Solutions

Solution 3.1

(See Exercise 3.1.) To decide whether we get a vector space, we need to check the axioms of a vector space (Definition 3.10), especially

• \((r+s)v=rv+sv\) (Definition 3.104.),

• \((rs)v=r(sv)\) (Definition 3.106.),

• and \(1v=v\) (item 7 in Definition 3.10).

• \(r\cdot(x,y,z)=(rx,y,rz)\). Let \(v=(x,y,z)\). Then

\[ (r+s)\cdot v=((r+s)x,y,(r+s)z), \]

while

\[ r\cdot v+s\cdot v=(rx,y,rz)+(sx,y,sz)=((r+s)x,2y,(r+s)z). \]

In general these are different (e.g. if \(y\neq 0\)), so

\[ (r+s)v\neq rv+sv. \]

Thus axiom Definition 3.104. fails. Hence this is not a vector space structure.

1. \(r\cdot(x,y,z)=(0,0,0)\). For any nonzero \(v\in V\), we have \(1\cdot v=(0,0,0)\neq v\). So the identity axiom \(1v=v\) from Definition 3.10 fails. Hence this is not a vector space structure.

2. \(r\cdot(x,y,z)=(r^2x,r^2y,r^2z)\).

   Here, for every \(v=(x,y,z)\in V\),

\[ 1\cdot v=(1^2x,1^2y,1^2z)=(x,y,z)=v, \]

so the axiom \(1v=v\) (item 7 in Definition 3.10) is satisfied. In addition, Definition 3.106. does hold here. We now consider the distributivity axiom:

\[ (r+s)\cdot v=((r+s)^2x,(r+s)^2y,(r+s)^2z), \]

while

\[ r\cdot v+s\cdot v=(r^2x,r^2y,r^2z)+(s^2x,s^2y,s^2z)=((r^2+s^2)x,(r^2+s^2)y,(r^2+s^2)z). \]

In general these are different (e.g. \(r=s=1\) and \(v\neq 0\)), so

\[ (r+s)v\neq rv+sv. \]

Hence Definition 3.104. fails, so this is not a vector space structure.

Therefore, in all three cases, \(V\) is not a vector space.

Solution 3.2

(See Exercise 3.2.) Recall Definition 3.17: a subspace \(V \subset {\bf R}^2\) must (i) contain the zero vector, (ii) be closed under addition, and (iii) be closed under scalar multiplication.

1. Correct. By condition (i), \(V\) contains the zero vector \((0,0)\), so \(V\) contains at least one element.

2. Incorrect. The trivial subspace \(V = \{(0,0)\}\) is a subspace of \({\bf R}^2\) (all three conditions are satisfied) and contains exactly one element.

3. Correct. Condition (i) of Definition 3.17 states that the zero vector must belong to \(V\).

4. Correct. Let \(v, w \in V\). By condition (iii), \((-1) \cdot w \in V\), i.e. \(-w \in V\). Then by condition (ii), \(v + (-w) = v - w \in V\).

Solution 3.3

(See Exercise 3.3.) Let \(D := \{ f : {\bf R} \to {\bf R} \mid f \text{ is differentiable} \}\), with sum and scalar multiple as in Equation (3.23) and Equation (3.24). Following the structure of Definition and Lemma 3.22, we check the conditions of being a subvector space of the space of all functions \(f : {\bf R} \to {\bf R}\). We note that the space of all functions does form a vector space. Its vector space axioms (commutativity and associativity of addition, distributivity, etc.) are inherited pointwise from \({\bf R}\): for any \(x \in {\bf R}\), the values \(f(x), g(x), h(x)\) are real numbers satisfying all these properties, hence so do the functions \(f, g, h\).

We verify the conditions of Definition 3.17, thereby confirming that \(D\) is a vector space.

• Closure under addition. If \(f, g \in D\), then both \(f\) and \(g\) are differentiable. By the sum rule from calculus, \(f + g\) is differentiable with \((f+g)' = f' + g'\). Hence \(f + g \in D\).

• Closure under scalar multiplication. If \(f \in D\) and \(r \in {\bf R}\), then by the constant multiple rule, \(rf\) is differentiable with \((rf)' = r f'\). Hence \(rf \in D\).

• Zero vector. The zero vector is the zero function \(f(x) = 0\), which is differentiable (with derivative \(0\)), so \(0 \in D\).

Solution 3.4

(See Exercise 3.4.) Union that is not a subspace. Take \(V = L((1,0))\) (the \(x\)-axis) and \(W = L((0,1))\) (the \(y\)-axis). Both are subspaces of \({\bf R}^2\). We have \((1,0) \in V \subset V \cup W\) and \((0,1) \in W \subset V \cup W\), but

\[ (1,0) + (0,1) = (1,1) \notin V \cup W, \]

since \((1,1)\) is not a scalar multiple of \((1,0)\) and not a scalar multiple of \((0,1)\). Hence \(V \cup W\) is not closed under addition, so it is not a subspace.

Union that is a subspace. Take \(V = \{(0,0)\}\) (the trivial subspace) and \(W\) any subspace of \({\bf R}^2\). Then \(V \cup W = W\), which is a subspace. Another possibility is to take \(V = W\) for any subspace \(V\), or also to take \(V = \bf R^2\) and any subspace \(W \subset \bf R^2\).

Solution 3.5

(See Exercise 3.5.) By definition of the span (Lemma 3.34), in each case we need to solve \(\lambda_1 v_1 + \lambda_2 v_2 = w\). Then \(w\) lies in the span of \(v_1\) and \(v_2\) precisely if there is a solution \((\lambda_1, \lambda_2)\) of that system.

1. \(w=(2,-1,0,1)\), \(v_1=(1,0,0,1)\), \(v_2=(0,1,0,1)\). Inspecting the four components of the equation \(\lambda_1 v_1 + \lambda_2 v_2 = w\) we get the linear system

\[ \lambda_1 = 2, \quad \lambda_2 = -1, \quad 0 = 0, \quad \lambda_1 + \lambda_2 = 1. \]

The last equation: \(2 + (-1) = 1\) holds. The system is consistent, so \(w \in L(v_1, v_2)\) with, namely \(w = 2v_1 - v_2\).

1. For \(w=(1,2,15,11)\), \(v_1=(2,-1,0,2)\), \(v_2=(1,-1,-3,1)\). The system \(\lambda_1 v_1 + \lambda_2 v_2 = w\) is:

\[ \begin{align*} 2\lambda_1 + \lambda_2 &= 1, \\ -\lambda_1 - \lambda_2 &= 2, \\ -3\lambda_2 &= 15, \\ 2\lambda_1 + \lambda_2 &= 11. \end{align*} \]

Equation (3) gives \(\lambda_2 = -5\); equation (1) then gives \(\lambda_1 = 3\). But equation (4) requires \(2(3)+(-5)=1\neq 11\). The system is inconsistent, so \(w \notin L(v_1,v_2)\).

1. \(w=(2,5,8,3)\), \(v_1=(2,-1,0,5)\), \(v_2=(-1,2,2,-3)\).

   The system \(\lambda_1 v_1 + \lambda_2 v_2 = w\) is:

\[ \begin{align*} 2\lambda_1 - \lambda_2 &= 2, \\ -\lambda_1 + 2\lambda_2 &= 5, \\ 2\lambda_2 &= 8, \\ 5\lambda_1 - 3\lambda_2 &= 3. \end{align*} \]

One solves this similarly as above and finds that the system has the solution \(\lambda_1 = 3\), \(\lambda_2 = 4\). Hence \(w \in L(v_1, v_2)\) with

\[ w = 3v_1 + 4v_2. \]

Solution 3.6

(See Exercise 3.6.) By Corollary 3.73, four vectors in \({\bf R}^4\) span \({\bf R}^4\) if and only if they are linearly independent, which by Method 3.56 is equivalent to the matrix formed by the vectors (as rows) having rank 4 after Gaussian elimination (Method 2.29).

1. Form the matrix with rows \((1,1,1,1)\), \((0,1,1,1)\), \((0,0,1,1)\), \((0,0,0,1)\):

\[ \begin{pmatrix} 1&1&1&1 \\ 0&1&1&1 \\ 0&0&1&1 \\ 0&0&0&1 \end{pmatrix}. \]

This matrix is already in row-echelon form (Definition 2.27) with 4 pivot rows, so its rank equals 4. Hence the four vectors are linearly independent and do span \({\bf R}^4\).

1. Form the matrix with rows \((1,3,-5,0)\), \((-2,1,0,0)\), \((0,2,1,-1)\), \((1,-4,5,0)\) and apply elementary row operations (Definition 2.28):

\[ \begin{pmatrix} 1&3&-5&0 \\ -2&1&0&0 \\ 0&2&1&-1 \\ 1&-4&5&0 \end{pmatrix} \xrightarrow{R_2+2R_1,\; R_4-R_1} \begin{pmatrix} 1&3&-5&0 \\ 0&7&-10&0 \\ 0&2&1&-1 \\ 0&-7&10&0 \end{pmatrix} \xrightarrow{R_4+R_2} \begin{pmatrix} 1&3&-5&0 \\ 0&7&-10&0 \\ 0&2&1&-1 \\ 0&0&0&0 \end{pmatrix} \xrightarrow{7R_3-2R_2} \begin{pmatrix} 1&3&-5&0 \\ 0&7&-10&0 \\ 0&0&27&-7 \\ 0&0&0&0 \end{pmatrix}. \]

The resulting row-echelon matrix has only 3 non-zero rows, so the rank equals 3 \(< 4\). Hence the four vectors are linearly dependent and do not span \({\bf R}^4\).

Solution 3.7

(See Exercise 3.7.) By Definition 3.49 and Method 3.56, we form the matrix with the given vectors as rows and apply Gaussian elimination (Method 2.29, Definition 2.28). The vectors are linearly independent if and only if the resulting row-echelon matrix (Definition 2.27) has no zero rows.

1. \(v_1=(1,-1,0)\), \(v_2=(3,2,-1)\), \(v_3=(3,5,-2)\) in \({\bf R}^3\):

\[ \begin{pmatrix} 1&-1&0 \\ 3&2&-1 \\ 3&5&-2 \end{pmatrix} \xrightarrow{R_2-3R_1,\; R_3-3R_1} \begin{pmatrix} 1&-1&0 \\ 0&5&-1 \\ 0&8&-2 \end{pmatrix} \xrightarrow{5R_3-8R_2} \begin{pmatrix} 1&-1&0 \\ 0&5&-1 \\ 0&0&-2 \end{pmatrix}. \]

There is no zero row, so the vectors are linearly independent.

1. \(v_1=(1,1,1)\), \(v_2=(1,-1,1)\), \(v_3=(0,0,1)\) in \({\bf R}^3\):

\[ \begin{pmatrix} 1&1&1 \\ 1&-1&1 \\ 0&0&1 \end{pmatrix} \xrightarrow{R_2-R_1} \begin{pmatrix} 1&1&1 \\ 0&-2&0 \\ 0&0&1 \end{pmatrix}. \]

Again, there is no zero row, so the vectors are linearly independent.

\[ \begin{pmatrix} 1&-1&1&-1 \\ 2&0&1&0 \\ 0&-2&1&-2 \end{pmatrix} \xrightarrow{R_2-2R_1} \begin{pmatrix} 1&-1&1&-1 \\ 0&2&-1&2 \\ 0&-2&1&-2 \end{pmatrix} \xrightarrow{R_3+R_2} \begin{pmatrix} 1&-1&1&-1 \\ 0&2&-1&2 \\ 0&0&0&0 \end{pmatrix}. \]

The presence of the zero row at the en means that the vectors are linearly dependent.

1. \((1,1,0,0)\), \((1,0,1,0)\), \((0,0,1,1)\), \((0,1,0,1)\) in \({\bf R}^4\):

\[ \begin{pmatrix} 1&1&0&0 \\ 1&0&1&0 \\ 0&0&1&1 \\ 0&1&0&1 \end{pmatrix} \xrightarrow{R_2-R_1} \begin{pmatrix} 1&1&0&0 \\ 0&-1&1&0 \\ 0&0&1&1 \\ 0&1&0&1 \end{pmatrix} \xrightarrow{R_4+R_2} \begin{pmatrix} 1&1&0&0 \\ 0&-1&1&0 \\ 0&0&1&1 \\ 0&0&1&1 \end{pmatrix} \xrightarrow{R_4-R_3} \begin{pmatrix} 1&1&0&0 \\ 0&-1&1&0 \\ 0&0&1&1 \\ 0&0&0&0 \end{pmatrix}. \]

Again, the vectors are linearly dependent.

Solution 3.8

(See Exercise 3.8.) Since \(\dim {\bf R}^2 = 2\), any three vectors in \({\bf R}^2\) are automatically linearly dependent: by Theorem 3.712., a linearly independent set can have at most \(\dim V = 2\) elements, so the last condition of the exercise is satisfied for any choice of three vectors in \({\bf R}^2\). It therefore suffices to find three vectors that are pairwise linearly independent. By Definition 3.49, two vectors are linearly independent if and only if neither is a scalar multiple of the other. A natural starting point is \(v_1 = (1,0)\) and \(v_2 = (0,1)\) (the standard basis vectors, clearly not scalar multiples of each other). For \(v_3\) we need a vector that is not a scalar multiple of \(v_1\) or of \(v_2\); one such choice is \(v_3 = v_1 + v_2 = (1,1)\), which also makes the dependence relation among the triple transparent.

Solution 3.9

(See Exercise 3.9.) We check the three conditions of Definition 3.17 for each subset of \(V = {\bf R}[x]^{\le 3}\) (Definition and Lemma 3.22).

1. \(S_1 = \{ f \in V \mid f(2) = 1 \}\). is not a subspace since the zero polynomial satisfies \(0(2) = 0 \neq 1\), so \(0 \notin S_1\). Condition (1) of Definition 3.17 fails.

2. \(S_2 = \{ x \cdot f \mid f \in {\bf R}[x]^{\le 2} \}\) is a subspace. Writing \(f = a_0 + a_1 x + a_2 x^2\), we get \(xf = a_0 x + a_1 x^2 + a_2 x^3\), so \(S_2 = L(x, x^2, x^3)\). By Lemma 3.34, every span is a subspace.

3. \(S_3 = \{ x \cdot f + (1-x) \cdot g \mid f, g \in {\bf R}[x]^{\le 2} \}\) also is a subspace. We verify Definition 3.17 directly:

   • \(0 \in S_3\): take \(f = g = 0\).

   • Closure under addition: \((xf_1 + (1-x)g_1) + (xf_2 + (1-x)g_2) = x(f_1+f_2) + (1-x)(g_1+g_2) \in S_3\).

   • Closure under scalar multiplication: \(c(xf + (1-x)g) = x(cf) + (1-x)(cg) \in S_3\).

   (In fact one has \(S_3 = V\): writing \(xf + (1-x)g = g + x(f-g)\), any \(h \in V\) can be achieved by choosing \(g\) as the constant-through-degree-2 part and \(f-g\) as needed.)

4. \(S_4 = \{ f \in {\bf R}[x]^{\le 3} \mid f(0) = 0 \}\). Is a subspace. We verify Definition 3.17:

   • \(0 \in S_4\): \(0(0) = 0\) holds true.

   • Closure under addition: \((f+g)(0) = f(0) + g(0) = 0 + 0 = 0\), so \(f + g \in S_4\) for \(f, g \in S_4\).

   • Closure under scalar multiplication: \((cf)(0) = c \cdot f(0) = 0\).

   (Equivalently, \(f(0)=0\) means the constant term of \(f\) is zero, so \(S_4 = L(x, x^2, x^3)\).)

Solution 3.10

(See Exercise 3.10.) We seek \(a, b, c \in {\bf R}\) such that \(a(x+1) + b(x-1) + c(x^2-1) = p(x)\) (Definition 3.31). Expanding the left side and collecting by degree:

\[ a(x+1) + b(x-1) + c(x^2-1) = cx^2 + (a+b)x + (a-b-c). \]

Thus, if \(p(x) = p_2 x^2 + p_1 x + p_0\), then we need \(c = p_2\), \(a+b = p_1\), and \(a-b-c = p_0\). I.e. \(a = \tfrac{1}{2}(p_1 + p_0 + c)\) and \(b = \tfrac{1}{2}(p_1 - p_0 - c)\). This leads to the following expressions for \(p\) as a linear combination of the three given polynomials:

\[ x^2+4x-2 = \tfrac{3}{2}(x+1) + \tfrac{5}{2}(x-1) + (x^2-1). \]

\[ x = \tfrac{1}{2}(x+1) + \tfrac{1}{2}(x-1). \]

\[ 40-x^2 = \tfrac{39}{2}(x+1) - \tfrac{39}{2}(x-1) - (x^2-1). \]

Solution 3.11

(See Exercise 3.11.) The sentence is incorrect, since the coefficient \(\tfrac{x}{4}\) is not a scalar. By Definition 3.31, a linear combination of vectors \(v_1, \dots, v_m\) is an expression \(a_1 v_1 + \dots + a_m v_m\) where the coefficients \(a_i\) are real numbers. The expression \(\tfrac{x}{4} \cdot x^2 + 3 \cdot x + 1\) uses \(\tfrac{x}{4}\) as the coefficient of \(x^2\), but \(\tfrac{x}{4}\) is a polynomial in \(x\), not a real number. It is therefore not a valid scalar coefficient.

Solution 3.12

(See Exercise 3.12.) From Example 3.63 the basis is \(v_1=(0,2,1)\), \(v_2=(1,0,2)\), \(v_3=(-1,1,1)\). We seek coefficients \(\alpha_i,\beta_i,\gamma_i\) with \(e_i = \alpha_i v_1 + \beta_i v_2 + \gamma_i v_3\) for each \(i\) (Definition 3.61, Definition 3.31). All three systems share the same coefficient matrix (columns \(v_1,v_2,v_3\)), so we row-reduce the augmented matrix \([v_1 \ v_2 \ v_3 \mid e_1 \ e_2 \ e_3]\) in one pass (Method 2.31):

\[ \begin{align*} \left ( \begin{array}{ccc|ccc} 0 & 1 & -1 & 1 & 0 & 0 \\ 2 & 0 & 1 & 0 & 1 & 0 \\ 1 & 2 & 1 & 0 & 0 & 1 \end{array} \right ) &\xrightarrow{R_1 \leftrightarrow R_3} \left ( \begin{array}{ccc|ccc} 1 & 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 1 & 0 & 0 \end{array} \right ) \\ &\xrightarrow{R_2 - 2R_1} \left ( \begin{array}{ccc|ccc} 1 & 2 & 1 & 0 & 0 & 1 \\ 0 & -4 & -1 & 0 & 1 & -2 \\ 0 & 1 & -1 & 1 & 0 & 0 \end{array} \right ) \\ &\xrightarrow{4R_3 + R_2} \left ( \begin{array}{ccc|ccc} 1 & 2 & 1 & 0 & 0 & 1 \\ 0 & -4 & -1 & 0 & 1 & -2 \\ 0 & 0 & -5 & 4 & 1 & -2 \end{array} \right ). \end{align*} \]

Back-substituting (\(R_3 \div (-5)\), then \(R_2\), then \(R_1\)) yields the reduced form

\[ \left ( \begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac{2}{5} & \tfrac{3}{5} & -\tfrac{1}{5} \\[2pt] 0 & 1 & 0 & \tfrac{1}{5} & -\tfrac{1}{5} & \tfrac{2}{5} \\[2pt] 0 & 0 & 1 & -\tfrac{4}{5} & -\tfrac{1}{5} & \tfrac{2}{5} \end{array} \right ). \]

(In terminology that will be introduced later on, the right hand part of the above matrix is a so-called base change matrix from the basis \(v_1,v_2,v_3\) to the standard basis \(e_1,e_2,e_3\), see Example 4.85.) Reading off the right-hand columns:

\[ \begin{align*} e_1 &= \tfrac{2}{5}\,v_1 + \tfrac{1}{5}\,v_2 - \tfrac{4}{5}\,v_3, \\ e_2 &= \tfrac{3}{5}\,v_1 - \tfrac{1}{5}\,v_2 - \tfrac{1}{5}\,v_3, \\ e_3 &= -\tfrac{1}{5}\,v_1 + \tfrac{2}{5}\,v_2 + \tfrac{2}{5}\,v_3. \end{align*} \]

Solution 3.13

(See Exercise 3.13.) A linear combination of \(A\) and \(B\) is of the form

\[ \alpha A + \beta B = \alpha \left ( \begin{array}{cc} 1 & 1 \\ 2 & 2 \end{array} \right ) + \beta \left ( \begin{array}{cc} 3 & 2 \\ 3 & 5 \end{array} \right ) \]

with \(\alpha, \beta \in {\bf R}\). Computing the left hand side, we need to find \(\alpha\) and \(\beta\) such that

\[ \left ( \begin{array}{cc} \alpha & \alpha \\ 2 \alpha & 2 \alpha \end{array} \right ) + \left ( \begin{array}{cc} 3 \beta & 2 \beta \\ 3 \beta & 5 \beta \end{array} \right ) = \left ( \begin{array}{cc} \alpha + 3\beta & \alpha + 2 \beta \\ 2\alpha+3\beta & 2\alpha+5\beta \end{array} \right ) = \left ( \begin{array}{cc} -1 & 0 \\ 2 & 4 \end{array} \right ). \]

Comparing the entries of the matrix, this gives the linear system

\[ \begin{align*} \alpha + 3 \beta & = -1 \\ \alpha + 2 \beta & = 0\\ 2\alpha + 3\beta & = 2\\ 2\alpha + 5 \beta & = 4. \end{align*} \]

The second gives \(\alpha = -2\beta\), inserting into the first gives \(-2 \beta+3 \beta = -1\), which means \(\beta = -1\). However, inserting into the third equation gives \(-4\beta + 3 \beta = 2\), so that \(\beta = -2\), contradicting the previous equation. Thus, there is no solution, so \(C\) is not a linear combination of \(A\) and \(B\).

Solution 3.14

(See Exercise 3.14.)

1. \(T\) is a subspace. The two defining constraints

\[ x_1 + x_4 + x_6 = 0, \qquad x_1 + x_4 + x_3 + x_5 = 0 \]

are homogeneous linear equations in the six entries of the matrix (Definition 2.13). Identifying \(\mathrm{Mat}_{2\times3}\) with \({\bf R}^6\), \(T\) is the solution set of a homogeneous linear system, hence a subspace by Proposition 3.18.

1. All matrices in \(T\). Solving the two equations for the dependent variables (Method 2.31):

\[ x_6 = -x_1 - x_4, \qquad x_5 = -x_1 - x_4 - x_3. \]

The free variables are \(x_1, x_2, x_3, x_4\). Setting \(x_1=a\), \(x_2=b\), \(x_3=c\), \(x_4=d\) gives

\[ T = \left\{ \begin{pmatrix} a & b & c \\ d & -a-d-c & -a-d \end{pmatrix} \;\middle|\; a,b,c,d \in {\bf R} \right\}. \]

1. Spanning set. Separating the four free parameters:

\[ \begin{align*} \begin{pmatrix} a & b & c \\ d & -a-d-c & -a-d \end{pmatrix} = a\,v_1 + b\,v_2 + c\,v_3 + d\,v_4, \end{align*} \]

where

\[ v_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & -1 \end{pmatrix}, \quad v_2 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad v_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}, \quad v_4 = \begin{pmatrix} 0 & 0 & 0 \\ 1 & -1 & -1 \end{pmatrix}. \]

Hence \(T = L(v_1, v_2, v_3, v_4)\) (Lemma 3.34). These four matrices are linearly independent (each has a non-zero entry in a position where the others are zero), so they also form a basis of \(T\) with \(\dim T = 4\).

Solution 3.15

(See Exercise 3.15.) The system \(x + y + z+t=0\) corresponds to the matrix

\[ ( 1 \ 1 \ 1 \ 1 ). \]

This matrix is already in reduced row echelon form (Definition 2.27): the leading one is for the variable \(x\), the variables \(y, z, t\) are free variables. Thus,

\[ S = \{(-\alpha - \beta - \gamma , \alpha, \beta, \gamma) \ | \ \alpha, \beta, \gamma \in {\bf R} \}. \]

We have

\[ \begin{align*} S & = \{(-\alpha, \alpha, 0,0) + (-\beta, 0, \beta, 0) + (-\gamma, 0, 0, \gamma) \ | \ \alpha, \beta, \gamma \in {\bf R}\} \\ & = \{\alpha(-1, 1, 0,0) + \beta (-1, 0, 1, 0) + \gamma(-1, 0, 0, 1) \ | \ \alpha, \beta, \gamma \in {\bf R}\} \\ & = L((-1,1,0,0), (-1,0,1,0), (-1,0,0,1)). \end{align*} \]

Solution 3.16

(See Exercise 3.16.) By definition (cf. Lemma 3.34), \(S\) consists of all the linear combinations of the three given vectors. These can be written as

\[ a (1,-1,0,1) + b(2,1,-2,0)+c(0,0,1,1) = (a+2b,-a+b,-2b+c,a+c) \]

for arbitrary \(a,b,c \in {\bf R}\). The intersection (cf. Lemma 3.19) is given by vectors as above satisfying the linear system determining \(T\), i.e.,

\[ \begin{align*} x_1 &= a+2b\\ x_2 & = -a+b \\ x_3 &= -2b+c \\ x_4 & = a+c \end{align*} \]

such that

\[ \begin{align*} 2(a+2b)-(-a+b)-3(a+c) & = 0\\ 2(a+2b)+(-2b+c)+(a+c) & = 0. \end{align*} \]

Simplifying these equations gives

\[ \begin{align*} 3b-3c & = 0\\ 3a+2b+2c & = 0. \end{align*} \]

Thus \(b=c\) and \(3a+4c=0\), i.e., \(a = -\frac 43c\), and \(c\) is a free variable. (Alternatively, the above system is associated to the matrix \(\left ( \begin{array}{ccc} 0 & 3 & -3 \\ 3 & 2 & 2 \end{array} \right )\), which can be brought into reduced row echelon form.) Thus,

\[ \begin{align*} S \cap T & = \{-\frac 43 c(1,-1,0,1) + c(2,1,-2,0)+c(0,0,1,1) \ | \ c \in {\bf R} \} \\ & = \{ c \left ( (-\frac 43, \frac 43, 0, -\frac 43) + (2,1,-2,0)+(0,0,1,1) \right ) \ | \ c \in {\bf R} \} \\ & = \{ c ( \frac 23, \frac 73, -1, - \frac 13) \ | \ c \in {\bf R} \} \\ & = L(( \frac 23, \frac 73, -1, - \frac 13)). \end{align*} \]

Solution 3.17

(See Exercise 3.17.) \(T\) is a subspace by Proposition 3.18 (Definition 2.13). To find \(T \cap W\) we parametrise \(W\) and then impose \(T\)’s equations (Method 2.31).

A general element of \(W\) is

\[ w = a(1,0,1,0) + b(2,0,0,0) + c(0,-3,-1,-1) = (a+2b,\,-3c,\,a-c,\,-c) \]

for \(a,b,c \in {\bf R}\). Requiring \(w \in T\):

\[ \begin{align*} x_1 - x_2 = 0 &\;\Longrightarrow\; (a+2b) - (-3c) = 0 \;\Longrightarrow\; a + 2b + 3c = 0, \\ x_1 + x_2 + x_3 = 0 &\;\Longrightarrow\; (a+2b) + (-3c) + (a-c) = 0 \;\Longrightarrow\; a + b - 2c = 0. \end{align*} \]

Row-reducing this \(2\times 3\) system in \(a,b,c\):

\[ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 1 & -2 \end{pmatrix} \xrightarrow{R_2 - R_1} \begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & -5 \end{pmatrix} \xrightarrow{R_1 + 2R_2} \begin{pmatrix} 1 & 0 & -7 \\ 0 & -1 & -5 \end{pmatrix}. \]

So \(c\) is free, \(b = -5c\), \(a = 7c\). Substituting back:

\[ w = 7c(1,0,1,0) - 5c(2,0,0,0) + c(0,-3,-1,-1) = c(-3,-3,6,-1). \]

Hence

\[ T \cap W = L\!\bigl((-3,-3,6,-1)\bigr), \]

which has Definition 3.61: \(\{(-3,-3,6,-1)\}\) is a basis, and \(\dim(T \cap W) = 1\).

Solution 3.18

(See Exercise 3.18.) By Corollary 3.73, two vectors in \({\bf R}^2\) span \({\bf R}^2\) if and only if they are linearly independent (Definition 3.49). Two vectors in \({\bf R}^2\) are linearly dependent if and only if one of them is a scalar multiple of the other.

• The vectors \((1,1)\) and \((2,-1)\) are not scalar multiples of each other, confirming that \({\bf R}^2 = L((1,1),(2,-1))\).

• The vectors \((0,-2)\) and \((1,1)\) are also not scalar multiples of each other, so again \({\bf R}^2 = L((0,-2),(1,1))\).

Solution 3.19

(See Exercise 3.19.) We ask whether there exist \(a_1, a_2, a_3 \in {\bf R}\) with \(a_1 v_1 + a_2 v_2 + a_3 v_3 = (1,5,0)\) (Definition 3.31). This is the linear system (Definition 2.7) with augmented matrix

\[ \left ( \begin{array}{ccc|c} 1 & 2 & 0 & 1 \\ 1 & 0 & 3 & 5 \\ 0 & 1 & -1 & 0 \end{array} \right ) \xrightarrow{R_2 - R_1} \left ( \begin{array}{ccc|c} 1 & 2 & 0 & 1 \\ 0 & -2 & 3 & 4 \\ 0 & 1 & -1 & 0 \end{array} \right ) \xrightarrow{2R_3 + R_2} \left ( \begin{array}{ccc|c} 1 & 2 & 0 & 1 \\ 0 & -2 & 3 & 4 \\ 0 & 0 & 1 & 4 \end{array} \right ). \]

Back-substitution gives \(a_3 = 4\), then \(a_2 = 4\), then \(a_1 = -7\). The system is consistent, so yes, \((1,5,0)\) is a linear combination:

\[ (1,5,0) = -7\,v_1 + 4\,v_2 + 4\,v_3. \]

Solution 3.20

(See Exercise 3.20.) As was noted in Example 3.75, the polynomials \(1,(x-1),(x-1)^2,(x-1)^3,(x-1)^4\) form a basis of \({\bf R}[x]^{\le 4}\). Hence every polynomial of degree \(\le 4\) has a unique such expansion. We use the substitution \(y = x-1\) (i.e. \(x = 1+y\)) and expand via the binomial theorem (alternatively, the exercise may also be solved similarly as in Solution 3.10).

1. \(f(x) = x^4 = (1+y)^4 = 1 + 4y + 6y^2 + 4y^3 + y^4\), so

\[ x^4 = 1 + 4(x-1) + 6(x-1)^2 + 4(x-1)^3 + (x-1)^4. \]

1. \(f(x) = x^3 = (1+y)^3 = 1 + 3y + 3y^2 + y^3\), so

\[ x^3 = 1 + 3(x-1) + 3(x-1)^2 + (x-1)^3. \]

1. \(f(x) = x^3 - 3x^2 + 4x + 2\). Using the expansions above together with \(x^2 = 1 + 2(x-1) + (x-1)^2\) and \(x = 1+(x-1)\):

\[ \begin{align*} f &= \bigl[1 + 3(x-1) + 3(x-1)^2 + (x-1)^3\bigr] \\ &\quad{} - 3\bigl[1 + 2(x-1) + (x-1)^2\bigr] \\ &\quad{} + 4\bigl[1 + (x-1)\bigr] + 2. \end{align*} \]

Collecting by degree: constant \(1-3+4+2 = 4\); degree 1: \(3-6+4 = 1\); degree 2: \(3-3 = 0\); degree 3: \(1\). Hence

\[ x^3-3x^2+4x+2 = 4 + (x-1) + (x-1)^3. \]

Solution 3.21

(See Exercise 3.21.) By Definition 3.61 we must show that \(\{x-a, x-b\}\) is linearly independent and spans \({\bf R}[x]^{\le 1}\).

Linear independence (Definition 3.49). Suppose \(\lambda(x-a) + \mu(x-b) = 0\) in \({\bf R}[x]^{\le 1}\). Collecting by degree:

\[ (\lambda+\mu)\,x + (-\lambda a - \mu b) = 0. \]

Since the zero polynomial has all coefficients zero:

\[ \lambda + \mu = 0 \quad\text{and}\quad \lambda a + \mu b = 0. \]

From the first equation \(\mu = -\lambda\); substituting into the second gives \(\lambda(a-b) = 0\). Since \(a \neq b\) we conclude \(\lambda = 0\) and hence \(\mu = 0\). Thus \(x-a\) and \(x-b\) are linearly independent.

Spanning. We have \(\dim {\bf R}[x]^{\le 1} = 2\) (the standard basis \(\{1, x\}\) has two elements). Since \(\{x-a, x-b\}\) consists of exactly 2 linearly independent vectors in a 2-dimensional space, it is a basis by Corollary 3.73, and in particular spans \({\bf R}[x]^{\le 1}\).

Solution 3.22

(See Exercise 3.22.) Basis and dimension of \(W_1\). \(W_1\) is the solution set of the homogeneous system (Definition 2.13) \(y+t=0\), \(y+z=0\), so it is a subspace by Proposition 3.18. Solving (Method 2.31): the dependent variables are \(z = -y\) and \(t = -y\), while \(x\) and \(y\) are free. Setting \((x,y) = (1,0)\) and \((0,1)\) in turn:

\[ W_1 = \bigl\{(x,y,-y,-y) \mid x,y \in {\bf R}\bigr\} = L\bigl((1,0,0,0),\,(0,1,-1,-1)\bigr). \]

The two generators are clearly linearly independent (Definition 3.49), so

\[ \mathcal{B}_1 = \{(1,0,0,0),\,(0,1,-1,-1)\} \]

is a basis (Definition 3.61) of \(W_1\) and \(\dim W_1 = 2\) (Definition 3.66).

\(W_1 \cap W_2\). A general element of \(W_2 = L((0,1,-1,0))\) is \(\lambda(0,1,-1,0)\) for \(\lambda \in {\bf R}\). Imposing the first condition of \(W_1\):

\[ y + t = \lambda + 0 = \lambda = 0. \]

Hence \(\lambda = 0\), and \(W_1 \cap W_2 = \{0\}\) is the trivial subspace.

Solution 3.23

(See Exercise 3.23.) Part 1: Basis and dimension of \(W_k\). We write the generators of \(W_k = L((1,0,-1,0),(1,1,0,1),(1,2,k,1))\) (cf. Definition 3.42) as rows of a matrix and apply the Gaussian algorithm (Method 2.29):

\[ \begin{pmatrix}1&0&-1&0\\1&1&0&1\\1&2&k&1\end{pmatrix} \xrightarrow{R_2-R_1,\,R_3-R_1} \begin{pmatrix}1&0&-1&0\\0&1&1&1\\0&2&k+1&1\end{pmatrix} \xrightarrow{R_3-2R_2} \begin{pmatrix}1&0&-1&0\\0&1&1&1\\0&0&k-1&-1\end{pmatrix}. \]

The third row \((0,0,k-1,-1)\) is nonzero for every \(k \in {\bf R}\) (since its last entry \(-1 \ne 0\), regardless of \(k\)). Hence the matrix is in row echelon form (Definition 2.27) with three pivot rows, so the rank equals 3 for all \(k\). Equivalently, the three vectors are linearly independent (Method 3.56), so \(\dim W_k = 3\) for all \(k \in {\bf R}\). Since the three vectors also span \(W_k\), they form a basis (Definition 3.61):

\[ \mathcal{B}_k = \{(1,0,-1,0),\,(0,1,1,1),\,(0,0,k-1,-1)\} \quad (k \ne 1), \]

and for \(k = 1\) the basis is \(\{(1,0,-1,0),(0,1,1,1),(0,0,0,-1)\}\).

Part 2: Membership of \((-1,1,1,1)\) in \(W_k\). We ask whether \((-1,1,1,1) = \alpha(1,0,-1,0)+\beta(1,1,0,1)+\gamma(1,2,k,1)\) for some \(\alpha,\beta,\gamma \in {\bf R}\). This corresponds to the augmented matrix ([def:augmented-matrix])

\[ \left ( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 2 & 1 \\ -1 & 0 & k & 1 \\ 0 & 1 & 1 & 1 \end{array} \right ) \xrightarrow{R_3+R_1} \left ( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & k+1 & 0 \\ 0 & 1 & 1 & 1 \end{array} \right ) \xrightarrow{R_3-R_2,\,R_4-R_2} \left ( \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & k-1 & -1 \\ 0 & 0 & -1 & 0 \end{array} \right ). \]

Row 4 gives \(-\gamma = 0\), i.e. \(\gamma = 0\). Substituting into row 3: \((k-1)\cdot 0 = -1\), i.e. \(0 = -1\), a contradiction. Therefore \((-1,1,1,1) \notin W_k\) for any \(k \in {\bf R}\).

Solution 3.24

(See Exercise 3.24.) Part 1: \(W\) is a subspace with \(\dim W = 2\). Every element of \(W\) decomposes as

\[ \begin{pmatrix}a & a+b & b \\ 0 & 0 & b\end{pmatrix} = a\underbrace{\begin{pmatrix}1 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}}_{=:E_1} + b\underbrace{\begin{pmatrix}0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}}_{=:E_2}, \]

so \(W = L(E_1, E_2)\) (Definition 3.42). We verify the subspace conditions (Definition 3.17): the zero matrix arises with \(a=b=0\); sums and scalar multiples correspond to replacing \((a,b)\) by \((a+a', b+b')\) and \((ra, rb)\), which stays in \(W\).

To find the dimension, we check that \(E_1, E_2\) are linearly independent (Definition 3.49): \(\alpha E_1 + \beta E_2 = 0\) gives \(\alpha = 0\) (from position \((1,1)\)) and then \(\beta = 0\) (from position \((1,3)\)). Hence \(\{E_1, E_2\}\) is a basis (Definition 3.61) and \(\dim W = 2\) (Definition 3.66).

Part 2: \(V \cap W = V\), so \(\dim(V \cap W) = 1\). A matrix belongs to \(V \cap W\) if and only if it has the form \(\begin{pmatrix}a&a+b&b\\0&0&b\end{pmatrix}\) (from \(W\)) and simultaneously \(\begin{pmatrix}c&0&-c\\0&0&-c\end{pmatrix}\) (from \(V\)). Matching entries gives \(a = c\), \(b = -c\), and \(a+b = c+(-c) = 0\) – which is automatically satisfied. Hence every element of \(V\) lies in \(W\), i.e., \(V \subset W\), so \(V \cap W = V\). A basis of \(V \cap W\) is

\[ \left\{ \begin{pmatrix}1 & 0 & -1 \\ 0 & 0 & -1\end{pmatrix} \right\}, \]

and \(\dim(V \cap W) = 1\).

Solution 3.25

(See Exercise 3.25.) We have to find a vector \(v \in W_1\) that is also contained in \(W_2\). This means that

\[ v = a (1,0,1)+b(2,1,0) = (a+2b,b,a) \]

(3.82)

for some \(a, b \in {\bf R}\) and at the same time

\[ v = \alpha(-1,-1,1) + \beta(0,3,0) = (-\alpha, -\alpha + 3 \beta, \alpha) \]

for some \(\alpha, \beta \in {\bf R}\). Comparing the two vectors gives the following linear system, where \(a, b, \alpha, \beta\) are the unknowns:

\[ \begin{align*} a + 2 b & = -\alpha \\ b & = -\alpha + 3\beta \\ a & = \alpha. \end{align*} \]

We solve this system: the last equation gives \(a = \alpha\) and, from the first equation, \(b = -\alpha\). The second equation implies \(\beta = 0\). There is no condition on \(\alpha\), this \(\alpha = r\) for an arbitrary real number \(r \in {\bf R}\).

Instead of solving the above system by hand, we may also use Gaussian elimination to solve this linear system. The matrix is the following (where the columns are for \(a, b, \alpha, \beta\), in that order):

\[ \begin{align*} \left ( \begin{array}{cccc} 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & -3 \\ 1 & 0 & -1 & 0 \end{array} \right ) \leadsto & \left ( \begin{array}{cccc} 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & -2 & -2 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 0 & -6 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{cccc} 1 & 2 & 1 & 0 \\ 0 & 1 & 1 & -3 \\ 0 & 0 & 0 & 1 \end{array} \right ). \end{align*} \]

The three leading ones are for the variables \(a, b, \beta\), and \(\alpha\) is a free variable, so let \(\alpha = r\), where \(r \in {\bf R}\) is an arbitrary real number. This gives again \(\beta = 0\), \(b + r - 3\beta = 0\), so that \(b = - r\) and \(a = r\).

Thus the intersection \(W_1 \cap W_2\) consists of the vectors

\[ v = \alpha (1,0,1) + (-\alpha)(2,1,0) = \alpha(-1,-1,1) + 0 (0,3,0) = (-\alpha, -\alpha, \alpha). \]

Thus,

\[ W_1 \cap W_2 = L((1,-1,1)), \]

so a basis of \(W_1 \cap W_2\) consists of (the single vector) \((1,-1,1)\), and in particular

\[ \dim W_1 \cap W_2 = 1. \]

We now consider \(W_1 + W_2\). According to Definition 3.36,

\[ W_1 + W_2 = \{w_1 + w_2 \ | \ w_1 \in W_1, w_2 \in W_2 \}, \]

i.e., of arbitrary sums whose two summands are in \(W_1\), respectively \(W_2\).

As was noted in the proof of Corollary 3.76, if \(V_1 = L(v_1, \dots, v_n)\) and \(V_2 = L(w_1, \dots, w_m)\) are two subspaces of a vector space \(V\), then the sum

\[ V_1 + V_2 = L(v_1, \dots, v_n, w_1, \dots, w_m). \]

For the subspaces \(W_1, W_2\) above, this means that we determine the span

\[ L(\underbrace{(1,0,1)}_{v_1}, \underbrace{(2,1,0)}_{v_2}, \underbrace{(-1,-1,1)}_{w_1}, \underbrace{(0,3,0)}_{w_2}). \]

By Definition 3.611., we obtain a basis of \(W_1 + W_2\) by (possibly) removing several of these four vectors. To determine which ones these are, we apply Method 3.56 and Method 3.46. The matrix built out of the four vectors is

\[ \left ( \begin{array}{c} v_1 \\ v_2 \\ w_1 \\ w_2 \end{array} \right ) = \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ \hline -1 & -1 & 1 \\ 0 & 3 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -2 \\ \hline 0 & -1 & 2 \\ 0 & 3 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -2 \\ \hline \underline{0} & \underline{0} & \underline{0} \\ 0 & 0 & 6 \end{array} \right ). \]

Note that in this process we only added multiples of some rows to another row, but did not interchange any rows. Since we have the zero vector (underlined) in the third row, the vector \(w_1\) is a linear combination of \(v_1\) and \(v_2\). The vectors \(v_1, v_2, w_2\) are however linearly independent. Thus, they form a basis of \(W_1 + W_2\). In particular, \(\dim (W_1 + W_2) = 3\).

An alternative way to determine at least the dimension of \(W_1 + W_2\) is to use Theorem 3.77:

\[ \dim (W_1 + W_2) = \dim W_1 + \dim W_2 - \dim (W_1 \cap W_2). \]

Using again Method 3.56, one can check that \(v_1, v_2\) is a basis of \(W_1\), so that \(\dim W_1 = 2\) and similarly that \(w_1, w_2\) form a basis of \(W_2\), so that \(\dim W_2 = 2\). Thus, using the first part of the exercise, we confirm \(\dim (W_1 + W_2) = 3\).

Solution 3.26

(See Exercise 3.26.) Denote \(v_1=(1,1,1,2)\), \(v_2=(2,0,3,5)\), \(w_1=(1,1,0,1)\), \(w_2=(0,2,-2,-2)\).

Bases and dimensions of \(W_1\), \(W_2\). Row-reducing the generator matrices (Method 2.29) gives

\[ \begin{pmatrix}1&1&1&2\\2&0&3&5\end{pmatrix} \xrightarrow{R_2-2R_1} \begin{pmatrix}1&1&1&2\\0&-2&1&1\end{pmatrix}, \qquad \begin{pmatrix}1&1&0&1\\0&2&-2&-2\end{pmatrix} \]

both already in row echelon form (Definition 2.27) with two nonzero rows, so \(\dim W_1 = \dim W_2 = 2\) and \(\{v_1,v_2\}\), \(\{w_1,w_2\}\) are bases (Definition 3.61).

\(W_1 + W_2\). Row-reducing the matrix of all four generators (Definition 3.36):

\[ \begin{pmatrix}1&1&1&2\\2&0&3&5\\1&1&0&1\\0&2&-2&-2\end{pmatrix} \xrightarrow{R_2-2R_1,\,R_3-R_1} \begin{pmatrix}1&1&1&2\\0&-2&1&1\\0&0&-1&-1\\0&2&-2&-2\end{pmatrix} \xrightarrow{R_4+R_2,\,R_4-R_3} \begin{pmatrix}1&1&1&2\\0&-2&1&1\\0&0&-1&-1\\0&0&0&0\end{pmatrix}. \]

The rank is 3, so \(\dim(W_1+W_2) = 3\) with basis \(\{(1,1,1,2),(0,-2,1,1),(0,0,-1,-1)\}\).

\(W_1 \cap W_2\). By the dimension formula (Theorem 3.77):

\[ \dim(W_1 \cap W_2) = \dim W_1 + \dim W_2 - \dim(W_1+W_2) = 2+2-3 = 1. \]

To find a generator, we solve \(\alpha v_1 + \beta v_2 = \gamma w_1 + \delta w_2\), i.e., \(\alpha v_1 + \beta v_2 - \gamma w_1 - \delta w_2 = 0\):

\[ \begin{pmatrix}1&2&-1&0\\1&0&-1&-2\\1&3&0&2\\2&5&-1&2\end{pmatrix} \xrightarrow{R_2-R_1,\,R_3-R_1,\,R_4-2R_1} \begin{pmatrix}1&2&-1&0\\0&-2&0&-2\\0&1&1&2\\0&1&1&2\end{pmatrix} \xrightarrow{R_4-R_3,\,2R_3+R_2} \begin{pmatrix}1&2&-1&0\\0&-2&0&-2\\0&0&2&2\\0&0&0&0\end{pmatrix}. \]

Setting the free variable \(\delta=1\): row 3 gives \(\gamma=-1\), row 2 gives \(\beta=-1\), row 1 gives \(\alpha=1\). The corresponding vector in \(W_1\) is

\[ \alpha v_1 + \beta v_2 = (1,1,1,2) - (2,0,3,5) = (-1,1,-2,-3). \]

Hence \(W_1 \cap W_2 = L((-1,1,-2,-3))\) with basis \(\{(-1,1,-2,-3)\}\) and \(\dim(W_1\cap W_2)=1\).

Solution 3.27

(See Exercise 3.27.) We will show that \(v_1, v_2, v_3\) are linearly independent (in \({\bf R}^4\) and therefore also in the subspace \(W\)) and therefore form a basis of \(W\). We use Method 3.56:

\[ \begin{align*} \left ( \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right ) = & \left ( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \\ 0 & 0 & 1 & 3 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 3 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 4 \end{array} \right )\\ \leadsto & \left ( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{array} \right ) \end{align*} \]

This matrix has three leading ones, so the vectors are linearly independent as claimed.

We “guess” \(v = (1, 2, 3, 4)\) and check that these vectors \(v_1, v_2, v_3, v\) are linearly independent. By Lemma 3.54, this will then imply that \(v\) is not a linear combination of the other vectors, so that \(W \subsetneq L(v_1, v_2, v_3, v)\). We use Method 3.56:

\[ \begin{align*} \left ( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \\ 0 & 0 & 1 & 3 \\ 1 & 2 & 3 & 4 \end{array} \right ) & \leadsto \left ( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 2 & 2 & 4 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 3 \\ 0 & 1 & 1 & 2 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{cccc} \underline1 & 0 & 1 & 0 \\ 0 & 0 & 0 & \underline4 \\ 0 & 0 & \underline1 & 3 \\ 0 & \underline1 & 1 & 2 \end{array} \right ) \end{align*} \]

After dividing the second row by 4, we can interchange rows and get a row echelon matrix with four leading ones (underlined). Thus, \(v_1, v_2, v_3, v\) are linearly independent. Therefore, they form in fact a basis of \({\bf R}^4\), and we know by Definition 3.613. that therefore

\[ W \subsetneq {\bf R}^4 = L(v_1, v_2, v_3, v). \]

Remark 3.83

A more systematic way of solving the second part of the exercise, without guessing, is to use Definition 3.61: we can take the standard basis of \({\bf R}^4\), and for (at least) one of the four standard basis vectors \(e_1, e_2, e_3, e_4\) we will have that this standard basis vector together with \(v_1, v_2, v_3\) form a basis of \({\bf R}^4\). We can then use Method 3.56 to see that, for example, \(v_1, v_2, v_3, e_1\) are linearly independent and therefore form a basis of \({\bf R}^4\), so that in particular \(W \subsetneq L(v_1, v_2, v_3, e_1)\).

Solution 3.28

(See Exercise 3.28.) We bring the matrix formed by these vectors in row-echelon form:

\[ \begin{align*} \left ( \begin{array}{cccc} 1 & 0 & -1 & 2 \\ 1 & 0 & 0 & 1 \\ 2 & 0 & -1 & 3 \\ 4 & t & -2 & 6 \end{array} \right ) & \leadsto \left ( \begin{array}{cccc} 1 & 0 & -1 & 2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & t & 2 & -2 \end{array} \right ) \\ &\leadsto \left ( \begin{array}{cccc} 1 & 0 & -1 & 2 \\ 0 & t & 2 & -2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & -1 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{cccc} 1 & 0 & -1 & 2 \\ 0 & t & 2 & -2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right ). \end{align*} \]

If \(t \ne 0\), then division by \(t\) is an elementary operation (Definition 2.28), which then yields a matrix with three leading ones. Thus, the space \(U_t\) which is spanned by these vectors has dimension 3 in this case. If \(t = 0\), we continue simplifying the matrix into row echelon form:

\[ \begin{align*} \left ( \begin{array}{cccc} 1 & 0 & -1 & 2 \\ 0 & t & 2 & -2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right ) & = \left ( \begin{array}{cccc} 1 & 0 & -1 & 2 \\ 0 & 0 & 2 & -2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right ) \\ & \leadsto \left ( \begin{array}{cccc} 1 & 0 & -1 & 2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right ). \end{align*} \]

This has two leading ones, thus \(\dim U_t = 2\) in this case.

We now consider \(t = 1\). The subspace \(U := U_1\) then has a basis consisting of the non-zero rows if the matrix above, i.e., it has a basis consisting of the vectors

\[ (1,0,-1,2), (0,1,2,-2), (0,0,1,-1). \]

In order to determine a basis of \(W\), we form the matrix associated to these homogeneous equations, and apply Gaussian elimination (Method 2.31):

\[ \left ( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & -3 \end{array} \right ) \leadsto \left ( \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & -1 & -1 & -3 \end{array} \right ). \]

This has two columns not having a leading one, namely the last two. These are the free variables, say \(x_3 = a\), \(x_4 = b\) for \(a , b \in {\bf R}\). To determine a basis of \(W\), we therefore have to consider the system

\[ \begin{align*} x_1 + x_2 + a & = 0\\ x_2 + a + 3b & = 0 \end{align*} \]

This gives \(x_2 = -a-3b\), and \(x_1 - 3b=0\) so that \(x_1 = 3b\). Thus, a basis of \(W\) is given by the two vectors

\[ (0,-1,1,0) \text{ and }(3,-3,0,1). \]

In order to determine the intersection \(U \cap W\) (cf. Lemma 3.19), consider a generic vector of \(U\), i.e., one of the form

\[ \begin{align*} v & = a(1,0,-1,2) + b(0,1,2,-2) + c(0,0,1,-1) \\ &= (a,b,-a+2b+c,2a-2b-c). \end{align*} \]

We require it to satisfy the equations describing \(W\):

\[ \begin{align*} a+b+(-a+2b+c) & = 0\\ a-3(2a-2b-c) & = 0. \end{align*} \]

Simplifying these expressions gives the system

\[ \begin{align*} 3b+c & = 0\\ -5a+6b+3c & = 0. \end{align*} \]

Therefore \(c = -3b\), plugging this into the second equation gives, after simplifying, \(-5a-3b = 0\) or \(a = -\frac35b\). Thus, our vector \(v \in U\) belongs to \(W\) precisely if it can be written as

\[ \begin{align*} & -\frac 35 b(1,0,-1,2)+b(0,1,2,-2)+(-3b)(0,0,1,-1) \\ = & (-\frac 3bb, b, \frac35b+2b-3b,-\frac 65b-2b+3b) \\ = & b(-\frac 35, 1, -\frac25, -\frac15), \end{align*} \]

where \(b \in {\bf R}\) is arbitrary. Thus, a basis of \(U \cap W\) is this vector

\[ (-\frac 35, 1, -\frac 25, -\frac 15). \]

In particular, \(\dim U \cap W = 1\).

Solution 3.29

(See Exercise 3.29.) Part 1: Values of \(t\) with \(\dim U_t = 2\). We row-reduce the matrix with \(v_1,v_2,v_3,v_4\) as rows (Method 2.29):

\[ \begin{pmatrix}1&0&0&1\\-1&1&2&3\\0&1&2&4\\t&2&4&8\end{pmatrix} \xrightarrow{R_2+R_1,\,R_4-tR_1} \begin{pmatrix}1&0&0&1\\0&1&2&4\\0&1&2&4\\0&2&4&8-t\end{pmatrix} \xrightarrow{R_3-R_2,\,R_4-2R_2} \begin{pmatrix}1&0&0&1\\0&1&2&4\\0&0&0&0\\0&0&0&-t\end{pmatrix}. \]

The last row \((0,0,0,-t)\) is zero if and only if \(t=0\). Hence \(\dim U_t = 2\) if and only if \(t = 0\). (For \(t \ne 0\) the rank is 3 and \(\dim U_t = 3\).)

Part 2: \(\dim U_1\). With \(t = 1 \ne 0\), we have \(\dim U_1 = 1\), as was noted above.

Part 3: Basis of \(W\) and of \(W \cap U_1\). \(W\) is the solution set of the homogeneous system (Proposition 3.18) \(x_1 - x_2 = 0\), \(x_2 - x_3 = 0\). These equations say \(x_1 = x_2 = x_3\), so the free variables are \(x_1\) and \(x_4\):

\[ (x_1, x_2, x_3, x_4) = x_1(1,1,1,0) + x_4(0,0,0,1). \]

Thus \(\{(1,1,1,0),(0,0,0,1)\}\) is a basis (Definition 3.61) of \(W\) and \(\dim W = 2\) (Definition 3.66).

For \(W \cap U_1\), we reduce the \(t=1\) row echelon form to RREF:

\[ \begin{pmatrix}1&0&0&1\\0&1&2&4\\0&0&0&1\\0&0&0&0\end{pmatrix} \xrightarrow{R_1-R_3,\,R_2-4R_3} \begin{pmatrix}1&0&0&0\\0&1&2&0\\0&0&0&1\\0&0&0&0\end{pmatrix}, \]

so \(U_1 = \{(\alpha,\beta,2\beta,\gamma) \mid \alpha,\beta,\gamma \in {\bf R}\}\). Imposing the conditions of \(W\) (i.e., \(x_1=x_2=x_3\)) gives \(\beta = \alpha\) and \(2\beta = \alpha\), hence \(2\alpha = \alpha\), so \(\alpha = 0\). The intersection is therefore \(\{(0,0,0,\gamma) \mid \gamma \in {\bf R}\}\), with basis \(\{(0,0,0,1)\}\) and \(\dim(W \cap U_1) = 1\).