Linear Algebra (2ed) Hoffman&Kunze 3.2

christangdt留下评论

这一节内容非常丰富。Theorem 4提出了linear transformation的线性组合是什么，以及所有的linear transformation是一个vector space。Theorem 5是说明L(V,W)和V,W的dimension之间的关系。Theorem 6定义了linear transformation的composition。在同一个space上的linear transformation称为operator，有翻译为算子的。EXAMPLE8-10都是很好的例子，其揭示了linear operator可以是多样的。之后定义invertible的概念，注意invertible是在两个不同的space中可以成为单位算子，Theorem 7说明inverse也是linear transformation。然后是non-singular的概念，很类似于函数中的injective，Theorem 8说明non-singular的线性变换可以保持线性无关性。EXAMPLE11说明non-singular和onto之间没有必然的关系，EXAMPLE12则是一个同时non-singular和onto（从而invertible）的例子。Theorem 9说明，如果\dim V=\dim W，那么从V到W的线性变换T满足non-singular、onto和invertible三者成立其一则另外两者也成立。最后给出了group和commutative group的定义。

Exercises

1. Let T and U be the linear operators on R^2 defined by

\displaystyle{T(x_1,x_2)=(x_2,x_1)\quad\text{and}\quad U(x_1,x_2)=(x_1,0)}

( a ) How would you describe T and U geometrically?
( b ) Give rules like the ones definning T and U for each of the transformations (U+T),UT,TU,T^2,U^2.

Solution:
( a ) T is a symmetric transformation with respect to the line y=x, U is the projection on x-axis.
( b ) We have

\displaystyle{\begin{aligned}(U+T)(x_1,x_2 )&=(x_1+x_2,x_1 ) \\ UT(x_1,x_2 )&=(x_2,0) \\ TU(x_1,x_2 )&=(0,x_1 ) \\ T^2 (x_1,x_2 )&=(x_1,x_2 ) \\ U^2 (x_1,x_2 )&=(x_1,0)\end{aligned}}

2. Let T be the (unique) linear operator on C^3 for which

\displaystyle{T\epsilon_1 =(1,0,i),\quad T\epsilon_2 =(0,1,1),\quad T\epsilon_3 =(i,1,0).}

Is T invertible?

Solution: No, since T({\epsilon}_3-i{\epsilon}_1 )=(i,1,0)-(i,0,-1)=(0,1,1)=T{\epsilon}_2, thus

\displaystyle{T({\epsilon}_3-i{\epsilon}_1-{\epsilon}_2 )=0}

but {\epsilon}_3-i{\epsilon}_1-{\epsilon}_2=(-i,-1,1)\neq 0.

3. Let T be the linear operator on R^3 defined by

\displaystyle{T(x_1,x_2,x_3)=(3x_1,x_1-x_2,2x_1+x_2+x_3)}

Is T invertible? If so, find a rule for T^{-1} like the one which defines T.

Solution: T is invertible, the rule for T^{-1} can be described as

\displaystyle{T^{-1} (x_1,x_2,x_3 )=\left(\frac{1}{3} x_1,\frac{1}{3} x_1-x_2,-x_1+x_2+x_3 \right)}

4. For the liear operator T of Exercise 3, prove that

\displaystyle{(T^2-I)(T-3I)=0.}

Solution: Since

\displaystyle{\begin{aligned}(T-3I)(x_1,x_2,x_3 )&=T(x_1,x_2,x_3 )-3(x_1,x_2,x_3 )\\&=(0,x_1-4x_2,2x_1+x_2-2x_3 )\end{aligned}}

and

\displaystyle{\begin{aligned}&{\quad}(T^2-I)(x_1,x_2,x_3 )\\&=T^2 (x_1,x_2,x_3 )-(x_1,x_2,x_3 )\\&=T(3x_1,x_1-x_2,2x_1+x_2+x_3 )-(x_1,x_2,x_3 )\\&=(9x_1,3x_1-(x_1-x_2 ),6x_1+(x_1-x_2 )+(2x_1+x_2+x_3 ))-(x_1,x_2,x_3 )\\&=(9x_1,2x_1+x_2 ,9x_1+x_3)-(x_1,x_2,x_3 )\\&=(8x_1,2x_1,9x_1)\end{aligned}}

so

\displaystyle{(T^2-I)(T-3I)(x_1,x_2,x_3)=(T^2-I)(0,x_1-4x_2,2x_1+x_2-2x_3)=(0,0,0)}

5. Let C^{2\times 2} be the complex vector space of 2\times 2 matrices with complex entries. Let

\displaystyle{B=\begin{bmatrix}1&-1\\-4&4\end{bmatrix}}

and let T be the linear operator on C^{2\times 2} defined by T(A)=BA. What is the rank of T? Can you describe T^2?

Solution: If A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, then

\displaystyle{T(A)=BA=\begin{bmatrix}1&-1\\-4&4\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a-c&b-d\\4c-4a&4d-4b\end{bmatrix}}

the rank of T is 2. Also we have

\displaystyle{\begin{aligned}T^2(A)=T(T(A))&=\begin{bmatrix}1&-1\\-4&4\end{bmatrix}\begin{bmatrix}a-c&b-d\\4c-4a&4d-4b\end{bmatrix}\\&=\begin{bmatrix}5a-5c&5b-5d\\20c-20a&20d-20b\end{bmatrix}\end{aligned}}.

6.Let T be a linear transformation from R^3 into R^2, and let U be a linear transformation from R^2 into R^3. Prove that the transformation UT is not invertible. Generalize the theorem.

Solution: Let a=T{\epsilon}_1,b=T{\epsilon}_2,c=T{\epsilon}_3, then a,b,c\in R^2, thus are linearly dependent, without loss of generality, we suppose c=k_1 a+k_2 b,k_1,k_2\in R, then k_1 a+k_2 b-c=0, or

\displaystyle{k_1 T{\epsilon}_1+k_2 T{\epsilon}_2-T{\epsilon}_3=T(k_1 {\epsilon}_1+k_2 {\epsilon}_2-{\epsilon}_3 )=0}

Notice k_1 {\epsilon}_1+k_2 {\epsilon}_2-{\epsilon}_3\neq 0, we can have

\displaystyle{UT(k_1 {\epsilon}_1+k_2 {\epsilon}_2-{\epsilon}_3 )=U(0)=0}

thus UT is not invertible.

7. Find two linear operators T and U on R^2 such that TU=0 but UT\neq 0.

Solution: Let U(x_1,x_2 )=(x_1,0),T(x_1,x_2 )=(x_2,0), then

\displaystyle{TU(x_1,x_2 )=T(x_1,0)=(0,0),\forall x_1,x_2\in R\\UT(0,1)=U(1,0)=(1,0)}

8. Let V be a vector space over the field F and T a linear operator on V. If T^2=0, what can you say about the relation of the range of T to the null space of T? Give an example of a linear operator T on R^2 such that T^2=0 but T\neq 0.

Solution: If T^2=0, then for any {\alpha} \in \text{range }T, we have {\alpha} =T{\beta},{\beta}\in V, thus T{\alpha} =T^2 {\beta}=0, so {\alpha} also belongs to the null space of T, it follows that \text{range }T is a subspace of the null space of T.
T(x_1,x_2 )=(x_2,0) is an example of T^2=0, while T\neq 0.

9. Let T be a linear operator on the finite-dimensional space V. Suppose there is a linear operator U on V such that TU=I. Prove that T is invertible and U=T^{-1}. Give an example which shows that this is false when V is not finite dimensional.

Solution: For any {\alpha} \in V, we have {\alpha} =(TU)({\alpha} )=T(U{\alpha}), this means {\alpha} \in \text{range }T, so the range of T is V, by Theorem 9, T is invertible. To see U=T^{-1}, notice that T^{-1} T=I and TU=I, thus

\displaystyle{U=IU=(T^{-1} T)U=T^{-1} (TU)=T^{-1} I=T^{-1}}

If T=D on the space of polynomial functions, then let U be the integrable operator on the space of polynomial functions, we have DU=I, but D is not invertible.

10. Let A be an m\times n with entries in F and let T be the linear transformation from F^{n\times 1} into F^{m\times 1} defined by T(X)=AX. Show that if mn we may have T non-singular but not onto.

Solution: If m<n, then the system AX=b can have a solution if rank A=m, so T can be onto, but AX=0 must have an nonzero solution, since the solution space has dimension n-m\geq 1.
If m>n, then let AX=x_1 A_1+\dots+x_n A_n, each A_i a m\times 1 vector, as long as those A_i are linearly independent, AX=0 will mean x_i=0 and X=0, so T is non-singular, but \dim F^{m\times 1}=m and A_1,\dots,A_n cannot span F^{m\times 1}, thus T is not onto.

11. Let V be a finite-dimensional vector space and let T be a linear operator on V. Suppose that \text{rank }(T^2)=\text{rank }(T). Prove that the range and null space of T are disjoint, i.e., have only the zero vector in common.

Solution: Assume {\alpha} \neq 0,{\alpha} \in \text{range }T\cap \text{null }T, then T{\alpha} =0, and {\alpha} is a linear independent set in \text{range }T, thus can be expanded to a basis of range T, namely ({\alpha} ,{\beta}_1,\dots,{\beta}_n ), so \text{rank }(T)=n+1. Notice that \text{range }T^2 is the set of vectors T^2 {\beta},{\beta}\in V, and T^2 {\beta}=T(T{\beta}), as T{\beta}\in \text{range }T, we have T{\beta}=y{\alpha} +\sum_{i=1}^ny_i {\beta}_i, in which y,y_1,\cdots,y_n are scalars. So T^2 {\beta}=T(y{\alpha} +\sum_{i=1}^ny_i {\beta}_i)=\sum_{i=1}^ny_i T{\beta}_i, which means T{\beta}_1,\dots,T{\beta}_n spans range T^2, so \text{rank }(T^2 )\leq n, a contradiction.

12. Let p,m and n be positive integers and F a field. Let V be the space of m\times n matrices over F and W the space of p\times n matrices over F. Let B be a fixed p\times m matrix and let T be the linear transformation from V into W defined by T(A)=BA. Prove that T is invertible if and only if p=m and B is an invertible m\times m matrix.

Solution: If p=m and B is an invertible m\times m matrix, then define U(A)=B^{-1} A, we have TU(A)=T(B^{-1} A)=B(B^{-1}A)=(BB^{-1})A=A and UT(A)=U(BA)=B^{-1} (BA)=(BB^{-1})A=A. Thus T is invertible and T^{-1} (A)=B^{-1} A.
Conversely, if T is invertible, then T is non-singular and \text{range }T=W, use Exercise 3.2.10 we know if p>m, then T cannot be onto, if p<m, then T cannot be non-singular. Thus p=m and B is a m\times m matrix, if B is not invertible then the column vectors of B are linearly dependent, thus there is x_1,\dots,x_m not all 0 such that x_1 b_1+\dots+x_m b_m=0, then let A be a matrix with x_1,\dots,x_m be in the first column and 0 otherwise, we have T(A)=0, this contradicts T being non-singular. Thus B must be invertible.

Linear Algebra (2ed) Hoffman&Kunze 3.1

christangdt留下评论

从这一章开始，就讨论本书的主题线性变换了。定义很像subspace的定义，线性变换是一个满足T(c\alpha+\beta)=cT(\alpha)+T(\beta)的函数。常见的例子有zero transformation，differentiation transformation，矩阵乘法的transformation，积分transformation等。
如果T是一个linear transformation，那么T(0)=0，且T保持linear combination。Theorem 1说明V的一组基可以用唯一的线性变换指到W的任何一组\dim V数量的向量上，特别地，任意一个linear transformation T 是唯一地由其 images of standard basis 决定。并且可以使用由 images of standard basis 为行组成的矩阵来显式表示出来。
接下来讨论的是range和null space，二者都是subspace，由定义可以证。range的dimension称为rank of T，null space的dimension称为nullity of T。Theorem 2是线性代数中一个很重要的定理：rank(T)+nullity(T)=\dim V。由此可以得到关于矩阵的Theorem 3：矩阵的行rank和列rank相同。证明中要使用到方程组解空间的维数是n-r，其中r为系数矩阵行空间dimension的结论。

Exercises

1. Which of the following functions T from R^2 into R^2 are linear transformations?
( a ) T(x_1,x_2)=(1+x_1,x_2)
( b ) T(x_1,x_2)=(x_2,x_1)
( c ) T(x_1,x_2)=(x_1^2,x_2)
( d ) T(x_1,x_2)=(\sin x_1,x_2)
( e ) T(x_1,x_2)=(x_1-x_2,0)

Solution:
( a ) No, since T(2(1,0))=T(2,0)=(1+2,0)=(3,0)\neq 2(2,0)=2T(1,0).
( b ) Yes, since

\begin{aligned}T(c(x_1,x_2)+(y_1,y_2))&=(cx_2+y_2,cx_1+y_1)\\&=c(x_2,x_1)+(y_2,y_1)\\&=cT(x_1,x_2)+T(y_1,y_2)\end{aligned}

( c ) No.
( d ) No.
( e ) Yes, since

\begin{aligned}T(c(x_1,x_2)+(y_1,y_2))&=(cx_1+y_1-cx_2-y_2,0)\\&=c(x_1-x_2,0)+(y_1-y_2,0)\\&=cT(x_1,x_2)+T(y_1,y_2)\end{aligned}

2. Find the range, rank, null space, and nullity for the zero transformation and the identity transformation on a finite-dimensional space V.

Solution: Let Z be the linear transformation, and I be the identity transformation, then
the range of Z is 0, the rank of Z is 0, the null space of Z is V, the nullity of Z is \dim V.
the range of I is V, the rank of I is \dim V, the null space of I is 0, the nullity of I is 0.

3. Describe the range and the null space for the differentiation transformation of Example 2. Do the same for the integration transformation of Example 5.

Solution:
For Example 2, range D=V, null D=\{f(x):f(x)=k,k\in F\}.
For Example 5, null T={0}, range T=\{f:f \text{ has a continuous first derivative}\}.

4. Is there a linear transformation T from R^3 into R^2 such that T(1,-1,1)=(1,0) and T(1,1,1)=(0,1)?

Solution: Let T be defined as

T(1,-1,1)=(1,0),\quad T(1,1,1)=(0,1),\quad T(0,0,1)=(0,0)

Since (1,-1,1),(1,1,1),(0,0,1) is a basis for R^3, T is well defined.

5. If

\alpha_1=(1,-1),\quad \beta_1=(1,0) \\ \alpha_2=(2,-1),\quad \beta_2=(0,1) \\ \alpha_3=(-3,2),\quad \beta_3=(1,1)

is there a linear transformation T from R^2 into R^2 such that T\alpha_i=\beta_i for i=1,2,3?

Solution: Assume such a T exists, then T(-\alpha_1-\alpha_2 )=-T\alpha_1-T\alpha_2, notice that -\alpha_1-\alpha_2=\alpha_3, we have

\beta_3=T\alpha_3=-T\alpha_1-T\alpha_2=-\beta_1-\beta_2=-(1,1)=-\beta_3

this means \beta_3=(0,0), a contradiction.

6. Describe explicitly (as in Exercises 1 and 2) the linear transformation T from F^2 into F^2 such that T\epsilon_1=(a,b),T\epsilon_2=(c,d).

Solution:
\begin{aligned}T(x_1,x_2 )&=x_1 T\epsilon_1+x_2 T\epsilon_2=x_1 (a,b)+x_2 (c,d)\\&=(x_1 a+x_2 c,x_1 b+x_2 d)\end{aligned}
If a=b=c=d=0, then T is the zero transformation.
If at least one of a,b,c,d is not 0, but ad-bc=0, then the rank of T and the nullity of T are both 1. The range of T is \{(0,y):y\in R\} if a=c=0, \{(x,0):x\in R\} if b=d=0, \{k(c,d):k\in R\} if a=b=0, \{k(a,b):k\in R\} if c=d=0, and \{k(c,d):k\in R\} otherwise.
If ad-bc\neq 0, then the rank of T is 2 and the nullity of T is 0, The range of T is R^2, the null space of T is \{0\}.

7. Let F be a subfield of the complex numbers and let T be the function from F^3 into F^3 defined by

T(x_1,x_2,x_3)=(x_1-x_2+2x_3,2x_1+x_2,-x_1-2x_2+2x_3).

( a ) Verify that T is a linear transformation.
( b ) If (a,b,c) is a vector in F^3, what are the conditions on a,b,c that the vectors be in the range of T? What is the rank of T?
( c ) What are the conditions on on a,b,c that (a,b,c) be in the null space of T? What is the nullity of T?

Solution:
( a ) Let x=(x_1,x_2,x_3),y=(y_1,y_2,y_3), then it’s easy to see
T(cx+y)=T(cx_1+y_1,cx_2+y_2,cx_3+y_3 )=cTx+Ty
( b ) The conditions are a-b=c, the rank of T is 2.
( c ) (a,b,c)=(2t,-4t,-3t),t\in R, the nullity of T is 1.

8. Describe explicitly a linear transformation from R^3 into R^3 which has as its range the subspace spaned by (1,0,-1) and (1,2,2).

Solution: One possible choice of T can be

T(x_1,x_2,x_3 )=(x_1+x_2,2x_2+2x_3,-x_1+2x_2+3x_3)

Since T\epsilon_1=(1,0,-1),T\epsilon_2=(1,2,2),T\epsilon_3=(0,2,3)=T\epsilon_2-T\epsilon_1, the condition is satisfied.

9. Let V be the vector space of all n\times n matrices over the field F, and let B be a fixed n\times n matrix. If

T(A)=AB-BA

verify that T is a linear transformation from V into V.

Solution: Let A,D be any matrix in V, then

\begin{aligned}T(cA+D)&=(cA+D)B-B(cA+D)\\&=cAB+DB-cBA-BD\\&=c(AB-BA)+(DB-BD)\\&=cT(A)+T(D)\end{aligned}

10. Let V be the set of all complex numbers regarded as a vector space over the field of real numbers (usual operations). Find a function from V into V which is a linear transformation on the above vector space, but which is not a linear transformation on C^1, i.e., which is not complex linear.

Solution: Let f be defined as f(a+bi)=a for a+bi\in C (in which a,b\in R), then f is a linear transformation from V into V. To see it’s not a transformation on C^1, we have f(1+i)=1, and f(i)=0, but for c=1+i we have

f(c(1+i)+i)=f(3i)=0\neq cf(1+i)+f(i)

11. Let V be the space of n\times 1 matrices over F and let W be the space of m\times 1 matrices over F. Let A be a fixed m\times n matrix over F and let T be the linear transformation from V into W defined by T(X)=AX. Prove that T is the zero transformation if and only if A is the zero matrix.

Solution: If A is the zero matrix, then obviously T is the zero transformation. Now suppose T is the zero transformation and assume A\neq 0, then at least one a_{ij}\neq 0, here 1\leq i\leq m,1\leq j\leq n, we let X=a_{ij} \epsilon_j in V, the j-th coordinate of X is a_{ij} and the other coordinates are 0, then the i-th coordinate of AX is a_{ij}^2\neq 0, thus T(X)=AX\neq 0, a contradiction.

12. Let V be an n-dimensional vector space over the field F and let T be a linear transformation from V into V such that the range and null space of T are identical. Prove that n is even. (Can you give an example of such a linear transformation T)?

Solution: We have rank(T)=nullity(T), since the range of T and the null space of T are equal, thus

n=\dim V=rank(T)+nullity(T)=2rank(T)

so n is even. One example may be T(x_1,x_2 )=(x_2,0).

13. Let V be a vector space over the field F and T a linear transformation from V into V. Prove that the following two statements about T are equivalent.
( a ) The intersection of the range of T and the null space of T is the zero subspace of V.
( b ) If T(T\alpha)=0, then T(\alpha)=0.

Solution: Let A be the range of T and B be the null space of T.
First suppose (a) is true, then A\cap B=\{0\}, if T(T\alpha)=0, then T\alpha\in B, but obviously T\alpha\in A, thus T\alpha\in A\cap B and T\alpha=0.
Conversely, suppose T(T\alpha)=0 means T\alpha=0, and let x\in A\cap B, then Tx=0, and there’s \beta\in V s.t. T\beta=x, so we have T(T\beta)=Tx=0, so T\beta=0=x, thus A\cap B=\{0\}.

Linear Algebra (2ed) Hoffman&Kunze 2.6

christangdt留下评论

这一节并没有定理内容，完全关于计算。核心是解决以下三个问题：

  1. 如何确定\alpha_1,\dots,\alpha_m张成的subspace W的维度？
  2. 怎么确定是否\beta\in span(\alpha_1,\dots,\alpha_m)？
  3. 如何显式地描述W？也就是W中的向量以最低自由度表示有什么形态？
    第一个问题可以通过化简\alpha_1,\dots,\alpha_m组成的矩阵至row-reduced echelon来解决；第三个问题

可以通过代表行变换的可逆矩阵P来得到；第二个问题则是将\alpha_1,\dots,\alpha_m作为列向量得到一个方程组，用增广矩阵方法确定方程组是否有解，这个实质是linear combination，比较好理解。
后面举了两个例子说明此问题。

Exercises

1. Let s<n and A an s\times n matrix with entries in the field F. Use Theorem 4 (not its proof) to show that there is a non-zero X in F^{n\times 1} such that AX=0.

Solution: Write A=[\alpha_1,\dots,\alpha_n], where each \alpha_i\in F^s, then AX=0 is equivalent to \sum_{i=1}^nx_i \alpha_i, by Theorem 4, any independent set of vectors in F^s contains no more than s elements, as s<n, the set of vectors {\alpha_1,\dots,\alpha_n } is linearly dependent, thus there must exists c_1, \dots,c_n not all zero such that \sum_{i=1}^nc_i \alpha_i=0, this (c_1,\dots,c_n)\in F^{n\times 1} is the X we need.

2. Let

\alpha_1=(1,1,-2,1),\quad \alpha_2=(3,0,4,-1),\quad \alpha_3=(-1,2,5,2).

Let

\alpha=(4,-5,9,-7),\quad \beta=(3,1,-4,4),\quad \gamma=(-1,1,0,1).

( a ) Which of the vectors \alpha,\beta,\gamma are in the subspaces of R^4 spanned by the \alpha_i?
( b ) Which of the vectors \alpha,\beta,\gamma are in the subspaces of C^4 spanned by the \alpha_i?
( c ) Does this suggest a theorem?

Solution:
( a ) We perform elementary row opeartions on [\alpha_1,\alpha_2,\alpha_3,\alpha,\beta,\gamma].

\begin{aligned}\begin{bmatrix}1&3&-1&y_1\\1&0&2&y_2\\-2&4&5&y_3\\1&-1&2&y_4\end{bmatrix}&\rightarrow\begin{bmatrix}1&0&2&y_2\\0&3&-3&y_1-y_2\\0&4&9&y_3+2y_2\\0&-1&0&y_4-y_2\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&2&y_2\\0&1&0&y_2-y_4\\0&0&-3&y_1-y_2+3(y_4-y_2 )\\0&0&9&y_3+2y_2+4(y_4-y_2 )\end{bmatrix}\\&\rightarrow\begin{bmatrix}1&0&2&y_2\\0&1&0&y_2-y_4\\0&0&-3&y_1-4y_2+3y_4\\0&0&0&3y_1-14y_2+y_3+13y_4\end{bmatrix}\end{aligned}

For \alpha we have 12+70+9-91=0, thus \alpha\in span\{\alpha_1,\alpha_2,\alpha_3 \}, since 9-14-4+42\neq 0 and -3-14+13\neq 0, both \beta and \gamma is not in the subspace.
( b ) The answer is the same with (a)
( c ) If F' is a subspace of F, then U is a subspace of F' if and only if U is a subspace of F.

3. Consider the vectors in R^4 defined by

\alpha_1=(-1,0,1,2),\quad \alpha_2=(3,4,-2,5),\quad \alpha_3=(1,4,0,9).

Find a system of homogeneous linear equations for which the space of solutions is exactly the subspace of R^4 spanned by the three given vectors.

Solution: First verify if \alpha_1,\alpha_2,\alpha_3 is linearly independent:

\begin{bmatrix}-1&0&1&2\\3&4&-2&5\\1&4&0&9\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-1&-2\\0&4&1&11\\0&4&1&11\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-1&-2\\0&4&1&11\\0&0&0&0\end{bmatrix}

thus \alpha_1,\alpha_3 can be a basis for span(\alpha_1,\alpha_2,\alpha_3 ), and one system may be

4x_1-x_2+4x_3=0 \\8x_1-11x_2+4x_4=0

4. In C^3, let

\alpha_1=(1,0,i),\quad \alpha_2=(1+i,1-i,1),\quad \alpha_3=(i,i,i).

Prove that these vectors form a basis for C^3. What are the coordinates of the vectors (a,b,c) in this basis?

Solution: Let c_1 \alpha_1+c_2 \alpha_2+c_3 \alpha_3=0, then

c_1+c_2 (1+i)+c_3 i=0 \implies c_1+c_2=0,c_2+c_3=0 \\c_2 (1-i)+c_3 i=0 \implies c_2=0 \implies c_1=c_3=0

thus the three vectors are linearly independent.
To show \alpha_1,\alpha_2,\alpha_3 may not be a basis for C^3, let x_1 \alpha_1+x_2 \alpha_2+x_3 \alpha_3=(a,b,c), then

\begin{bmatrix}1&1+i&i&a\\0&1-i&i&b\\-i&1&i&c\end{bmatrix}\rightarrow\begin{bmatrix}1&1+i&i&a\\0&1-i&i&b\\0&i&i-1&c+ai\end{bmatrix}\\ \rightarrow\begin{bmatrix}1&2i&0&a-b\\0&1&2i-1&b+c+ai\\0&0&1+2i&c-a+(a-b-c)i\end{bmatrix}

If (a,b,c)=(1+2i,2i,0), then from the first two equations we have x_1=1,x_2=0, and x_3=2,
thus the last equation means -i+2i=c, or i=0, a contradiction.

5. Given an explicit description of the type (2-25) for the vectors \beta=(b_1,b_2,b_3,b_4,b_5) in R^5 which are linear combinations of the vectors

\alpha_1=(1,0,2,1,-1),\quad \alpha_2=(-1,2,-4,2,0),\\ \alpha_3=(2,-1,5,2,1)\quad \alpha_4=(2,1,3,5,2).

Solution: We have

\begin{aligned}\begin{bmatrix}1&-1&2&2&b_1\\0&2&-1&1&b_2\\2&-4&5&3&b_3\\1&2&2&5&b_4\\-1&0&1&2&b_5 \end{bmatrix}&\rightarrow\begin{bmatrix}1&-1&2&2&b_1\\0&2&-1&1&b_2\\0&-2&1&-1&b_3-2b_1\\0&3&0&3&b_4-b_3\\0&-1&3&4&b_1+b_5 \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&-1&2&2&b_1\\0&-1&3&4&b_1+b_5\\0&3&0&3&b_4-b_3\\0&2&-1&1&b_2\\0&0&0&0&b_3+b_2-2b_1 \end{bmatrix}\\&\rightarrow\begin{bmatrix}1&-1&2&2&b_1\\0&-1&3&4&b_5+b_1\\0&0&5&9&2b_1+b_2+2b_5\\0&0&9&15&3b_1-b_3+b_4+3b_5\\0&0&0&0&b_3+b_2-2b_1 \end{bmatrix}\end{aligned}

thus those vectors which are linear combinations of \alpha_1,\alpha_2,\alpha_3,\alpha_4 are those \beta which satisfy b_3+b_2=2b_1.

6. Let V be the real vector space spaned by the rows of the matrix

A=\begin{bmatrix}3&21&0&9&0\\1&7&-1&-2&-1\\2&14&0&6&1\\6&42&-1&13&0\end{bmatrix}

( a ) Find a basis for V.
( b ) Tell which vectors (x_1,x_2,x_3,x_4,x_5) are elements of V.
( c ) If (x_1,x_2,x_3,x_4,x_5) is in V what are its coordinates in the basis chosen in part (a)?

Solution:
( a ) We have

\begin{bmatrix}3&21&0&9&0\\1&7&-1&-2&-1\\2&14&0&6&1\\6&42&-1&13&0\end{bmatrix}\rightarrow\begin{bmatrix}1&7&0&3&0\\0&0&-1&-5&-1\\0&0&0&0&1\\0&0&-1&-5&0\end{bmatrix}\rightarrow\begin{bmatrix}1&7&0&3&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&1&5&0\end{bmatrix}

So a basis for V is (1,7,0,3,0),(0,0,0,0,1),(0,0,1,5,0).
( b ) Those (x_1,x_2,x_3,x_4,x_5) which satisfy x_2=7x_1 and x_4=3x_1+5x_3.
( c ) The coordinates are (x_1,x_5,x_3).

7. Let A be an m\times n matrix over the field F, and consider the system of equations AX=Y. Prove that this system of equations has a solution if and only if the row rank of A is equal to the row rank of the augmented matrix of the system.

Solution: Perform elementary row operations on A to get a row-reduced echelon matrix R, the same operations will lead A'=[A,Y] to a matrix [R,Y'], the system has a solution if and only if the nonzero rows of [R,Y'] is the same as the nonzero rows of R, which means their row ranks are the same. The conclusion follows since row rank of A equals row rank of R, and row rank of A' equals row rank of [R,Y'], by Theorem 9.