This set of exercises is retrieved from the third chapter of Linear Algebra by Robert J. Valenza. Note that these solutions are not fully elaborated; You have to fill the descriptions by yourself.
Problem 3.1
Show that the solution set \(W\) of vectors \((x_1 ,\,x_2 )\) in \(\mathbb{R}^2\) satisfying the equation
\[x_1 + 8x_2 = 0\]
is a subspace of \(\mathbb{R}^2 .\)
Solution. The solution set is \[S = \left\{ (-8s,\, s) \,\vert\, s\in\mathbb{R} \right\}.\] This set is closed under addition and scalar multiplication, thus it is a subspace of \(\mathbb{R}^2 .\)
Problem 3.2
Determine whether the solution set to the equation
\[x_1 ^2 + x_2 ^2 + x_3 ^2 = 1\]
is a subspace of \(\mathbb{R}^3.\)
Solution. The solution set is not a vector space, because it does not contain \((0,\,0,\,0).\)
Problem 3.3
Let \(V = \mathbb{R}^2 .\) Exhibit a subset of \(V\) which is an additive subgroup \(V,\) but not a subspace of \(V.\)
Solution. Take \(S =\mathbb{Z} \times \mathbb{Z} .\)
Problem 3.4
Show that a subspace of \(\mathbb{R}^2\) containing both \((1,\,0)\) and \((1,\,-1)\) must be all of \(\mathbb{R}^2\) itself.
Solution. Let \(S\) be a subspace containing both \((1,\,0)\) and \((1,\,-1).\) Let \((x,\,y)\) be any element of \(\mathbb{R}^2 .\) Take \(a=x+y\) and \(b = -y\) then \[(x,\,y) = a(1,\,0) + b(1,\,-1),\] thus \((x,\,y)\) belongs to \(S,\) that is, \(\mathbb{R}^2 \subseteq S .\) Therefore \(\mathbb{R}^2 = S.\)
Problem 3.5
Let \(V\) be a vector space over a field \(K\) and suppose that \(U\) and \(W\) are subspaces of \(V.\) Show that \(U \cap W\) is likewise a subspace of \(V.\)
Solution. Let \(x\) and \(y\) be elements of \(U\cap W\) and \(\lambda\) be a scalar. Then \(x\in U\) and \(y\in U,\) thus \[x+y \in U \,\,\text{and} \,\, \lambda x \in U. \tag{*}\] Likewise, we have \[x+y \in W \,\,\text{and} \,\, \lambda x \in W, \tag{**}\] therefore, combining (*) and (**), we have \[x+y \in U\cap W \,\,\text{and} \,\, \lambda x \in U\cap W,\] which implies that \(U\cap W\) is a subspace of \(V.\)
Problem 3.6
Show by example, using \(V = \mathbb{R}^2\) for instance, that even though \(U\) and \(W\) are subspaces of \(V,\) the union \(U \cup W\) is not necessarily a subspace of \(V.\)
Solution. Take \[\begin{gather} U = \left\{ (x,\,0) \,\vert\, x \in \mathbb{R} \right\} ,\\[7pt] W = \left\{ (0,\,y) \,\vert\, y \in \mathbb{R} \right\} . \end{gather}\] Then \(U,\) \(W\) are subspaces of \(V.\)
Since \((1,\,0)\) and \((0,\,1)\) belong to \(U \cup W,\) while their sum doesn't, \(U \cup W\) is not a subspace of \(V.\)
Problem 3.7
Continuing in the context of the previous problem, show that \(U\cup W\) is a subspace of \(V\) if and only if either \(U \subseteq W\) or \(W \subseteq U .\)
Solution. If one of \(U\) and \(W\) is a subset of the other, then \(U \cup W\) is either \(U\) or \(W\) and is a subspace of \(V.\)
Conversely, suppose that \(U \cup W\) is a subspace of \(V.\) If neither \(U\) nor \(W\) is a subset of the other, then we can take \(u \in U \setminus W\) and \(w \in W \setminus U .\) Observe that \(u+w \) does not belong to \(U \cup W,\) while \(u\) and \(w\) do, which is a contradiction.
Problem 3.8
Again let \(U\) and \(W\) be subspaces of \(V.\) Let \(U+W\) denote the set of all vectors of the form \(u+w\) where \(u\in U\) and \(w\in W.\) Show that \(U+W\) is a subspace of\(V.\)
Solution. Let \(x_1\) and \(x_2\) be elements of \(U+W\) and \(\lambda\) be a scalar. Then there exist \(u_1 ,\) \(u_2 \in U\) and \(w_1 ,\) \(w_2 \in W\) such that \[x_1 = u_1 + w_1 \quad \text{and}\quad x_2 = u_2 + w_2 .\] Thus we have \[x_1 + x_2 = (u_1 + u_2) + (w_1 + w_2) \in U+W\] and \[\lambda x_1 = ( \lambda u_1 ) + (\lambda u_2 ) \in U+W .\] Therefore \(U+W\) is a subspace of \(V.\)
Problem 3.9
Continuing in the context of the previous problem, show that \(U+W\) is the smallest subspace of \(V\) containing both \(U\) and \(W;\) that is, if \(X\) is any subspace of \(V\) containing both \(U\) and \(W,\) then \(U+W \subseteq X.\)
Solution. Suppose \(x \in U+W\) then there exist \(u\in U\) and \(w\in W\) such that \(x=u+w.\) Since \(u\in X,\) \(w\in X\) and \(X\) is a vector space, \(u+w \in X,\) that is, \(x\in X.\) Therefore \(U+W \subseteq X.\)
Problem 3.10
Let \(V = \mathbb{R}^2 .\) What is the span of the vectors \((2,\,4)\) and \((-5,\,-10) ? \) Describe this set geometrically.
Solution. Since \((2,\,4) = 2(1,\,2)\) and \((-5, \, -10) = -5(1,\,2),\) we have \[a_1 (2,\,4) + a_2 (-5 ,\,-10) = (2a_1 -5a_2 )(1,\,2)\] for scalars \(a_1 ,\) \(a_2 .\) Take \(\lambda = 2a_1 -5 a_2\) then we have \[\operatorname{Span}\left\{ (2,\,4) ,\, (-5,\,10)\right\} = \left\{ \lambda(1,\,2) \,\vert\, \lambda \in \mathbb{R} \right\}. \] This set consists of the points on the line \(y=2x\) in the coordinate plane.
Problem 3.11
Let \(V = \mathbb{R}^2.\) What is the span of the vectors \((1,\,2)\) and \((2,\,1)?\) Describe this set geometrically.
Solution. Let \((x,\,y)\) be any point in \(\mathbb{R}^2.\) Take \[a = -\frac{1}{3} x + \frac{2}{3} y \quad \text{and}\quad b = \frac{2}{3} x - \frac{1}{3} y\] then \[(x,\,y) = a(1,\,2) + b(2,\,1),\] that is, any point in \(\mathbb{R}^2\) belongs to the span. Therefore \[\operatorname{Span}\left\{ (1,\,2) ,\, (2,\,1) \right\} = \mathbb{R}^2 .\]
Problem 3.12
Let \(V = \mathbb{R}^2.\) State geometrically under what conditions the span of two nonzero vectors \(x\) and \(y\) is all of \(V.\)
Solution. \(x \ne \lambda y\) for any scalar \(\lambda ,\) that is, \(x\) and \(y\) are not parallel.
Problem 3.13
Is it possible to find a subspace of \(\mathbb{R}^2\) which is neither a point, not a line, not all of \(\mathbb{R}^2\) itself? Explain geometrically.
Solution. Suppose that \(V\) is such a subspace. Since \(V\) consists of more than a single point, \(V\) contain a nonzero element \(v.\) Geometrically, \(V\) has to include the line \(\ell\) through both origin and \(v.\) Since \(V\) is not a line, \(V\) has to contain a point \(w\) which is not on the line \(\ell.\) Thus \(V\) include two lines - one is the line through both origin and \(v,\) the other is the line through origin and \(w.\) But the subspace of \(\mathbb{R}^2\) that include these two lines becomes \(\mathbb{R}^2\) itself. Therefore there is no such a subspace.
Problem 3.14
Show that the functions \(e^x ,\) \(\cos x,\) \(\sin x\) are not in the span of the infinite family of monomials \(1,\) \(x,\) \(x^2 ,\) \(\cdots .\)
Solution. The functions \(\cos x,\) \(\sin x\) are all bounded. But every polynomial function other than constant function is not bounded. The function \(e^x\) tends to zero as \(x \rightarrow - \infty,\) but every polynomial function other than constant function does not have such a property.
Problem 3.15
Show that at least two vectors are required to span the vector space \(\mathbb{R}^2.\)
Solution. It is trivial that \(\left\{ 0 \right\}\) does not span \(\mathbb{R}^2.\) Suppose now that a nonzero single vector \(v = (v_1 ,\, v_2 )\) spans \(\mathbb{R}^2.\) WLOG, suppose further that \(v_1 \ne 0.\) Consider the vector \(x = (v_1 ,\, v_2 + 1 ).\) This vector has to be expressible as a linear combination of \(v,\) that is, there exists a scalar \(\lambda\) such that \(x= \lambda v.\) This equation yields \[ (v_1 ,\, v_2 +1 ) = x = \lambda v = (\lambda v_1 ,\, \lambda v_2 ).\] But no scalar \(\lambda\) satisfies this equation. Therefore, there is no single vector that spans \(\mathbb{R}^2.\)
Problem 3.16
Let \(a\in\mathbb{R}\) and consider the function
\[f : \mathbb{R} \rightarrow \mathbb{R} ,\quad x \mapsto ax .\]
Show that \(f\) is a linear transformation of real vector spaces. What is the kernel of \(f?\) The image?
Solution. If \(a=0,\) then \(f\) is trivially linear transformation; Its kernel is \(\mathbb{R}\) and its image is \(\left\{ 0 \right\}.\)
Suppose now that \(a \ne 0.\) Then for all \(x,\,y \in \mathbb{R}\) and scalar \(\lambda,\) \[f(x+ \lambda y) = a(x+ \lambda y) = ax + \lambda (ay) = f(x) + \lambda f(y),\] that is, \(f\) is a linear transformation. In this case, the kernel of \(f\) is \(\left\{ 0 \right\}\) and the image is \(\mathbb{R}.\)
Problem 3.17
Define a function \(A\) from \(\mathbb{R}^n\) to \(\mathbb{R}\) as follows:
\[A(x) = \frac{1}{n} \sum_{j=1}^n x_j \]
for \(x = (x_1 ,\, x_2 ,\, \cdots ,\, x_n ).\)
Thus \(A(x)\) is just the average of the components of \(x.\) Show that \(A\) is a linear transformation.
Solution. Suppose \(x = (x_1 ,\, \cdots ,\, x_n )\) and \(y = (y_1 ,\, \cdots ,\, y_n )\) are vectors and \(\lambda\) is a scalar, then we have \[\begin{align} A(x+\lambda y) &= \frac{1}{n} \sum_{j=1}^n (x_j + \lambda y_j ) \\[4pt] &= \frac{1}{n} \sum_{j=1}^n x_j + \lambda \frac{1}{n} \sum_{j=1}^{n} y_j \\[6pt] &= A(x) + \lambda A(y), \end{align}\] that is, \(A\) is a linear transformation.
Problem 3.18
Express the kernel of the linear transformation \(T : \mathbb{R}^2 \rightarrow \mathbb{R}\) defined by
\[T(x_1 ,\,x_2 ) = 2x_1 - 5x_2\]
as the span of a single vector.
Solution. The kernel consists of the vectors \((x_1 ,\,x_2 )\) that satisfy \[2x_1 + 5x_2 = 0.\] Substituting \(5t\) to \(x_1 ,\) we have \[ (x_1 ,\,x_2 ) = (5t ,\, -2t) = t(5,\,-2) .\] Therefore, the kernel is generated by the a single vector \((5,\,-2 ).\)
Problem 3.19
What is the kernel of the second derivative operator
\[C^2 (\mathbb{R} ) \rightarrow C^0 (\mathbb{R} ) ,\quad f \mapsto \frac{d^2 f}{dx^2}\]
on the real vector space of twice differentiable functions with continuous second derivative?
Solution. The set of first-order polynomial functions and constant functions is the kernel.
Problem 3.20
What is the kernel of the \(n\)th derivative operator
\[C^n (\mathbb{R} ) \rightarrow C^0 (\mathbb{R} ) ,\quad f \mapsto \frac{d^n f}{dx^n}\]
on the real vector space of \(n\)-times differentiable functions with continuous \(n\)th derivative?
Solution. The set of constant functions and polynomial functions whose orders are less than \(n\) is the kernel.
Problem 3.21
Explain succinctly why the solution space of the differential equation
\[y ' ' - 2 y ' + y = 0 \tag{*}\]
is a subspace of \(C^2 (\mathbb{R} ).\)
Solution. Let \(W\) be the set of all functions that satisfy (*). If \(y=f(x)\) be a solution for (*), then \(f\) has to be twice-differentiable, and \(f ' \) has to be continuous. Besides, from the equation \[y ' ' = 2 y ' - y ,\] we have that \(f ' ' \) also has to be continuous. Therefore \(f\) must be in \(C^2 (\mathbb{R} ),\) that is, \(W\) is a subset of \(C^2 ( \mathbb{R} ).\) Furthermore, \(x \mapsto e^x\) is a solution for (*); Hence \(W\) is not empty.
We now suppose that two functions \(f_1 ,\) \(f_2 \) belong to \(W,\) and that \(k\) is a scalar(real number). Since \[\begin{gather} (f_1 + f_2 ) ' ' = 2 ( f_1 + f_2 ) ' - (f_1 + f_2 ), \\[5pt] (kf_1 ) ' ' = 2 ( k f_1 ) - (kf_1 ), \end{gather} \] \(W\) is closed under vector addition and scalar multiplication. Therefore \(W\) is a subspace of \(C^2 (\mathbb{R} ).\)
Problem 3.22
Show that a linear map \(T : \mathbb{R} \rightarrow \mathbb{R}^2\) cannot be surjective.
Solution. Suppose that \(T : \mathbb{R} \rightarrow \mathbb{R}^2\) is a surjective linear map. Take \(v_1 \) and \(v_2\) in \(\mathbb{R}\) such that \[T(v_1 ) = (1,\,0) \quad \text{and}\quad T(v_2 ) = (0,\,1).\] Since \(v_1\) and \(v_2\) are nonzero real numbers, \(\lambda = v_2 / v_1\) is also a nonzero real number. Since \(T\) is linear, we have \[(1,\,0) = T(v_1) = \frac{1}{\lambda} T(\lambda v_1) = \frac{1}{\lambda} T(v_2) = \frac{1}{\lambda} (0,\,1),\] that is, \[\lambda (1,\,0) = (0,\,1) .\] But no \(\lambda\) satisfies this equation, therefore we conclude that \(T\) cannot be a surjective linear map.
Problem 3.23
Let \(a\in\mathbb{R}\) be fixed and consider the function \(\nu_a\) defined by
\[\nu_a : C^0 ( \mathbb{R} ) \rightarrow \mathbb{R} ,\quad f \mapsto f(a) .\]
This map is called evaluation at \(a\). Show that for all real numbers \(a,\) the evaluation map \(\nu_a\) is a linear transformation and describe \(\operatorname{Ker} (\nu _a ).\)
Solution. Let \(f\) and \(g\) be continuous functions, \(\lambda\) be a scalar. Then we have \[\nu_a ( f + \lambda g) = (f + \lambda g)(a) = f(a) + \lambda g(a) = \nu_a (f) + \lambda \nu_a (g),\] which shows the linearity of \(\nu_a .\) The kernel is the set of functions \(f\) with \(f(a) = 0.\)
Problem 3.24
Let \(a\) and \(b\) be real numbers. Define a function \(I : C^0 (\mathbb{R}) \rightarrow \mathbb{R}\) on the space of continuous real-valued functions on \(\mathbb{R}\) as follows:
\[I(f) = \int_a^b f(x) dx.\]
Show that \(I\) is a linear transformation. Deduce from this that the set
\[\left\{ f \in C^0 ( \mathbb{R} ) \,\bigg\vert \int_a ^b f(x) dx =0 \right\}\]
is a subspace of \(C^0 (\mathbb{R} ) .\) Assuming that \(a \ne b,\) what is the image of \(I?\)
Solution. If \(f\) and \(g\) are continuous functions and \(\lambda\) is a scalar, then \[I(f + \lambda g) = \int_a^b ( f(x) + \lambda g(x)) dx = \int_a^b f(x) dx + \lambda \int_a^b g(x) dx = I(f) + \lambda I(g),\] which proves the linearity of \(I.\) The image of \(I\) is \(\mathbb{R}.\)
Problem 3.25
Let \(T : V \rightarrow V\) be a linear transformation of vector spaces over \(K\) and let \(\lambda \in K\) be a scalar such that there exists a nonzero \(v\in V\) satisfying the equation
\[T(v) = \lambda v.\tag{*}\]
Thus \(T\) maps \(v\) onto a scalar multiple of itself. Then \(\lambda\) is called an eigenvalue of \(T\) and any \(v\) (even \(\mathbf{0}\)) satisfying the equation above is called an eigenvector belonging to \(\lambda\). Show that the set of all eigenvectors belonging to the eigenvalue \(\lambda\) is a subspace of \(V.\) This is called the eigenspace belonging to \(\lambda.\)
Solution. Let \(W\) be the set of vectors \(v\) that satisfy (*).
First, since \(T(\mathbf{0}) = \lambda \mathbf{0} = \mathbf{0},\) the zero vector belongs to \(W.\)
Next, suppose that \(u\) and \(v\) are vectors in \(W\) and \(k\) is a scalar. Since \[T(u) = \lambda u \quad\text{and}\quad T(v) = \lambda v ,\] we have \[T(u+v) = T(u) + T(v) = \lambda u + \lambda v = \lambda (u+v),\] which shows that \(u+v \in W.\) Besides, since \[T(ku) = kT(u) = k \lambda u = \lambda (ku) ,\] \(ku\) also belongs to \(W.\) Therefore \(W\) is a subspace of \(V.\)
Problem 3.26
Given a vector space \(V\) over \(K,\) assume that there exist subspace \(W_0\) and \(W_1\) such that \(V = W_0 + W_1 .\) Show that if both \(W_0\) and \(W_1\) are finitely generated, then \(V\) is likewise finitely generated.
Solution. If \(u_1 ,\) \(\cdots ,\) \(u_m\) generate \(W_0\) and \(v_1 ,\) \(\cdots ,\) \(v_n\) generate \(W_1 ,\) then \[u_1 ,\, \cdots ,\, u_m ,\, v_1 ,\, \cdots ,\, v_n\] generate \(W_0 + W_1 .\)
Problem 3.27
Define subspaces \(W_0\) and \(W_1\) in \(\mathbb{R}^2\) as follows:
\[W_0 = \operatorname{Span} \left\{ (0,\,1) \right\} \quad \text{and} \quad W_1 = \operatorname{Span} \left\{ (1,\,1) \right\} .\]
Show that \(\mathbb{R}^2 = W_0 \oplus W_1 .\)
Solution. Let \((x,\,y) \in \mathbb{R}^2\) and take \(a = -x+y\) and \(b=x,\) then \[(x,\,y) = a(0,\,1) + b(1,\,1) \in W_0 + W_1 .\] It remains to show that \(W_0 \cap W_1 = \left\{ (0,\,0) \right\}.\) Suppose \((x_0 ,\,y_0 ) \in W_0 \cap W_1 ,\) then there exist scalars \(s\) and \(t\) such that \[(x_0 ,\,y_0 ) = s(0,\,1) = t(1,\,1).\] This equality holds only when \(s=t=0,\) that is, \((x_0 ,\,y_0 ) = (0,\,0).\) Therefore \(W_0 \cap W_1 = \left\{ (0,\,0) \right\}.\)
Problem 3.28
Let \(U,\) \(W_0\) and \(W_1\) be vector spaces over \(K\) and let there be given linear transformations \(T_0 : U \rightarrow W_0\) and \(T_1 : U \rightarrow W_1 .\) Consider the linear transformation \(T_0 \times T_1 : U \rightarrow W_0 \times W_1\) defined by
\[(T_0 \times T_1)(u) = (T_0 (u) ,\, T_1 (u)).\]
Prove that \(\operatorname{Ker}(T_0 \times T_1 ) = \operatorname{Ker}(T_0 ) \cap \operatorname{Ker}(T_1 ).\)
Solution. The proof is straightforward, that is, \[\begin{align} \operatorname{Ker}(T_0 \times T_1 ) &= \left\{ u\in U \,\,\vert\,\, (T_0 (u) ,\, T_1 (u)) = (0,\,0) \right\} \\[7pt] &= \left\{ u\in U \,\,\vert\,\, T_0 (u)=0 \,\,\text{and}\,\, T_1 (u)=0 \right\} \\[7pt] &= \left\{ u\in U \,\,\vert\,\, T_0 (u)=0 \right\} \cap \left\{ u\in U \,\,\vert\,\, T_1 (u)=0 \right\} \\[7pt] &= \operatorname{Ker}(T_0 ) \cap \operatorname{Ker}(T_1 ). \end{align}\]
Problem 3.29
Let \(T : V \rightarrow V\) be a linear transformation from a vector space \(V\) to itself such that \(T \circ T = T.\) (Such a transformation is called idempotent.) Show that
\[V = \operatorname{Ker} (T) \oplus \operatorname{Im} (T) .\]
Solution. Let \(v \in V\) and take \(w = v-T(v).\) Since \(T\) is a linear transformation and \(T(T(v)) = T(v),\) we have \[T(w) = T(v-T(v)) = T(v) - T(T(v)) = T(v) - T(v) = \mathbf{0},\] that is, \(w\in \operatorname{Ker}(T).\) Observe that \[v = (v-f(v))+f(v) = w+f(v)\] where \(w\in \operatorname{Ker}(T)\) and \(f(v) \in \operatorname{Im}(T),\) thus we have \(v\in \operatorname{Ker}(T) + \operatorname{Im}(T).\) This proves \(V = \operatorname{Ker}(T) + \operatorname{Im}(T).\)
Now it remains to prove that \(\operatorname{Ker}(T) \cap \operatorname{Im}(T) = \left\{ \mathbf{0} \right\}.\) Suppose \(x\in \operatorname{Ker}(T) \cap \operatorname{Im}(T).\) Since \(x\in \operatorname{Ker}(T),\) we have \(T(x)=0,\) and since \(x\in \operatorname{Im}(T),\) there exists \(u\in V\) such that \(T(u)=x.\) From this, we deduce \(T(T(u)) = T(x) ,\) which yields \(T(u) = T(x) = \mathbf{0},\) and consequently we have \(x = T(u) = \mathbf{0}.\) This proves that \(\operatorname{Ker}(T) \cap \operatorname{Im}(T) = \left\{ \mathbf{0} \right\}.\)