{"id":5010,"date":"2020-08-25T13:24:53","date_gmt":"2020-08-25T04:24:53","guid":{"rendered":"https:\/\/sasamath.com\/blog\/?p=5010"},"modified":"2021-04-17T17:05:18","modified_gmt":"2021-04-17T08:05:18","slug":"exercises-vector-spaces-and-linear-transformations","status":"publish","type":"post","link":"https:\/\/sasamath.com\/blog\/articles\/exercises-vector-spaces-and-linear-transformations\/","title":{"rendered":"Exercises: Vector Spaces and Linear Transformations"},"content":{"rendered":"
\n

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.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.1<\/span>
\nShow that the solution set \\(W\\) of vectors \\((x_1 ,\\,x_2 )\\) in \\(\\mathbb{R}^2\\) satisfying the equation
\n\\[x_1 + 8x_2 = 0\\]
\nis a subspace of \\(\\mathbb{R}^2 .\\)\n<\/p>\n

\n

Solution.<\/span>
\nThe solution set is
\n\\[S = \\left\\{ (-8s,\\, s) \\,\\vert\\, s\\in\\mathbb{R} \\right\\}.\\]
\nThis set is closed under addition and scalar multiplication, thus it is a subspace of \\(\\mathbb{R}^2 .\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.2<\/span>
\nDetermine whether the solution set to the equation
\n\\[x_1 ^2 + x_2 ^2 + x_3 ^2 = 1\\]
\nis a subspace of \\(\\mathbb{R}^3.\\)\n<\/p>\n

\n

Solution.<\/span>
\nThe solution set is not a vector space, because it does not contain \\((0,\\,0,\\,0).\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.3<\/span>
\nLet \\(V = \\mathbb{R}^2 .\\) Exhibit a subset of \\(V\\) which is an additive subgroup \\(V,\\) but not a subspace of \\(V.\\)\n<\/p>\n

\n

Solution.<\/span>
\nTake \\(S =\\mathbb{Z} \\times \\mathbb{Z} .\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.4<\/span>
\nShow that a subspace of \\(\\mathbb{R}^2\\) containing both \\((1,\\,0)\\) and \\((1,\\,-1)\\) must be all of \\(\\mathbb{R}^2\\) itself.\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(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
\n\\[(x,\\,y) = a(1,\\,0) + b(1,\\,-1),\\]
\nthus \\((x,\\,y)\\) belongs to \\(S,\\) that is, \\(\\mathbb{R}^2 \\subseteq S .\\) Therefore \\(\\mathbb{R}^2 = S.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.5<\/span>
\nLet \\(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.\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(x\\) and \\(y\\) be elements of \\(U\\cap W\\) and \\(\\lambda\\) be a scalar. Then \\(x\\in U\\) and \\(y\\in U,\\) thus
\n\\[x+y \\in U \\,\\,\\text{and} \\,\\, \\lambda x \\in U. \\tag{*}\\]
\nLikewise, we have
\n\\[x+y \\in W \\,\\,\\text{and} \\,\\, \\lambda x \\in W, \\tag{**}\\]
\ntherefore, combining (*) and (**), we have
\n\\[x+y \\in U\\cap W \\,\\,\\text{and} \\,\\, \\lambda x \\in U\\cap W,\\]
\nwhich implies that \\(U\\cap W\\) is a subspace of \\(V.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.6<\/span>
\nShow 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.\\)\n<\/p>\n

\n

Solution.<\/span>
\nTake
\n\\[\\begin{gather}
\nU = \\left\\{ (x,\\,0) \\,\\vert\\, x \\in \\mathbb{R} \\right\\} ,\\\\[7pt]
\nW = \\left\\{ (0,\\,y) \\,\\vert\\, y \\in \\mathbb{R} \\right\\} .
\n\\end{gather}\\]
\nThen \\(U,\\) \\(W\\) are subspaces of \\(V.\\)<\/p>\n

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.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.7<\/span>
\nContinuing 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 .\\)\n<\/p>\n

\n

Solution.<\/span>
\nIf 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.\\)<\/p>\n

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
\n\\(u+w \\) does not belong to \\(U \\cup W,\\) while \\(u\\) and \\(w\\) do, which is a contradiction.\n<\/p><\/div>\n

<\/p>\n

Problem 3.8<\/span>
\nAgain 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.\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(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
\n\\[x_1 = u_1 + w_1 \\quad \\text{and}\\quad x_2 = u_2 + w_2 .\\]
\nThus we have
\n\\[x_1 + x_2 = (u_1 + u_2) + (w_1 + w_2) \\in U+W\\]
\nand
\n\\[\\lambda x_1 = ( \\lambda u_1 ) + (\\lambda u_2 ) \\in U+W .\\]
\nTherefore \\(U+W\\) is a subspace of \\(V.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.9<\/span>
\nContinuing 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.\\)\n<\/p>\n

\n

Solution.<\/span>
\nSuppose \\(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.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.10<\/span>
\nLet \\(V = \\mathbb{R}^2 .\\) What is the span of the vectors \\((2,\\,4)\\) and \\((-5,\\,-10) ? \\) Describe this set geometrically.\n<\/p>\n

\n

Solution.<\/span>
\nSince \\((2,\\,4) = 2(1,\\,2)\\) and \\((-5, \\, -10) = -5(1,\\,2),\\) we have
\n\\[a_1 (2,\\,4) + a_2 (-5 ,\\,-10) = (2a_1 -5a_2 )(1,\\,2)\\]
\nfor scalars \\(a_1 ,\\) \\(a_2 .\\) Take \\(\\lambda = 2a_1 -5 a_2\\) then we have
\n\\[\\operatorname{Span}\\left\\{ (2,\\,4) ,\\, (-5,\\,10)\\right\\} = \\left\\{ \\lambda(1,\\,2) \\,\\vert\\, \\lambda \\in \\mathbb{R} \\right\\}. \\]
\nThis set consists of the points on the line \\(y=2x\\) in the coordinate plane.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.11<\/span>
\nLet \\(V = \\mathbb{R}^2.\\) What is the span of the vectors \\((1,\\,2)\\) and \\((2,\\,1)?\\) Describe this set geometrically.\n<\/p>\n

\n

Solution.<\/span>
\nLet \\((x,\\,y)\\) be any point in \\(\\mathbb{R}^2.\\) Take
\n\\[a = -\\frac{1}{3} x + \\frac{2}{3} y \\quad \\text{and}\\quad b = \\frac{2}{3} x – \\frac{1}{3} y\\]
\nthen
\n\\[(x,\\,y) = a(1,\\,2) + b(2,\\,1),\\]
\nthat is, any point in \\(\\mathbb{R}^2\\) belongs to the span. Therefore
\n\\[\\operatorname{Span}\\left\\{ (1,\\,2) ,\\, (2,\\,1) \\right\\} = \\mathbb{R}^2 .\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.12<\/span>
\nLet \\(V = \\mathbb{R}^2.\\) State geometrically under what conditions the span of two nonzero vectors \\(x\\) and \\(y\\) is all of \\(V.\\)\n<\/p>\n

\n

Solution.<\/span>
\n\\(x \\ne \\lambda y\\) for any scalar \\(\\lambda ,\\) that is, \\(x\\) and \\(y\\) are not parallel.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.13<\/span>
\nIs 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.\n<\/p>\n

\n

Solution.<\/span>
\nSuppose 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.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.14<\/span>
\nShow 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 .\\)\n<\/p>\n

\n

Solution.<\/span>
\nThe 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.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.15<\/span>
\nShow that at least two vectors are required to span the vector space \\(\\mathbb{R}^2.\\)\n<\/p>\n

\n

Solution.<\/span>
\nIt 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
\n\\[ (v_1 ,\\, v_2 +1 ) = x = \\lambda v = (\\lambda v_1 ,\\, \\lambda v_2 ).\\]
\nBut no scalar \\(\\lambda\\) satisfies this equation. Therefore, there is no single vector that spans \\(\\mathbb{R}^2.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.16<\/span>
\nLet \\(a\\in\\mathbb{R}\\) and consider the function
\n\\[f : \\mathbb{R} \\rightarrow \\mathbb{R} ,\\quad x \\mapsto ax .\\]
\nShow that \\(f\\) is a linear transformation of real vector spaces. What is the kernel of \\(f?\\) The image?\n<\/p>\n

\n

Solution.<\/span>
\nIf \\(a=0,\\) then \\(f\\) is trivially linear transformation; Its kernel is \\(\\mathbb{R}\\) and its image is \\(\\left\\{ 0 \\right\\}.\\)\n<\/p>\n

Suppose now that \\(a \\ne 0.\\) Then for all \\(x,\\,y \\in \\mathbb{R}\\) and scalar \\(\\lambda,\\)
\n\\[f(x+ \\lambda y) = a(x+ \\lambda y) = ax + \\lambda (ay) = f(x) + \\lambda f(y),\\]
\nthat is, \\(f\\) is a linear transformation. In this case, the kernel of \\(f\\) is \\(\\left\\{ 0 \\right\\}\\) and the image is \\(\\mathbb{R}.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.17<\/span>
\nDefine a function \\(A\\) from \\(\\mathbb{R}^n\\) to \\(\\mathbb{R}\\) as follows:
\n\\[A(x) = \\frac{1}{n} \\sum_{j=1}^n x_j \\]
\nfor \\(x = (x_1 ,\\, x_2 ,\\, \\cdots ,\\, x_n ).\\)
\nThus \\(A(x)\\) is just the average of the components of \\(x.\\) Show that \\(A\\) is a linear transformation.\n<\/p>\n

\n

Solution.<\/span>
\nSuppose \\(x = (x_1 ,\\, \\cdots ,\\, x_n )\\) and \\(y = (y_1 ,\\, \\cdots ,\\, y_n )\\) are vectors and \\(\\lambda\\) is a scalar, then we have
\n\\[\\begin{align}
\nA(x+\\lambda y)
\n&= \\frac{1}{n} \\sum_{j=1}^n (x_j + \\lambda y_j ) \\\\[4pt]
\n&= \\frac{1}{n} \\sum_{j=1}^n x_j + \\lambda \\frac{1}{n} \\sum_{j=1}^{n} y_j \\\\[6pt]
\n&= A(x) + \\lambda A(y),
\n\\end{align}\\]
\nthat is, \\(A\\) is a linear transformation.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.18<\/span>
\nExpress the kernel of the linear transformation \\(T : \\mathbb{R}^2 \\rightarrow \\mathbb{R}\\) defined by
\n\\[T(x_1 ,\\,x_2 ) = 2x_1 – 5x_2\\]
\nas the span of a single vector.\n<\/p>\n

\n

Solution.<\/span>
\nThe kernel consists of the vectors \\((x_1 ,\\,x_2 )\\) that satisfy
\n\\[2x_1 + 5x_2 = 0.\\]
\nSubstituting \\(5t\\) to \\(x_1 ,\\) we have
\n\\[ (x_1 ,\\,x_2 ) = (5t ,\\, -2t) = t(5,\\,-2) .\\]
\nTherefore, the kernel is generated by the a single vector \\((5,\\,-2 ).\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.19<\/span>
\nWhat is the kernel of the second derivative operator
\n\\[C^2 (\\mathbb{R} ) \\rightarrow C^0 (\\mathbb{R} ) ,\\quad f \\mapsto \\frac{d^2 f}{dx^2}\\]
\non the real vector space of twice differentiable functions with continuous second derivative?\n<\/p>\n

\n

Solution.<\/span>
\nThe set of first-order polynomial functions and constant functions is the kernel.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.20<\/span>
\nWhat is the kernel of the \\(n\\)th derivative operator
\n\\[C^n (\\mathbb{R} ) \\rightarrow C^0 (\\mathbb{R} ) ,\\quad f \\mapsto \\frac{d^n f}{dx^n}\\]
\non the real vector space of \\(n\\)-times differentiable functions with continuous \\(n\\)th derivative?\n<\/p>\n

\n

Solution.<\/span>
\nThe set of constant functions and polynomial functions whose orders are less than \\(n\\) is the kernel.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.21<\/span>
\nExplain succinctly why the solution space of the differential equation
\n\\[y ‘ ‘ – 2 y ‘ + y = 0 \\tag{*}\\]
\nis a subspace of \\(C^2 (\\mathbb{R} ).\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(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
\n\\[y ‘ ‘ = 2 y ‘ – y ,\\]
\nwe 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.<\/p>\n

We now suppose that two functions \\(f_1 ,\\) \\(f_2 \\) belong to \\(W,\\) and that \\(k\\) is a scalar(real number). Since
\n\\[\\begin{gather}
\n(f_1 + f_2 ) ‘ ‘ = 2 ( f_1 + f_2 ) ‘ – (f_1 + f_2 ), \\\\[5pt]
\n(kf_1 ) ‘ ‘ = 2 ( k f_1 ) – (kf_1 ),
\n\\end{gather}
\n\\]
\n\\(W\\) is closed under vector addition and scalar multiplication. Therefore \\(W\\) is a subspace of \\(C^2 (\\mathbb{R} ).\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.22<\/span>
\nShow that a linear map \\(T : \\mathbb{R} \\rightarrow \\mathbb{R}^2\\) cannot be surjective.\n<\/p>\n

\n

Solution.<\/span>
\nSuppose that \\(T : \\mathbb{R} \\rightarrow \\mathbb{R}^2\\) is a surjective linear map. Take \\(v_1 \\) and \\(v_2\\) in \\(\\mathbb{R}\\) such that
\n\\[T(v_1 ) = (1,\\,0) \\quad \\text{and}\\quad T(v_2 ) = (0,\\,1).\\]
\nSince \\(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
\n\\[(1,\\,0) = T(v_1) = \\frac{1}{\\lambda} T(\\lambda v_1) = \\frac{1}{\\lambda} T(v_2) = \\frac{1}{\\lambda} (0,\\,1),\\]
\nthat is,
\n\\[\\lambda (1,\\,0) = (0,\\,1) .\\]
\nBut no \\(\\lambda\\) satisfies this equation, therefore we conclude that \\(T\\) cannot be a surjective linear map.\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.23<\/span>
\nLet \\(a\\in\\mathbb{R}\\) be fixed and consider the function \\(\\nu_a\\) defined by
\n\\[\\nu_a : C^0 ( \\mathbb{R} ) \\rightarrow \\mathbb{R} ,\\quad f \\mapsto f(a) .\\]
\nThis map is called evaluation at \\(a\\)<\/span>. Show that for all real numbers \\(a,\\) the evaluation map \\(\\nu_a\\) is a linear transformation and describe \\(\\operatorname{Ker} (\\nu _a ).\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(f\\) and \\(g\\) be continuous functions, \\(\\lambda\\) be a scalar. Then we have
\n\\[\\nu_a ( f + \\lambda g) = (f + \\lambda g)(a) = f(a) + \\lambda g(a) = \\nu_a (f) + \\lambda \\nu_a (g),\\]
\nwhich shows the linearity of \\(\\nu_a .\\) The kernel is the set of functions \\(f\\) with \\(f(a) = 0.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.24<\/span>
\nLet \\(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:
\n\\[I(f) = \\int_a^b f(x) dx.\\]
\nShow that \\(I\\) is a linear transformation. Deduce from this that the set
\n\\[\\left\\{ f \\in C^0 ( \\mathbb{R} ) \\,\\bigg\\vert \\int_a ^b f(x) dx =0 \\right\\}\\]
\nis a subspace of \\(C^0 (\\mathbb{R} ) .\\) Assuming that \\(a \\ne b,\\) what is the image of \\(I?\\)\n<\/p>\n

\n

Solution.<\/span>
\nIf \\(f\\) and \\(g\\) are continuous functions and \\(\\lambda\\) is a scalar, then
\n\\[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),\\]
\nwhich proves the linearity of \\(I.\\) The image of \\(I\\) is \\(\\mathbb{R}.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.25<\/span>
\nLet \\(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
\n\\[T(v) = \\lambda v.\\tag{*}\\]
\nThus \\(T\\) maps \\(v\\) onto a scalar multiple of itself. Then \\(\\lambda\\) is called an eigenvalue of \\(T\\)<\/span> and any \\(v\\) (even \\(\\mathbf{0}\\)) satisfying the equation above is called an eigenvector belonging to \\(\\lambda\\)<\/span>. Show that the set of all eigenvectors belonging to the eigenvalue \\(\\lambda\\) is a subspace of \\(V.\\) This is called the eigenspace<\/span> belonging to \\(\\lambda.\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(W\\) be the set of vectors \\(v\\) that satisfy (*).<\/p>\n

First, since \\(T(\\mathbf{0}) = \\lambda \\mathbf{0} = \\mathbf{0},\\) the zero vector belongs to \\(W.\\)\n<\/p>\n

Next, suppose that \\(u\\) and \\(v\\) are vectors in \\(W\\) and \\(k\\) is a scalar. Since
\n\\[T(u) = \\lambda u \\quad\\text{and}\\quad T(v) = \\lambda v ,\\]
\nwe have
\n\\[T(u+v) = T(u) + T(v) = \\lambda u + \\lambda v = \\lambda (u+v),\\]
\nwhich shows that \\(u+v \\in W.\\) Besides, since
\n\\[T(ku) = kT(u) = k \\lambda u = \\lambda (ku) ,\\]
\n\\(ku\\) also belongs to \\(W.\\) Therefore \\(W\\) is a subspace of \\(V.\\)\n<\/p><\/div>\n

<\/p>\n

Problem 3.26<\/span>
\nGiven 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.\n<\/p>\n

\n

Solution.<\/span>
\nIf \\(u_1 ,\\) \\(\\cdots ,\\) \\(u_m\\) generate \\(W_0\\) and \\(v_1 ,\\) \\(\\cdots ,\\) \\(v_n\\) generate \\(W_1 ,\\) then
\n\\[u_1 ,\\, \\cdots ,\\, u_m ,\\, v_1 ,\\, \\cdots ,\\, v_n\\]
\ngenerate \\(W_0 + W_1 .\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.27<\/span>
\nDefine subspaces \\(W_0\\) and \\(W_1\\) in \\(\\mathbb{R}^2\\) as follows:
\n\\[W_0 = \\operatorname{Span} \\left\\{ (0,\\,1) \\right\\} \\quad \\text{and} \\quad W_1 = \\operatorname{Span} \\left\\{ (1,\\,1) \\right\\} .\\]
\nShow that \\(\\mathbb{R}^2 = W_0 \\oplus W_1 .\\)\n<\/p>\n

\n

Solution.<\/span>
\nLet \\((x,\\,y) \\in \\mathbb{R}^2\\) and take \\(a = -x+y\\) and \\(b=x,\\) then
\n\\[(x,\\,y) = a(0,\\,1) + b(1,\\,1) \\in W_0 + W_1 .\\]
\nIt remains to show that \\(W_0 \\cap W_1 = \\left\\{ (0,\\,0) \\right\\}.\\)
\nSuppose \\((x_0 ,\\,y_0 ) \\in W_0 \\cap W_1 ,\\) then there exist scalars \\(s\\) and \\(t\\) such that
\n\\[(x_0 ,\\,y_0 ) = s(0,\\,1) = t(1,\\,1).\\]
\nThis 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\\}.\\)\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.28<\/span>
\nLet \\(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
\n\\[(T_0 \\times T_1)(u) = (T_0 (u) ,\\, T_1 (u)).\\]
\nProve that \\(\\operatorname{Ker}(T_0 \\times T_1 ) = \\operatorname{Ker}(T_0 ) \\cap \\operatorname{Ker}(T_1 ).\\)\n<\/p>\n

\n

Solution.<\/span>
\nThe proof is straightforward, that is,
\n\\[\\begin{align}
\n\\operatorname{Ker}(T_0 \\times T_1 )
\n&= \\left\\{ u\\in U \\,\\,\\vert\\,\\, (T_0 (u) ,\\, T_1 (u)) = (0,\\,0) \\right\\} \\\\[7pt]
\n&= \\left\\{ u\\in U \\,\\,\\vert\\,\\, T_0 (u)=0 \\,\\,\\text{and}\\,\\, T_1 (u)=0 \\right\\} \\\\[7pt]
\n&= \\left\\{ u\\in U \\,\\,\\vert\\,\\, T_0 (u)=0 \\right\\} \\cap \\left\\{ u\\in U \\,\\,\\vert\\,\\, T_1 (u)=0 \\right\\} \\\\[7pt]
\n&= \\operatorname{Ker}(T_0 ) \\cap \\operatorname{Ker}(T_1 ).
\n\\end{align}\\]\n<\/p>\n<\/div>\n

<\/p>\n

Problem 3.29<\/span>
\nLet \\(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<\/span>.) Show that
\n\\[V = \\operatorname{Ker} (T) \\oplus \\operatorname{Im} (T) .\\]\n<\/p>\n

\n

Solution.<\/span>
\nLet \\(v \\in V\\) and take \\(w = v-T(v).\\) Since \\(T\\) is a linear transformation and \\(T(T(v)) = T(v),\\) we have
\n\\[T(w) = T(v-T(v)) = T(v) – T(T(v)) = T(v) – T(v) = \\mathbf{0},\\]
\nthat is, \\(w\\in \\operatorname{Ker}(T).\\) Observe that
\n\\[v = (v-f(v))+f(v) = w+f(v)\\]
\nwhere \\(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).\\)<\/p>\n

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
\n\\(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\\}.\\)\n<\/p>\n<\/div>\n

<\/p>\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

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…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_lmt_disableupdate":"","_lmt_disable":"","footnotes":""},"categories":[57],"tags":[402,400,412,411],"class_list":["post-5010","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","tag-exercises","tag-linear-algebra","tag-linear-transformation","tag-vector-space"],"_links":{"self":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5010","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/comments?post=5010"}],"version-history":[{"count":87,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5010\/revisions"}],"predecessor-version":[{"id":6171,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/posts\/5010\/revisions\/6171"}],"wp:attachment":[{"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/media?parent=5010"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/categories?post=5010"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sasamath.com\/blog\/wp-json\/wp\/v2\/tags?post=5010"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}