Exercises: Dimension

Solution. Suppose not, that is, suppose that \(v_i = v_j\) with \(i \ne j\) for some \(i,\) \(j.\) Then \[0v_1 + \cdots + 1 v_i + (-1)v_j + \cdots + 0v_n = \mathbf{0},\] which means \(v_1 ,\) \(\cdots ,\) \(v_n \) are not linearly independent.

Solution. Suppose that \[a (1,\,1,\,1) + b(0,\,2,\,5) = (0,\,0,\,0)\] for scalars \(a\) and \(b.\) This equation yields \[(a,\,a+2b ,\, a+5b) = (0,\,0,\,0)\] or \[\begin{cases} a=0 \\[6pt] a+2b = 0 \\[6pt] a+5b = 0 . \end{cases}\] This system only has the solution \(a=0 ,\) \(b=0.\)

Solution. Suppose that \[a \sin x + b \cos x = 0 \tag{*}\] for scalars \(a\) and \(b,\) and for all real numbers \(x.\)

Suppose further that \(a \ne 0\) or \(b\ne 0.\) Then there exists \(\theta\) uniquely such that \[\cos \theta = \frac{a}{\sqrt{a^2 + b^2}} ,\,\, \sin \theta = \frac{b}{\sqrt{a^2 + b^2}} ,\,\, 0 \le \theta < 2\pi .\] Thus (*) becomes \[\cos\theta \sin x + \sin\theta \cos x = 0\] or \[\sin(x+\theta ) = 0 .\] But this equation holds only for some \(x,\) not all \(x.\) Therefore both \(a\) and \(b\) have to equal zero.

Solution. Take \((1,\,1),\) \((1,\,2),\) for example.

Solution. Let \(x = (x_1 ,\, x_2 ,\, x_3 )\) be given. Take \[\begin{align} a &= x_3 , \\[7pt] b &= x_2 - x_3 ,\\[7pt] c &= \frac{1}{2} x_1 + \frac{1}{2} x_2 - x_3 , \end{align}\] then \(x\) is expressed as \[x = au + bv + cw, \tag{*}\] that is, \(u,\) \(v\) and \(w\) span \(\mathbb{R}^3.\)

It remains to prove the uniqueness of the expression (*). Suppose \(a ' ,\) \(b ' \) and \(c ' \) be scalars such that \[a ' u + b ' v + c ' w = 0 .\tag{**}\] Combining (*) and (**) produces \[(a - a ' ) u + (b - b ' ) v + ( c - c ' ) w = 0;\] Solving this equation as a linear system, we have \[a - a ' = b - b ' = c - c ' = 0,\] which yields the uniqueness of the expression (*).

Solution. Let \(a,\) \(b\) and \(c\) be scalars satisfying \[1-2x+5x^2 = a1 +b(1+x) + c(1-x^2 ).\] From this equation, we have \[1-2x+5x^2 = (a+b+c) + bx + (-c)x^2\] or \[\begin{cases} a+b+c=1 \\[6pt] b=-2 \\[6pt] c =-5, \end{cases}\] that is, \(a=8,\) \(b=-2\) and \(c=-5.\)

Solution. We only have to show that \(e^x\) and \(e^{2x}\) are linearly independent. Suppose \(a\) and \(b\) are scalars satisfying \[ae^x + be^{2x}=0. \tag{*}\] Suppose further that \(a \ne 0\) or \(b \ne 0.\) If one of \(a\) and \(b\) equals to zero, the so the other, thus in this case we have \(a \ne 0\) AND \(b \ne 0.\) From (*), \[ae^x = -be^x\] or \[a = -be^x.\tag{**}\] But LHS of (**) is constant while RHS is not, which is a contradiction. Therefore \(a = b = 0\) and we conclude that \(e^x\) and \(e^{2x}\) are linearly independent.

The value of the associated coordinate map for the function \(-2e^x + 5e^{2x}\) is trivially \((-2 ,\, 5).\)

Solution. Let \(x=(x_1 ,\,x_2 ,\,x_3,\,x_4)\in\mathbb{R}^4\) be given. Take \[a = x_1 - x_2 ,\,\, b = x_2 - x_3 ,\,\, c = x_3 - x_4 ,\,\, d = x_4,\] then \[x = av_1 + bv_2 + cv_3 + dv_4 .\] Therefore \(v_1 ,\) \(v_2 ,\) \(v_3 \) and \(v_4\) span \(\mathbb{R}^4.\) Linear independence of these vectors is trivial.

Solution. Take \(v_3 = (0,\,1,\,1),\) then it is trivial that \(v_1 ,\) \(v_2 \) and \(v_3\) are linearly independent. Let \(x=(x_1 ,\,x_2,\,x_3) \in \mathbb{R}^3\) be given. Take \[\begin{gather} a = \frac{1}{2} (x_1 - x_2 + x_3 ), \\[5pt] b = \frac{1}{2} (x_1 + x_2 - x_3 ), \\[5pt] c = \frac{1}{2} (-x_1 + x_2 + x_3 ) \end{gather}\] then we have \[x= av_1 + bv_2 + cv_3,\] that is, \(v_1,\) \(v_2\) and \(v_3\) span \(\mathbb{R}^3.\)

Solution. The first example is the space consists of polynomial functions of degree less than or equal to \(3.\) This space has the basis \(\left\{ 1,\, x ,\, x^2 ,\, x^3 \right\}.\)

The second one is the space spanned by \(e^x ,\) \(e^{2x},\) \(e^{3x},\) \(e^{4x}.\)

Solution. Let \(S\) be the space consists of real sequences with all but finite zero terms. Define the sequence \(b^{(n)} = \left\{ b^{(n)}_k \right\} \) by \(b^{(n)}_k = \delta_k^n,\) the Kronecker delta. Then \[b^{(1)},\, b^{(2)},\, b^{(3)},\,\cdots\] substitute a basis for \(S.\)

Solution. The given equation is a second-order linear homogeneous differential equation. Take \(y=e^{\lambda x}\) then \[y ' = \lambda e^{\lambda x} \,\,\text{and}\,\, y ' ' = \lambda^2 e^{\lambda x}.\] Substitute these into differential equation, we obtain \[\lambda^2 e^{\lambda x} - 2\lambda e^{\lambda x} + e^{\lambda x} = 0.\] Factorization gives \[e^{\lambda x} (\lambda^2 - 2\lambda +1 ) = 0,\] that is, \(\lambda = 1.\) Therefore the solution is the function of form \[y = c_1 e^{\lambda x} + c_2 x e^{\lambda x} = c_1 e^x +c_2 xe^x\] for constants \(c_1\) and \(c_2 .\) Since \(e^x\) and \(xe^x\) are independent, it completes the proof.

Solution. The solution set is exactly the kernel of \(T.\) If the dimension of the kernel of \(T\) equals to zero, then the set must consist of zero vector only. If the dimension of the kernel of \(T\) is positive, then there must be a nonzero vector \(v\) that satisfies \(T(v) = \mathbf{0},\) and all the vectors of the form \(\lambda v\) where \(\lambda\) is a scalar belongs to the kernel, that is, the kernel consists of infinitely many vectors.

Solution. If \(v ' \) is not in the image of \(T,\) then the solution set is empty.

Now we suppose that \(T(v) = v ' \) has at least one solution, say \(u.\) If \(\operatorname{dim}(\operatorname{Ker}(T))=0,\) then \(u\) is the only solution for the equation. If \(\operatorname{dim}(\operatorname{Ker}(T)) > 0,\) then all the vectors of the form \(u+w\) where \(w\) belongs to the kernel of \(T\) are solutions, that is, \(T(v)=\mathbf{0}\) has infinitely many solutions.

Solution. Let \(v ' \in V ' \) be given. Since \(T\) is surjective, there exists \(v\in V\) such that \(T(v) = v ' .\) For \(v_1 ,\) \(\cdots,\) \(v_n\) span \(V,\) we have \[v = a_1 v_1 + \cdots + a_n v_n\] for some scalars \(a_1,\) \(\cdots,\) \(v_n.\) Hence, from the linearity of \(T,\) we have \[v ' = T(v) = T(a_1 v_1 + \cdots + a_n v_n) = a_1 T(v_1) + \cdots + a_n T(v_n),\] that is, \(v '\) belongs to the span of \(T(v_1),\) \(\cdots ,\) \(T(v_n).\)

Solution. Suppose \(a_1,\) \(\cdots ,\) \(a_n\) are scalars satisfying \[a_1 T(v_1) + \cdots + a_n T(v_n ) = \mathbf{0}.\] Since \(T\) is linear, we have \[T(a_1 v_1 + \cdots + a_n v_n) = a_1 T(v_1) + \cdots + a_n T(v_n ) = \mathbf{0} = T(\mathbf{0}),\] and since \(T\) is injective, we have \[a_1 v_1 + \cdots + a_n v_n = \mathbf{0}.\] But for \(v_1,\) \(\cdots,\) \(v_n\) are linearly independent, this equality holds only if \[a_1 = \cdots = a_n = 0.\]

Solution. An isomorphism is injective and surjective. Therefore the result directly comes from the precious two problems 15 and 16.

Solution(alt). Since \(V\) is finitely-generated, there are finite vectors \(w_1,\) \(\cdots,\) \(w_n\) that span \(V.\) For \(S\) spans \(V,\) each \(w_j\) can be expressed as a linear combination of finite elements of \(S,\) that is, there exist scalars \(a_1,\) \(\cdots,\) \(a_m\) and vectors \(v_1,\) \(\cdots,\) \(v_m\) in \(S\) such that \[w_j = a_1 v_1 + \cdots + a_m v_m .\] Let \(W_j\) be the set of such vectors \(v_1,\) \(\cdots,\) \(v_m\) in \(S,\) then each set \(W_j\) is finite. Take \(W = W_1 \cup \cdots \cup W_n ,\) then \(W\) is a finite subset of \(S\) and all \(w_j\)s are generated by \(W.\) Hence \(W\) spans \(V.\)

Solution. If \(T\) were injective, then \(\operatorname{dim}(\operatorname{Ker}(T))=0.\) Thus, by Rank-Nullity Theorem, we have \[\begin{align} 2 &= \operatorname{dim}(V ' ) \\[7pt] &\ge \operatorname{dim}(\operatorname{Im}(T)) \\[7pt] &= \operatorname{dim}(V) - \operatorname{dim}(\operatorname{Ker}(T)) = 3, \end{align}\] which is impossible. Therefore \(T\) cannot be injective.

Solution. If \(T\) were surjective, by Rank-Nullity Theorem, we have \[\begin{align} 4 &= \operatorname{dim}(V ' ) \\[7pt] &= \operatorname{dim}(\operatorname{Im}(T)) \\[7pt] &= \operatorname{dim}(V) - \operatorname{dim}(\operatorname{Ker}(T)) \\[7pt] &\le \operatorname{dim}(V) = 3, \end{align}\] which is impossible.

Solution. Let \(p = ax^2 + bx +c \in V,\) then a direct calculation shows that \[T(p) = -ax^2 - bx + 2a-c.\] Suppose \[T(p) = -ax^2 - bx +2a-c = 0,\] then we have \[\begin{cases} -a =0 \\[5pt] -b =0 \\[5pt] 2a-c = 0 \end{cases}\] or, equivalently, \(a=b=c=0,\) that is, \(p=0.\) Therefore the kernel of \(L\) is trivial and we conclude that \(L\) is injective.

Solution. By the Rank-Nullity Theorem, we have \(2 \le \operatorname{dim}(\operatorname{Ker}(T)) \le 6.\) Henceforth \(T\) cannot be injective, since if it could, the nullity had to be zero.

Solution. Let \(p = ax^2 +bx +c \in V,\) then a direct calculation shows that \[T(p) = \frac{2}{3} a + 2c.\tag{*}\] Let \(p_1 = a_1 x^2 + b_1 x + c_1\) and \(p_2 = a_2 x^2 + b_2 x + c_2\) be vectors in \(V\) and \(\lambda\) be a scalar. Then \[\begin{align} T(p_1 + \lambda p_2) &= T((a_1 + \lambda a_2)x^2 + (b_1 + \lambda b_2 )x + (c_1 + \lambda c_2)) \\[6pt] &= \frac{2}{3}(a_1 + \lambda a_2) + 2(c_1 + \lambda c_2) \\[6pt] &= \left( \frac{2}{3} a_1 + 2 c_1 \right) + \lambda \left( \frac{2}{3} a_2 + 2c_2 \right) \\[6pt] &= T(p_1 ) + \lambda T(p_2). \end{align}\] Therefore \(T\) is linear.

From (*), we see that every element of kernel of \(T\) is a function of the form \[p(x) = -3cx^2 + bx +c .\] Hence the kernel is generated by two vectors \[p_1 (x) = -3x^2 +1 ,\quad p_2 (x) = x .\] These vectors are linearly independent; Therefore the dimension of kernel is \(2.\)

Solution. Let the dimensions of \(V\) and \(V ' \) be \(m\) and \(n\) respectively. Suppose \[v_1 ,\, \cdots ,\, v_m\] be a basis for \(V.\) Define a linear transformation \(T : V \rightarrow V ' \) by \[T(v) = a_1 T(v_1) + a_2 T(v_2) + \cdots + a_n T(v_n )\] for \(v = a_1 v_1 + \cdots + a_m v_m\) in \(V,\) that is, \(T\) cuts off last \(n-m\) terms. Then \(T\) is linear and surjective.

Solution. If \(v\in V,\) then \(v= u+w\) for some \(u\in W_0\) and \(w\in W_1.\) Since \(u\) is a linear combination of \(u_1 ,\) \(\cdots,\) \(u_n\) and \(w\) is a linear combination of \(w_1,\) \(\cdots,\) \(w_m,\) \(v=u+w\) is a linear combination of \(u_1,\) \(\cdots,\) \(u_n,\) \(w_1,\) \(\cdots,\) \(w_m.\)

It remains to show that \(u_1,\) \(\cdots,\) \(u_n,\) \(w_1,\) \(\cdots,\) \(w_m\) are linearly independent. Suppose \[a_1 u_1 + \cdots + a_n u_n + a_{n+1}w_1 + \cdots + a_{n+m}w_m = 0.\] Since \[0+0=0 ,\quad 0\in W_0 ,\quad 0\in W_1,\] by the uniqueness of the expression of direct sum, we have \[a_1 u_1 + \cdots + a_n u_n = 0\tag{*}\] and \[a_{n+1}w_1 + \cdots + a_{n+m}w_m = 0.\tag{**}\] Since \(a_1,\) \(\cdots,\) \(a_n\) are linearly independent, from (*), we have \[a_1 = \cdots = a_n = 0;\] Likewise, from (*), we have \[a_{n+1} = \cdots = a_{n+m} = 0.\] Hence \(u_1,\) \(\cdots,\) \(u_n,\) \(w_1,\) \(\cdots,\) \(w_m\) are linearly independent.

Solution. Linear independence of \(v_1,\) \(\cdots,\) \(v_n\) shows that \[Kv_i \cap Kv_j = \left\{ \mathbf{0} \right\} \,\,\text{if}\,\, i\ne j.\] Now we only have to show that \[V = Kv_1 + Kv_2 + \cdots + Kv_n.\tag{*}\] Let \(v\in V\) then \[v = a_1 v_1 + \cdots + a_n v_n \tag{**}\] for some scalars \(a_1,\) \(\cdots,\) \(a_n.\) Since \(a_j v_j \in Kv_j ,\) we obtain the desired result directly from (*).

Solution. Define \(T(kv_0 ) = kw_0\) for any scalar \(k.\) Now is the time for you to prove the uniqueness of the function with this property.

Solution. Define \[T(a_1 v_1 + \cdots + a_n v_n ) = a_1 w_1 + \cdots + a_n w_n\] for any scalars \(a_1,\) \(\cdots,\) \(a_n .\)

Solution. In this problem, we need one more assumption: \[ 1\le \operatorname{dim}(W_0) < \operatorname{dim}(V),\] which is omitted in Valenza's book.

Let \(n=\operatorname{dim}(V)\) and \(m=\operatorname{dim}(W_0 ).\) Suppose \[v_1 ,\, \cdots ,\, v_n\] is a basis for \(V.\) Then, among \(v_1,\) \(\cdots,\) \(v_n,\) exactly \(m\) elements belong to \(W_0,\) say \(v_1,\) \(\cdots,\) \(v_m.\) (Prove it!) Take \(W_1\) be the span of \(v_{m+1},\) \(\cdots,\) \(v_n\) then we obtain the desired result.

Solution. Let \(n = \operatorname{dim}(V)\) and \(m=\operatorname{dim}(V ' ),\) then by the Rank-Nullity Theorem we have \[\operatorname{dim}(\operatorname{Ker}(T)) = n-m.\] Suppose \(v_1 ,\) \(\cdots,\) \(v_n\) is a basis for \(V,\) then only \(n-m\) elements among these basis elements belong to \(\operatorname{Ker}(T),\) say \(v_{m+1},\) \(\cdots,\) \(v_n .\) Let \(W\) be the span of \(v_1,\) \(\cdots,\) \(v_n,\) then, by Problem 29, we have \[V = \operatorname{Ker}(T) \oplus W.\] Let \(u_1 ,\) \(\cdots,\) \(u_m\) be a basis for \(V ' ,\) and define a linear transformation \(L : W \rightarrow V '\) by \[L(a_1 v_1 + \cdots + a_m v_m) = a_1 u_1 + \cdots + a_m u_m\] for scalars \(a_1,\) \(\cdots,\) \(a_m,\) then \(L\) is an isomorphism, that is, \(W \cong V ' .\) (In general, two vector spaces over a common field with the same finite dimension are always isomorphic.)

Solution. Take \(V = \mathbb{R}[x]\) and define \(T\) by \[T(a_0 + a_1 x^1 + a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots) = a_0 + a_2 x^2 + a_4 x^4 + \cdots\] where \(a_j = 0\) for all but a finite number of \(j\)s.