vector space over the field F and let W be an m-dimensional vector space over F. For each pair of ordered bases \mathfrak B,\mathfrak B' for V and W respectively, the function which assigns to a linear transformation T its matrix relative to \mathfrak B,\mathfrak B' is an isomorphism between the space L(V,W) and the space of all m\times n matrices over the field F.

Theorem 13. Let V,W,Z be finite dimensional vector spaces over the field F; let T be a linear transformation from V into W and U a linear transformation from W into Z. If \mathfrak B,\mathfrak B' and \mathfrak B'' are ordered bases for the spaces V,W and Z, respectively, if A is the matrix of T relative to the pair \mathfrak B,\mathfrak B' and B is the matrix of U relative to the pair \mathfrak B',\mathfrak B'', then the matrix of the composition UT relative to the pair \mathfrak B,\mathfrak B'' is the product matrix C=BA.

Theorem 14. Let V be a finite-dimensional vector space over the field F, and let

\displaystyle{\mathfrak B=\{\alpha_1,\dots,\alpha_n\}\text{ and }\mathfrak B'=\{\alpha'_1,\dots,\alpha'_n\}}

be ordered bases for V. Suppose T is a linear operator on V. If P=[P_1,\dots,P_n] is the n\times n matrix with columns P_j=[\alpha'_j]_{\mathfrak B}, then

\displaystyle{[T]_{\mathfrak B'}=P^{-1}[T]_{\mathfrak B}P}

Alternatively, if U is the invertible operator on V defined by U\alpha_j=\alpha'_j,j=1,\dots,n, then

\displaystyle{[T]_{\mathfrak B'}=[U]_{\mathfrak B}^{-1}[T]_{\mathfrak B}[U]_{\mathfrak B}}

Definition. Let A and B be n\times n (square) matrices over the field F. We say B is similar to A over F if there is an invertible n\times n matrix P over F such that B=P^{-1}AP.

Theorem 15. Let V be a finite-dimensional vector space over the field F, and let \mathfrak B=\{\alpha_1,\dots,\alpha_n\} be a basis for V. Then there is a unique dual basis \mathfrak B^*=\{f_1,\dots,f_n\} for V^* such that f_i(\alpha_j)=\delta_{ij}. For each linear functional f on V we have

\displaystyle{f=\sum_{i=1}^nf(\alpha_i)f_i}

and for each vector \alpha in V we have

\displaystyle{\alpha =\sum_{i=1}^nf_i(\alpha)\alpha_i}

Definition. If V is a vector space over the field F and S is a subset of V, the annihilator of S is the set S^0 of linear functionals f on V such that f(\alpha)=0 for every \alpha in S.

Theorem 16. Let V be a finite-dimensional vector space over the field F, and let W be a subspace of V. Then \dim W+\dim W^0=\dim V.
Corollary. If W is a k-dimensional subspace of an n-dimensional vector space V, then W is the intersection of (n-k) hyperspaces in V.
Corollary. If W_1 and W_2 are subspaces of a finite-dimensional vector space, then W_1=W_2 if and only if W_1^0=W_2^0.

Theorem 17. Let V be a finite-dimensional vector space over the field F. For each vector \alpha in V define

\displaystyle{L_{\alpha}(f)=f(\alpha),\qquad f\text{ in }V^*.}

The mapping \alpha \to L_{\alpha} is then an isomorphism of V onto V^{**}.
Corollary. Let V be a finite-dimensional vector space over the field F. If L is a linear functional on the dual space V^* of V, then there is a unique vector \alpha in V such that L(f)=f(\alpha),\forall f\in V^*.
Corollary. Let V be a finite-dimensional vector space over the field F. Each basis for V^* is the dual of some basis for V.

Theorem 18. If S is any subset of a finite-dimensional vector space V, then (S^0)^0 is the subspace spanned by S.

Definition. If V is a vector space, a hyperspace in V is a maximal proper subspace of V.

Theorem 19. If f is a non-zero linear functional on the vector space V, then the null space of f is a hyperspace in V. Conversely, every hyperspace in V is the null space of a (not unique) non-zero linear functional on V.

Lemma. If f and g are linear functionals on a vector space V, then g is a scalar multiple of f if and only if the null space of g contains the null space of f, that is, if and only if f(\alpha)=0 implies g(\alpha)=0.

Theorem 20. Let g,f_1,\dots,f_r be linear functionals on a vector space V with respective null spaces N,N_1,\dots,N_r. Then g is a linear combination of f_1,\dots,f_r if and only if N contains the intersection N_1\cap{\cdots}\cap N_r.

Theorem 21. Let V and W be vector spaces over the field F. For each linear transformation T from V into W, there is a unique linear transformation T^t from W^* into V^* such that

\displaystyle{(T^tg)(\alpha)=g(T\alpha),\qquad \forall g\in W^*,\alpha \in V}

Theorem 22. Let V and W be vector spaces over the field F, and let T be a linear transformation from V into W. The null space of T^t is the annihilator of the range of T. If V and W are finite-dimensional, then
( i ) rank (T^t)=rank (T).
( ii ) the range of T^t is the annihilator of the null space of T.

Theorem 23. Let V and W be finite-dimensional vector spaces over the field F. Let \mathfrak B be an ordered basis for V with dual basis \mathfrak B^*, and let \mathfrak B' be an ordered basis for W with dual basis \mathfrak B^{'*}. Let T\in L(V,W) and A be the matrix of T relative to \mathfrak B,\mathfrak B', and let B be the matrix of T^t relative to \mathfrak B^{'*},\mathfrak B^*. Then B_{ij}=A_{ji}.

Definition. If A is an m\times n matrix over the field F, the transpose of A is the n\times m matrix A^t defined by A^t_{ij}=A_{ji}.

Theorem 24. Let A be any m\times n matrix over the field F. Then the row rank of A is equal to the column rank of A.

Linear Algebra (2ed) Hoffman & Kunze 3.6

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在已经有了V的dual space（V^*=L(V,F)）之后，还有一个V^*的dual space（V^{*}=L(V^*,F)），利用Theorem 17和其两个推论可以证明V^{**}中的所有元素都有L_{\alpha}(f)=f(\alpha)的形式，并且V^{**}和V是isomorphism的，因此dual了两次之后其实回到了自身，故V和V^*可以是in duality with each other。这种很好的性质一是可以证明V^*的每一个basis都有V的一个basis与之dual，同时还有对于V的subspaceW，有结论W=W^{00}，有点像负负得正。Theorem 18推广了一个一般情形，任何子集S在annihilate两次后能得到该子集张成的subspace。最后是引入hyperspace的概念（即最大的真子空间）和关于linear functional的两个结论：Theorem 19说明V上的非零linear functional的null space和hyperspace有对应关系（但不是一一对应），Theorem 20说明null space的包含关系与linear functionals的线性组合关系可以互推。

Exercises

Exercises 3.6.1. Let n be a positive integer and F a field. Let W be the set of all vectors (x_1,\dots,x_n) in F^n such that x_1+\cdots +x_n=0.
( a ) Prove that W^0 consists of all linear functionals f of the form

\displaystyle{f(x_1,\dots,x_n)=c\sum_{j=1}^nx_j.}

( b ) Show that the dual space W^* of W can be ‘naturally’ identified with the linear functionals f(x_1,\dots,x_n)=c_1x_1+\cdots+c_nx_n on F^n which satisfy c_1+\cdots +c_n=0.

Solution:
( a ) Suppose f is of the form

\displaystyle{f(x_1,\dots, x_n )=c\sum_{j=1}^n x_j}

then let (a_1,\dots,a_n )\in W, we have \sum_{j=1}^n a_j =0 and thus f(a_1,\dots,a_n)=c0=0.
Also let \{\epsilon_1,\dots,\epsilon_n \} be the standard basis for F^n and \{f_1,\dots,f_n \} be its dual, then any f\in V^* can be written as f=\sum_{j=1}^n f(\epsilon_j ) f_j, if f\in W^0, then for any (x_1,\dots,x_n) such that \sum_{j=1}^n x_j =0, we have

\displaystyle{f(x_1,\dots,x_n )=\sum_{j=1}^nf(\epsilon_j ) f_j (x_1,\dots,x_n )=\sum_{j=1}^n f(\epsilon_j ) x_j=0}

so let c=f(\epsilon_1) and (x_1,\dots,x_n )=(1,0,\dots,-1,\dots,0), we can see f(\epsilon_i )=c,2\leq i\leq n, so f has the desired form.

( b ) Notice that {\dim (W^*)} ={\dim {W}}=n-1. Define

\begin{aligned}f_1 (x_1,\dots,x_n )&=x_1-x_n \\ &\ldots \\f_{n-1} (x_1,\dots,x_n )&=x_{n-1}-x_n\end{aligned}

then it is easy to verify f_i\in W^*,i=1,\dots,n-1, since for \alpha,\beta\in W we have

\displaystyle{\begin{aligned}f_i (c\alpha+\beta)&=f_i (c\alpha_1+\beta_1,\dots,c\alpha_n+\beta_n )\\&=c\alpha_i+\beta_i-c\alpha_n-\beta_n\\&=c(\alpha_i-\alpha_n )+(\beta_i-\beta_n )\\&=cf_i (\alpha)+f_i (\beta)\end{aligned}}

Now suppose we have k_1,\dots,k_{n-1}\in F s.t. \sum_{i=1}^{n-1} k_i f_i=0 on W, then \sum_{i=1}^{n-1}k_i f_i\in W^0, by ( a ) we have

\displaystyle{\sum\limits_{i=1}^{n-1}k_i f_i (x_1,\dots,x_n )=\sum\limits_{i=1}^{n-1}k_i (x_i-x_n)=c\sum\limits_{i=1}^nx_i ,\quad c\in F}

thus k_1 x_1+\cdots+k_{n-1} x_{n-1}-x_n \sum_{i=1}^{n-1} k_i =c(x_1+\cdots+x_n), a comparison will show that k_1=\cdots=k_{n-1}=c,\sum_{i=1}^{n-1}k_i =-c, thus (n-1)c=-c or nc=0, since n is positive, we have c=0 and k_1=\cdots=k_{n-1}=0, so f_1,\dots,f_{n-1} are linearly independent, thus spans W^*, thus any f\in W^* will have the form

\displaystyle{f=c_1 f_1+\cdots+c_{n-1} f_{n-1},\qquad c_1,\dots,c_{n-1}\in F}

If we let c_n=-\sum_{i=1}^{n-1}c_i , then c_1+\cdots+c_n=0 and

\displaystyle{f(x_1,\dots,x_n )=\sum_{i=1}^{n-1}c_i (x_i-x_n)=\sum_{i=1}^{n-1}c_i x_i-x_n \sum_{i=1}^{n-1}c_i =\sum_{i=1}^nc_i x_i}

Exercises 3.6.2. Use Theorem 20 to prove the following. If W is a subspace of a finite-dimensional vector space V and if \{g_1,\dots,g_r\} is any basis for W^0, then

\displaystyle{W=\bigcap_{i=1}^rN_{g_i}.}

Solution: As g_i\in W^0,1\leq i\leq r, we know W\subseteq N_{g_i },1\leq i\leq r, thus W\subseteq \cap_{i=1}^r N_{g_i}, to prove the other direction, assume \cap_{i=1}^r N_{g_i}\not\subseteq W, then \exists a\in \cap_{i=1}^r N_{g_i} but a\notin W. Since V is finite dimensional, so is W, thus we can find a basis for W to be \{w_1,\dots,w_k \}, as a\in W, \{w_1,\dots,w_k,a\} is linearly independent, and can be expanded to a basis for V, namely \{w_1,\dots,w_k,a,w_{k+2},\dots,w_n \}, define g\in V^* by

\displaystyle{g(x_1 w_1+\dots+x_k w_k+x_{k+1} a+x_{k+2} w_{k+2}+\dots+x_n w_n )=x_{k+1}}

then g(a)=1 and if b\in W, then we must have g(b)=0, so g\in W^0, which means g is a linear combination of \{g_1,\dots,g_r \}, by Theorem 20, \cap_{i=1}^r N_{g_i}\subseteq N_g, so a\in N_g, which leads to g(a)=0 and a contradiction. Thus \cap_{i=1}^r N_{g_i}\subseteq W and the conclusion follows.

Exercises 3.6.3. Let S be a set, F a field, and V(S;F) the space of all functions from S into F:

\displaystyle{(f+g)(x)=f(x)+g(x),\qquad (cf)(x)=cf(x).}

Let W be any n-dimensional subspace of V(S;F). Show that there exist points x_1,\dots,x_n in S and functions f_1,\dots,f_n in W such that f_i(x_j)={\delta}_{ij}.

Solution: First we prove if there exists a n-dimensional subspace of V(S,F), then S has at least n elements. Assume not, then S shall have only k elements with k<n, denote them x_1,\dots,x_k and define f_1,\dots,f_k by f_i(x_j )=\delta_{ij}, thenf_i\in V(S,F) and for any f\in V(S,F), f is determined by its evaluations on the points x_1,\dots,x_k, thus f=\sum_{i=1}^kf(x_i)f_i, so any subspace of V(S,F) will have dimension less than k, a contradiction.

Now we prove the main conclusion by induction. For the case n=1, we can find f\in W s.t. f\neq 0, so \exists x_1\in S,f(x_1 )\neq 0, let f_1=\dfrac{f(x)}{f(x_1)}, then f_1 (x_1)=1.
Suppose the conclusion is true for subspaces of V(S,F) with dimension n-1, and let W\subseteq V(S,F) has dimension n, by the proofs in the beginning, S has at least n elements. By the induction hypothesis, for a subspace W' of W with dimension n-1, we can find x_1,\dots,x_{n-1}\in S and f_1,\dots,f_{n-1}\in W' s.t. f_i (x_j)=\delta_{ij}, if we choose c_1,\dots,c_{n-1} s.t. \sum_{i=1}^{n-1}c_i f_i =0, then evaluation at x_i would show c_i=0 for i=1,\dots,n-1, thus \{f_1,\dots,f_{n-1}\} are linearly independent, and can be extended to a basis \{f_1,\dots,f_{n-1},g\} of W.
Notice the set S'=S\backslash \{x_1,\dots,x_{n-1}\} is not empty, if we assume for any x\in S', the relation g(x)=\sum_{i=1}^{n-1}f_i (x)g(x_i) holds, then combined with the fact that g(x_j )=f_j (x_j )g(x_j )=\sum_{i=1}^{n-1}f_i (x_j )g(x_i), we can see

\displaystyle{g(x)=\sum\limits_{i=1}^{n-1}f_i (x)g(x_i),\quad \forall x\in S}

this means g\in \text{span}(f_1,\dots,f_{n-1} ), a contradiction since f_1,\dots,f_{n-1},g are linearly independent. Thus we are able to find x_n\in S (in particular x_n\in S') s.t. g(x_n )\neq \sum_{i=1}^{n-1}f_i (x_n )g(x_i).
Consider the matrix

\displaystyle{A=\begin{bmatrix}{f_1(x_1)}&{\cdots}&{f_{n-1}(x_1)}&{g(x_1)}\\{\cdots}&{\cdots}&{\cdots}&{\cdots}\\{f_1(x_n)}&{\cdots}&f_{n-1} (x_n)&g(x_n) \end{bmatrix}=\begin{bmatrix}1&{\cdots}&0&g(x_1)\\{\vdots}&{\ddots}&{\vdots}&{\vdots}\\0&{\cdots}&1&g(x_{n-1})\\ f_1(x_n)&{\cdots}&f_{n-1}(x_n )&g(x_n) \end{bmatrix}}

by elementary row operations we can have

\displaystyle{A=\left[\begin{matrix} 1&{\cdots}&0&g(x_1) \\ {\vdots}&{\ddots}&{\vdots}&{\vdots} \\0&{\cdots}&1&g(x_{n-1})\\0&{\cdots}&0&g(x_n)-\sum_{i=1}^{n-1}f_i(x_n)g(x_i) \end{matrix}\right]}

so A is invertible, let A^{-1}=(A_{ij}), we see that

\displaystyle{AA^{-1}=\begin{bmatrix} 1&{\cdots}&0&g(x_1) \\ {\vdots}&{\ddots}&{\vdots}&{\vdots} \\0&{\cdots}&1&g(x_{n-1})\\ f_1(x_n)&{\cdots}&f_{n-1}(x_n)&g(x_n) \end{bmatrix} \left[\begin{matrix} A_{11}&{\cdots}&A_{1n} \\ {\vdots}&{\ddots}&{\vdots} \\ A_{n1}&{\cdots}&A_{nn} \end{matrix}\right] = I}

which means the functions defined by

\displaystyle{h_i=\sum\limits_{j=1}^{n-1}A_{ji}f_j +A_{ni}g,\quad i=1,\dots,n}

has the property h_i (x_j)=\delta_{ij}, so the conclusion is true.

Linear Algebra (2ed) Hoffman & Kunze 3.5

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从V到F的线性变换称为linear functional，学习linear functional的意义在于其有助于对有限维空间上的子空间、线性方程和坐标等问题的讨论。EXAMPLE18-21给出了很多有意义的linear functional，包括n维传统空间F^n上的、矩阵的迹trace、函数在固定点t的取值（evaluation at t）和连续函数在固定区间的积分。在V上所有的linear functional的集合形成一个vector space，称为V的dual space，记作V^*=L(V,F)。由Theorem 5可以立刻得到\dim V^*=\dim V，由Theorem 1可知，对1\leq i\leq n有一个唯一的linear functional f_i使得f_i(\alpha_j)=\delta_{ij}，这n个linear functional \{f_1,\dots,f_n\}即是\{\alpha_1,\dots,\alpha_n\}的dual basis。Theorem 15正式阐述了上述结论，dual basis的本质是其中每个f_i可以得到任一\alpha在其dual的原basis中的第i个坐标。
接下来讨论linear functional和subspaces的联系。在一个n维空间中，n-1维的子空间称为hyperspace（或hyperplane），其与linear functional 的联系是：每一个hyperspace都是某一个linear functional的null space。而后对V的subspace S引入annihilator的概念，即所有在S上取值是0的linear functional的集合。Theorem 16说明subspace的维数加上annihilator的维数是原来spaceV的维数。这一定理及证明都很类似于本章的Theorem 2。这一定理的两个重要推论，一是k维子空间是n-k个hyperspaces的交集，二是两个子空间相同当且仅当annihilator相同。

Exercises

1. In R^3,let {\alpha}_1=(1,0,1),{\alpha}_2=(0,1,-2),{\alpha}_3=(-1,-1,0).
( a ) If f is a linear functional on R^3 such that

\displaystyle{f({\alpha}_1)=1,\quad f({\alpha}_2)=-1,\quad f({\alpha}_3)=3,}

and if {\alpha}=(a,b,c), find f(\alpha).
( b ) Describe explicitly a linear functional f on R^3 such that

\displaystyle{f({\alpha}_1)=f({\alpha}_2)=0\quad \text{but}\quad f({\alpha}_3)\neq 0.}

( c ) Let f be any linear functional such that

\displaystyle{f({\alpha}_1)=f({\alpha}_2)=0\quad \text{and}\quad f({\alpha}_3)\neq 0.}

If \alpha =(2,3,-1), show that f(\alpha)\neq 0.

Solution:
( a ) It is apparent that {\alpha_1,\alpha_2,\alpha_3} is a basis for R^3, and we can represent \alpha as

\displaystyle{\alpha=(2a-2b-c)\alpha_1+(a-b-c)\alpha_2+(a-2b-c)\alpha_3}

thus

\displaystyle{\begin{aligned}f(\alpha)&=(2a-2b-c)f(\alpha_1)+(a-b-c)f(\alpha_2)+(a-2b-c)f(\alpha_3)\\ &=(2a-2b-c)-(a-b-c)+3(a-2b-c)\\&=4a-5b-3c\end{aligned}}

( b ) Such a f can be f(x_1,x_2,x_3)=x_1-2x_2-x_3
( c ) As \alpha=-\alpha_1-3\alpha_3,we can see f(\alpha)=-3f(\alpha_3)\neq0.

2. Let \mathfrak B =\{{\alpha}_1,{\alpha}_2,{\alpha}_3\} be the basis for C^3 defined by

\displaystyle{{\alpha}_1=(1,0,-1),\quad{\alpha}_2=(1,1,1),\quad{\alpha}_3=(2,2,0)}.

Find the dual basis of \mathfrak B.

Solution: Let the dual basis be \mathfrak{B} = \{f_1,f_2,f_3\}, since each linear functional on C^3 shall satisfy

\displaystyle{f(x_1,x_2,x_3)=x_1c_1+x_2c_2+x_3c_3}

and f_1 shall satisfy:

\displaystyle{f_1(\alpha_1)=1,\qquad f_1(\alpha_2)=0,\qquad f_1(\alpha_3)=0}

so we can have

\displaystyle{c_1-c_3=1 \\c_1+c_2+c_3=0 \\2c_1+2c_2=0}

Solve it we get c_1=0,c_2=-1,c_3=0, thus f_1(x_1,x_2,x_3)=x_1-x_2.
Similarly we can get

\displaystyle{f_2(x_1,x_2,x_3)=x_1-x_2+x_3\\f_3(x_1,x_2,x_3)=-\dfrac{1}{2}x_1+x_2-\dfrac{1}{2}x_3}

3. If A and B are n\times n matrices over the field F, show that \text{trace }(AB)=\text{trace }(BA). Now show that similar matrices have the same trace.

Solution: Let A=(A_{ij}) and B=(B_{ij}), then

\displaystyle{AB_{ii}=\sum_{j=1}^{n}A_{ij}B_{ji}, \qquad BA_{jj}=\sum_{i=1}^{n}B_{ji}A_{ij}}

thus

\displaystyle{trace(AB)=\sum_{i=1}^{n} AB_{ii}=\sum_{i=1}^{n}\sum_{j=1}^{n} A_{ij}B_{ji}=\sum_{j=1}^{n} \sum_{i=1}^{n}B_{ji} A_{i j}=\sum_{j=1}^{n} B A_{j j}=trace(BA)}

Now suppose B is similar to A, then there is P invertible and B=P^{-1} AP, so

\displaystyle{trace(B)=trace\left(P^{-1} A P\right)=trace\left(A P P^{-1}\right)=trace(A)}

4. Let V be the vector space of all polynomial functions p from R ino R which have degree 2 or less:

\displaystyle{p(x)=c_0+c_1x+c_2x^2.}

Define three linear functionals on V by

\displaystyle{f_1(p)=\int_0^1p(x)dx,\quad f_2(p)=\int_0^2p(x)dx,\quad f_3(x)=\int_0^{-1}p(x)dx.}

Show that \{f_1,f_2,f_3\} is a basis for V^* by exhibiting the basis for V of which it is the dual.

Solution: To find the basis of V such that \{f_1,f_2,f_3 \} is the dual basis for V^*, the three vectors \{p,q,r\} in the basis shall satisfy:

\displaystyle{f_{1}(p)=1, \quad f_{2}(p)=0, \quad f_{3}(p)=0}

If we suppose p=c_0+c_1 x+c_2 x^2, then f_1 (p)=a,f_2 (p)=b,f_3 (p)=c is equivalent to

\displaystyle{\left\{\begin{array}{l}{c_{0}+ \displaystyle\frac{1}{2} c_{1}+ \displaystyle \frac{1}{3} c_{2}=a} \\ {2 c_{0}+2 c_{1}+\dfrac{8}{3} c_{2}=b} \\ {-c_{0}+\dfrac{1}{2} c_{1}-\dfrac{1}{3} c_{2}=c}\end{array}\right.}

solve it we get

\displaystyle{\left\{\begin{array}{c}{c_{0}=a-\dfrac{b}{6}-\dfrac{c}{3}} \\ {c_{1}=a+c} \\ {c_{2}=\dfrac{b}{2}-\dfrac{3}{2} a-\dfrac{c}{2}}\end{array}\right.}

thus for (a,b,c)=(1,0,0) we get

\displaystyle{p(x)=1+x-\dfrac{3}{2} x^{2}}

To verify \{p,q,r\} is indeed a basis for V, we only need to prove they’re linearly independent, suppose kp+mq+nr=0, then

\displaystyle{\left\{\begin{array}{c}{k-\dfrac{1}{6} m-\dfrac{1}{3} n=0} \\ {k+n=0} \\ {-\dfrac{3}{2} k+\dfrac{1}{2} m-\dfrac{1}{2} n=0}\end{array}\right.}

which means k=m=n=0.

5. If A and B are n\times n complex matrices, show that AB-BA=I is impossible.

Solution: Assume AB-BA=I, then trace(AB-BA)=trace(I)=n, but

\displaystyle{trace(AB-BA)=trace(AB)-trace(BA)=0}

this is a contradiction.

6. Let m and n be positive integers and F a field, let f_1,\dots,f_m be linear functionals on F^n. For \alpha in F^n define

\displaystyle{T\alpha =(f_1(\alpha),\dots,f_m(\alpha)).}

Show that T is a linear transformation from F^n into F^m. Then show that every linear tranformation from F^n into F^m is of the above form, for some f_1,\dots,f_m.

Solution: To show T is a linear transformation, notice that f_{1}, \dots, f_{m} are linear, thus

\displaystyle{\begin{aligned}T(c \alpha+\beta)&=\bigl(f_{1}(c \alpha+\beta), \ldots, f_{m}(c \alpha+\beta)\bigr)\\&=\bigl(c f_{1}(\alpha)+f_{1}(\beta), \ldots, c f_{m}(\alpha)+f_{m}(\beta)\bigr)\\&= c\bigl(f_{1}(\alpha), \ldots, f_{m}(\alpha)\bigr)+\bigl(f_{1}(\beta), \ldots, f_{m}(\beta)\bigr)\\&=c T \alpha+T \beta \end{aligned}}

Now let T be any linear transformation from T^n to T^m, let A be the m\times n matrix of T with respect to the standard basis \{\epsilon_{1}, \dots, \epsilon_{n}\} and \{\varepsilon_{1}, \dots, \varepsilon_{m}\} , i.e., A satisfies

\displaystyle{T \epsilon_{j}=\sum_{i=1}^{m} A_{ij} \varepsilon_{i}, \quad j=1, \dots, n}

then if \alpha=\sum_{j=1}^{n} x_{j} \epsilon_{j}, we have

\displaystyle{T \alpha=\sum_{j=1}^{n} x_{j} T \epsilon_{j}=\sum_{j=1}^{n} x_{j} \sum_{i=1}^{m} A_{ij} \varepsilon_{i}=\sum_{i=1}^{m}\left(\sum_{j=1}^{n} A_{ij} x_{j}\right) \varepsilon_{i}}

so it is easy to see by comparison that

\displaystyle{f_{i}\left(x_{1}, \ldots, x_{n}\right)=\sum_{j=1}^{n} A_{ij} x_{j}, \quad i=1, \dots, m}

these m linear funtionals will make T \alpha=\left(f_{1}(\alpha), \ldots, f_{m}(\alpha)\right).

7. Let {\alpha}_1=(1,0,1,-2) and {\alpha}_2=(2,3,1,1), and let W be the subspace of R^4 spanned by \alpha_1 and \alpha_2. Which linear functionals f:

\displaystyle{f(x_1,x_2,x_3,x_4)=c_1x_1+c_2x_2+c_3x_3+c_4x_4}

are in the annihilator of W?

Solution: Since

\displaystyle{A=\left[\begin{matrix}{1} & {0} & {-1} & {2} \\ {2} & {3} & {1} & {1}\end{matrix}\right] \rightarrow R=\left[\begin{matrix}{1} & {0} & {-1} & {2} \\ {0} & {1} & {1} & {-1}\end{matrix}\right]}

if f is a linear functional on R^4:

\displaystyle{f(x_{1}, x_{2}, x_{3}, x_{4})=\sum_{j=1}^{4} c_{j} x_{j}}

then f is in the annihilator of W if and only if f(\alpha_{1})=0, f(\alpha_{2})=0, i.e., if and only if

\displaystyle{\sum_{j=1}^{4}A_{ij}c_{j}=0, \quad i=1,2}

this is equivalent to

\displaystyle{\sum_{j=1}^{4}R_{ij}c_{j}=0, \quad i=1,2}

or

c_{1}-c_{3}+2 c_{4}=0\\c_{2}+c_{3}-c_{4}=0

Solve it we have

\displaystyle{\left\{\begin{aligned}&{c_{1}=a-2 b} \\& {c_{2}=b-a} \\& {c_{3}=a} \\& {c_{4}=b}\end{aligned}\right.}

So if f \in W^0, then f (x_1, x_{2}, x_{3}, x_{4})=(a-2 b) x_{1}+(b-a) x_{2}+a x_{3}+b x_{4}, a, b \in R.

8. Let W be the subspace of R^5 which is spanned by the vectors

{\alpha}_1={\epsilon}_1+2{\epsilon}_2+{\epsilon}_3,\quad {\alpha}_2={\epsilon}_2+3{\epsilon}_3+3{\epsilon}_4+{\epsilon}_5 \\{\alpha}_3={\epsilon}_1+4{\epsilon}_2+6{\epsilon}_3+4{\epsilon}_4+{\epsilon}_5.

Find a basis for W^0.

Solution: We shall write \alpha_i in the coordinates of \{\epsilon_1,\ldots, \epsilon_5 \}, namely \alpha_1 =(1,2,1,0,0), \alpha_ 2=(0,1,3,3,1), \alpha _3=(1,4,6,4,1), form the 3\times5 matrix A whose rows are \alpha_i and we get

\displaystyle{A=\left[\begin{matrix}{1} & {2} & {1} & {0} & {0} \\ {0} & {1} & {3} & {3} & {1} \\ {1} & {4} & {6} & {4} & {1}\end{matrix}\right] \rightarrow R=\left[\begin{matrix}{1} & {0} & {0} & {4} & {3} \\ {0} & {1} & {0} & {-3} & {-2} \\ {0} & {0} & {1} & {2} & {1}\end{matrix}\right]}

then f is in the annihilator of W if and only if f(\alpha_i )=0,i=1,2,3, i.e., if and only if

\displaystyle{\begin{aligned}c_{1}+4 c_{4}+3 c_{5}&=0 \\c_{2}-3 c_{4}-2 c_{5}&=0\\c_{3}+2 c_{4}+c_{5}&=0\end{aligned}}

If we let c_4=a,c_5=b, then c_1=-4a-3b,c_2=3a+2b,c_3=-2a-b, so if f\in W^0, then f(x_{1}, \ldots, x_{5})=(-4 a-3 b) x_{1}+(3 a+2 b) x_{2}-(2 a+b) x_{3}+a x_{4}+b x_{5}, a basis for W^0 can be found by taking (a,b)=(1,0), and then (0,1), or

f_1 (x_{1}, \ldots, x_{5})= -4x_1+3x_2-2x_3+x_4\\ f_2 (x_{1}, \ldots, x_{5})= -3x_1+2x_2-x_3+x_5

9. Let V be the vector space of all 2\times 2 matrices over the field of real numbers, and let

\displaystyle{B=\begin{bmatrix}2&-2\-1&1\end{bmatrix}.}

Let W be the subspace of V consisting of all A such that AB=0. Let f be a linear functional on V which is in the annihilator of W. Suppose that f(I)=0 and f(C)=3, where I is the 2\times 2 identity matrix and

\displaystyle{C=\begin{bmatrix}0&0\\0&1\end{bmatrix}.}

Find f(B).

Solution: If A\in W,AB=0,then A shall satisfy 2A_{11}-A_{12}=0,-2A_{21}+A_{22}=0, thus we can write

\displaystyle{A=\begin{bmatrix}{a} & {2 a} \\ {c} & {2 c}\end{bmatrix}, \quad a, c \in R}

If f\in W^0, then f(A)=0,\forall A\in W, since

\displaystyle{B=\begin{bmatrix}{2} & {-2} \\ {-1} &{1}\end{bmatrix}=\begin{bmatrix}{-1} & {-2} \\ {-1} &{-2}\end{bmatrix}+\begin{bmatrix}{3} & {0} \\ {0} & {3}\end{bmatrix}=\begin{bmatrix}{-1} & {-2} \\ {-1} & {-2}\end{bmatrix}+3 I}

while D :=\begin{bmatrix}{-1} & {-2} \\ {-1} & {-2}\end{bmatrix}\in W, so f(B)=f(D+3I)=f(D)+3f(I)=0.

10. Let F be a subfield of the complex numbers. We define n linear functionals on F^n(n\geq 2) by

\displaystyle{f_k(x_1,\dots,x_n)=\sum_{j=1}^n(k-j)x_j,\quad 1\leq k\leq n.}

What is the dimension of the subspace annihilated by f_1,\dots,f_n?

Solution: Let W be the subspace annihilated by f_1,\cdots,f_n,then if (x_1,\cdots,x_n )\in W, we shall have

\displaystyle{f_{k}(x_{1}, \ldots, x_{n})=0, \quad 1 \leq k \leq n}

equivalently, W is the solutions space of the system

f_{1}(x_{1}, \ldots, x_{n})=0 \\ \vdots \\ f_{n}(x_{1}, \ldots, x_{n})=0

the coefficient matrix of the system A is row equivalent to R, which are

\displaystyle{A=\left[\begin{matrix}{0} & {-1} & {\cdots} & {-(n-1)} \\ {1} & {0} & {\cdots} & {-(n-2)} \\ {\cdots} & {\cdots} & {\cdots} & {\cdots} \\ {n-1} & {n-2} & {\cdots} & {0}\end{matrix}\right] \rightarrow R=\left[\begin{matrix}{0} & {1} & {\cdots} & {(n-1)} \\ {1} & {1} & {\cdots} & {1} \\ {\cdots} & {\cdots} & {\cdots} & {\cdots} \\ {0} & {0} & {\cdots} & {0}\end{matrix}\right]}

It is easy to see the dimension of the row space generated by R is 2, thus \dim W=n-2.

11. Let W_1 and W_2 be subspaces of a finite-dimensional vector space V.
( a ) Prove that (W_1+W_2)^0=W_1^0\cap W_2^0.
( b ) Prove that (W_1\cap W_2)^0=W_1^0+ W_2^0.

Solution:
( a ) First suppose f\in (W_1+W_2 )^0, then \forall\alpha\in W_1,\alpha+0\in W_1+W_2, so f(\alpha)=0, which means f\in W_1^0, similarly f\in W_2^0, so (W_1+W_2 )^0\subseteq W_1^0\cap W_2^0. Conversely, if f\in W_1^0\cap W_2^0, then \forall\alpha\in W_1 and \forall \beta \in W_2, f(\alpha)=f(\beta)=0, so given \forall\gamma\in(W_1+W_2), we can write

\displaystyle{\gamma=\alpha+\beta, \quad \alpha \in W_{1}, \beta \in W_{2}}

thus f(\gamma)=f(\alpha+\beta)=f(\alpha)+f(\beta)=0, so f\in (W_1+W_2)^0 and the proof is complete.

( b ) First let f\in W^0_1+W^0_2, then \exists f_1\in W_1^0, f_2\in W_2^0, s.t. f=f_1+f_2, so \forall \gamma \in W_1\cap W_2, we have

\displaystyle{f(\gamma)=f_1(\gamma)+f_2(\gamma)=0+0=0}

thus f\in (W_1\cap W_2)^0.
Conversely, let f\in (W_1\cap W_2)^0, we can find a basis of (W_1\cap W_2) to be \{\alpha_1,\ldots,\alpha_k\}, extend it to a basis of W_1: \{\alpha_1,\ldots,\alpha_k,\beta_1,\ldots,\beta_m\} and a basis of W_2: \{\alpha_1,\ldots,\alpha_k,\gamma_1,\ldots,\gamma_n\}, then

\displaystyle{\{\alpha_{1}, \ldots, \alpha_{k}, \beta_{1}, \ldots, \beta_{m}, \gamma_{1}, \ldots, \gamma_{n}\}}

is a basis for (W_1\cup W_2), we can further extend it to a basis of V, namely

\displaystyle{\{\alpha_{1}, \ldots, \alpha_{k}, \beta_{1}, \ldots, \beta_{m}, \gamma_{1}, \ldots, \gamma_{n}, \delta_1,\ldots,\delta_k\}}

of course in the process, k,m,n,l can be 0, in such cases the corresponding basis vanishes. Since f\in(W_1\cap W_2 )^0, we have

\displaystyle{f(\alpha_{j})=0, \quad 1 \leq j \leq k}

Now define f_1 and f_2 as

\displaystyle{f_{1}\left(\sum_{j=1}^kx_j{\alpha}_j+\sum_{j=1}^my_j{\beta}_j+\sum_{j=1}^nz_j{\gamma}_j+\sum_{j=1}^lw_j{\delta}_j\right)=\sum_{j=1}^nz_jf({\gamma}_j)+\sum_{j=1}^lw_jf({\delta}_j)}
\displaystyle{f_{2}\left(\sum_{j=1}^kx_j{\alpha}_j+\sum_{j=1}^my_j{\beta}_j+\sum_{j=1}^nz_j{\gamma}_j+\sum_{j=1}^lw_j{\delta}_j\right)=\sum_{j=1}^my_jf({\beta}_j)+\sum_{j=1}^lw_jf({\delta}_j)}

then we can see f_1\in W_1^0,f_2\in W_2^0, and f=f_1+f_2.

12. Let V be a finite-dimensional vector space over the field F and let W be a subspace of V. If f is a linear functional on W, prove that there is a linear functional g on V such that g(\alpha)=f(\alpha) for each \alpha in the subspace W.

Solution: Let \{\alpha_1,\ldots,\alpha_k\} be a basis for W, then we can extend it to a basis for V, namely \{\alpha_1,\ldots,\alpha_n\}. Now for \alpha=\sum_{i=1}^{n} x_{i} \alpha_{i} define g:V\rightarrow F to be

\displaystyle{g(\alpha)=g\left(\sum\limits_{i=1}^{n} x_{i} \alpha_{i}\right)=\sum\limits_{i=1}^{k} x_{i} f(\alpha_{i})}

it is easy to see g(\alpha)=f(\alpha) if \alpha \in W. To see g is linear, let \alpha=\sum_{i=1}^{n} x_{i} \alpha_{i} and \beta=\sum_{i=1}^{n} y_{i} \alpha_{i} in V, let c\in F, then

\displaystyle{\begin{aligned}g(c\alpha+\beta)&=g\left(c\sum\nolimits_{i=1}^{n} x_{i} \alpha_{i}+\sum\nolimits_{i=1}^{n} y_{i} \alpha_{i}\right)\\&=g\left(\sum_{i=1}^{n}\left(c x_{i}+y_{i}\right) \alpha_{i}\right)=\sum_{i=1}^{k}\left(c x_{i}+y_{i}\right) f\left(\alpha_{i}\right) \\&=c \sum_{i=1}^{k} x_{i} f\left(\alpha_{i}\right)+\sum_{i=1}^{k} y_{i} f\left(\alpha_{i}\right)=c g(\alpha)+g(\beta)\end{aligned}}

13. Let F be a subfield of the field of complex numbers and let V be any vector space over F. Suppose that f and g are lienar functionals on V such that the function h defined by h(\alpha)=f(\alpha)g(\alpha) is also a linear functional on V. Prove that either f=0 or g=0.

Solution: Assume f\neq 0 and g\neq 0, then \exists a,b\in V s.t. f(a)\neq 0,g(b)\neq 0, since h is a linear functional, for any c\in F we have

\displaystyle{h(ca+b)=ch(a)+h(b)}

on the other hand, we have

\displaystyle{\begin{aligned}h(ca+b)&=f(ca+b)g(ca+b)\\&=(cf(a)+f(b))(cg(a)+g(b))\\&=c^2 h(a)+c(f(a)g(b)+f(b)g(a))+h(b)\end{aligned}}

thus

\displaystyle{ch(a)-c^2 h(a)=c(f(a)g(b)+f(b)g(a))}

since it is true for all c, we must have

\displaystyle{h(a)-ch(a)=(1-c)f(a)g(a)=f(a)g(b)+f(b)g(a)}

which gives

\displaystyle{f(a)g(b)=(1-c)f(a)g(a)-f(b)g(a)=[(1-c)f(a)-f(b)]g(a)}

If we let c= \dfrac{f(a)-f(b)}{f(a)}, then f(a)g(b)=0 \cdot g(a)=0, which is a contradiction.

14. Let F be a field of chareacteristic zero and let V be a finite-dimensional vector space over F. If {\alpha}_1,\dots,{\alpha}_m are finitely many vectors in V, each different from the zero vector, prove that there is a linear functional f on V such that

\displaystyle{f(\alpha_i)\neq 0,\qquad i=1,\dots,m.}

Solution: I will prove by induction. For case m=1, as \alpha_1\neq 0, let f\in L(V,F) be any linear functional such that f(\alpha_1 )\neq 0 and we are done. Now suppose for nonzero vectors \{\alpha_1,\dots,\alpha_{m-1}\} in V, we can find a linear functional such that f(\alpha_i )\neq 0,i=1,\dots,m-1, consider the set \{\alpha_1,\dots,\alpha_m\}, if f(\alpha_m )\neq0, then f is what we want. Suppose f(\alpha_m )=0, expand \alpha_m to a basis \{\alpha_m,\beta_1,\cdots,\beta_n \} and define f':V\to F by

\displaystyle{f'\left(k \alpha_{m}+\sum\limits_{i=1}^{n} x_{i} \beta_{i}\right)=k P, \quad k, P, x_{1}, \ldots, x_{n} \in F}

Obviously f' is a linear functional, and we can have f' (\alpha_i )=k_i P,i=1,\dots,m-1, now for the linear functional g=f+f', we have

\displaystyle{g(\alpha_i)=f(\alpha_{i})+f^{\prime}\left(\alpha_{i}\right)=f\left(\alpha_{i}\right)+k_{i} P, \quad i=1, \ldots, m-1\\g(\alpha_m)=f'(\alpha_m)=P}

Since f(\alpha_i) and k_i are known elements in F, to make g(\alpha_i)\neq 0,i=1,\cdots,m, we only need the condition that P\neq 0, and when k_i\neq 0,P\neq -\dfrac{f(\alpha_i)}{k_i} for i=1,\dots,m-1, this is possible since P only needs to avoid at most m values.
So the induction is valid and then the conclusion holds.

15. According to Exercise 3, similar matrices have the same trace. Thus we can define the trace of a linear operator on a finite-dimensional space to be the trace of any matrix which represents the operator in an ordered basis. This is well defined since all such representing matrices for one operator are similar.
Now let V be the space of all 2\times 2 matrices over the field F and let P be a fixed 2\times 2 matrix. Let T be the linear operator on V defined by T(A)=PA. Prove that \text{trace }(T)=2\text{trace }(P).

Solution: Suppose P=\left[\begin{matrix}{P_{11}} & {P_{12}} \\ {P_{21}} & {P_{22}}\end{matrix}\right], and let one standard basis of V be

E_1=\left[\begin{matrix}{1} & {0} \\ {0} & {0}\end{matrix}\right], \quad E_{2}=\left[\begin{matrix}{0} & {1} \\ {0} & {0}\end{matrix}\right], \quad E_{3}=\left[\begin{matrix}{0} & {0} \\ {1} & {0}\end{matrix}\right], \quad E_{4}=\left[\begin{matrix}{0} & {0} \\ {0} & {1}\end{matrix}\right]

then let the matrix of T with respect to this basis be (A_{ij}), we shall have

\displaystyle{T(E_j)=PE_j=\sum^4_{i=1}A_{ij}E_i,\quad j=1,2,3,4}

A simple calculation will show

\displaystyle{A=\left[\begin{matrix}P_{11}&{0}&P_{12}&{0} \\ {0}&P_{11}&{0}&P_{12} \\ P_{21}&0&P_{22}&0 \\0&P_{21}&0&P_{22}\end{matrix}\right]}

thus trace(T)=trace(A)=2(P_{11}+P_{22} )=2trace(P).

16. Show that the trace functional on n\times n matrices is unique in the following sence. If W is the space of n\times n matrices over the field F and if f is a linear functional on W such that f(AB)=f(BA) for each A and B in W, then f is a scalar multiple of the trace function. If, in addition, f(I)=n, then f is the trace function.

Solution: Let A=(A_{ij}), and define \varepsilon_{ij} to be the n\times n matrix which has only 1 in the i-th row and j-th column, then \{\varepsilon_{ij}:1\leq i,j \leq n\} is a basis for W, and if a_{ij}:= f(\varepsilon_{ij}), then as f is a linear functional we can write

f \left(\sum\limits^{n}_{i=1}\sum\limits^n_{j=1}x_{ij}{\varepsilon}_{ij}\right)=\sum\limits^{n}_{i=1}\sum\limits^n_{j=1}x_{ij}a_{ij}

Notice that for any 1\leq i\leq n,

\displaystyle{A\varepsilon_{ii}=\left[\begin{matrix} {\quad}&A_{1i}&{\quad}\\{\quad}&\vdots &{\quad}\\{\quad}&A_{ni}&{\quad} \end{matrix}\right], \quad \varepsilon_{ii}A=\left[\begin{matrix} {\quad}&{\quad}&{\quad}\\{A_{i1}}&\cdots &{A_{in}}\\{\quad}&{\quad}&{\quad} \end{matrix}\right]}

thus f(A\varepsilon_{ii})=\sum_{j=1}^n A_{ji}a_{ji}=f(\varepsilon_{ii}A)=\sum_{j=1}^n A_{ij}a_{ij}, since A_{ij} can be arbitrary values, we have

\displaystyle{a_{ij}=a_{ji}=0,\quad j\neq i}

thus

f \left(\sum\limits^{n}_{i=1}\sum\limits^n_{j=1}x_{ij}\varepsilon_{ij}\right)=\sum\limits^{n}_{i=1}x_{ii}a_{ii}

Next let E_j,2\leq j\leq n be the matrix which is the identity matrix plus E_{1j}=1, then

\displaystyle{AE_j = \left[\begin{matrix} A_{11}&\cdots&{A_{1j}+A_{11}}&\cdots&A_{1n} \\ \vdots&\cdots&\vdots&\cdots&\vdots \\ A_{n1}&\cdots&{A_{nj}+A_{n1}}&\cdots&A_{nn} \end{matrix}\right],\quad E_jA=\left[\begin{matrix} {A_{11}+A_{j1}}&\cdots&{A_{1n}+A_{jn}}\\ \cdots&\cdots&\cdots \\ {A_{j1}}&\cdots&A_{jn} \\ \cdots&\cdots&\cdots \\ {A_{n1}}&\cdots&A_{nn} \end{matrix}\right]}

which means f\left(AE_j\right)=\sum_{i=1}^n A_{ii}a_{ii}+A_{j1}a_{jj}=f\left(E_jA\right)=\sum_{i=1}^n A_{ii}a_{ii}+A_{j1}a_{11}, thus

\displaystyle{a_{jj}=a_{11},\quad 2\leq j \leq n}

so let a=a_{11}, we have f(A)=a\sum_{i=1}^nA_{ii}, i.e., f is the scalar multiple of the trace function.
If f(I)=n, then a\sum_{i=1}^nA_{ii}=an=n, thus a=1 and f is the trace function.

17. Let W be the space of n\times n matrices over the field F, and let W_0 be the subspace spanned by the matrices C of the form C=AB-BA. Prove that W_0 is exactly the subspace of matrices which have trace zero.

Solution: Let the space of traceless matrices be S. First for C\in W_0, we have C=AB-BA, so by Exercise 3.5.3 we have

\displaystyle{trace(C)=trace(AB-BA)=trace(AB)-trace(BA)=0}

thus W_0 \subseteq S.
Conversely, it is easy to see \dim S=n^2-1, since only one condition is needed to make a matrix A traceless, namely trace(A)=0. Next, define \varepsilon_{ij} to be the n\times n matrix which has only 1 in the i-th row and j-th column, and define

\displaystyle{E_i=\varepsilon_{in}\varepsilon_{ni}-\varepsilon_{ni}\varepsilon_{in},\quad 1\leq i\leq n-1}

then it is easy to see E_i only has two nonzero entries: 1 at the ii-th position and -1 at the nn-th position. So E_i\in S, also we can verify

\displaystyle{\varepsilon_{in}\varepsilon_{ni}-\varepsilon_{ni}\varepsilon_{in}=\varepsilon_{ij},\quad i\neq j }

Obviously \varepsilon_{ij}\in S,i\neq j and for i\neq j, there are n^2-n such matrices, adding them with E_1,\dots,E_{n-1}, we get n^2-1 matrices in S, to show they are linearly independent, notice that if some linear combination of them equals the n\times n zero matrix, then the ij-th entry of the sum is uniquely determined by the coefficient of \varepsilon_{ij} when i\neq j, thus is 0. And the ii-th entry of the sum is uniquely determined by the coefficient of E_i when 1\leq i\leq n-1, thus 0, too. So the n^2-1 matrices are linearly independent, thus a basis of S, so S\subseteq W_0 as W_0 is a subspace.
Combined we get S=W_0.