Span & Generating Sets - Linear Algebra

1. The set $S = \{\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}$ generates a plane $W$ in $\mathbb{R}^3$. Which of the following vectors $\vec{v}$, if added to $S$, would create a new set that generates all of $\mathbb{R}^3$?
   Answer: $\vec{v} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$ To generate all of $\mathbb{R}^3$, the new vector $\vec{v}$ must not lie in the plane $W$ spanned by $S$. A vector $(x,y,z)$ in $W$ can be written as $a(1,0,1) + b(0,1,1) = (a, b, a+b)$. So, any vector in $W$ must satisfy the condition $z = x+y$. We test each option: A) For $(1,1,2)$, $2 = 1+1$. It is in $W$. B) For $(2,-1,1)$, $1 = 2+(-1)$. It is in $W$. C) For $(-1,-1,-2)$, $-2 = -1+(-1)$. It is in $W$. D) For $(3,2,4)$, $4 \neq 3+2$. This vector is not in $W$. Adding it to the set will span a 3-dimensional space, which is all of $\mathbb{R}^3$.
2. Consider the set of vectors $S = \{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\}$ in $\mathbb{R}^3$. Which of the following is a minimal generating set for $\mathrm{span}(S)$?
   Answer: $\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\}$ First, we check for linear dependencies in $S$. Let the vectors be $\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4$. We can see that $\vec{v}_3 = \vec{v}_1 - \vec{v}_2$ and $\vec{v}_4 = \vec{v}_1 + \vec{v}_2$. This means $\vec{v}_3$ and $\vec{v}_4$ are redundant and can be removed without changing the span. The set $\{\vec{v}_1, \vec{v}_2\}$ generates $\mathrm{span}(S)$. Since $\vec{v}_1$ and $\vec{v}_2$ are not scalar multiples of each other, they are linearly independent. A minimal generating set is a basis, which must be linearly independent. Therefore, $\{\vec{v}_1, \vec{v}_2\}$ is a minimal generating set. Choice A is too small (spans a line, not a plane). Choice C is not minimal because it's linearly dependent. Choice D spans the xy-plane, which is a different subspace from the one spanned by $S$.
3. Consider the set $T = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ where $\mathbf{v}_1 = (1,2,0)$, $\mathbf{v}_2 = (0,1,1)$, and $\mathbf{v}_3 = (2,3,-1)$. For which vector $\mathbf{b}$ is the equation $\mathbf{b} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3$ inconsistent?
   Answer: $(0,1,2)$ We need to find which vector is not in span(T). First, let's check if the vectors in T are linearly independent by row reducing the matrix $\begin{pmatrix} 1&2&0 \\ 0&1&1 \\ 2&3&-1 \end{pmatrix}$. Applying $R_3 \leftarrow R_3 - 2R_1$: $\begin{pmatrix} 1&2&0 \\ 0&1&1 \\ 0&-1&-1 \end{pmatrix}$. Then $R_3 \leftarrow R_3 + R_2$: $\begin{pmatrix} 1&2&0 \\ 0&1&1 \\ 0&0&0 \end{pmatrix}$. Since the rank is 2, the vectors span a 2-dimensional subspace. From the reduced form, we see that $\mathbf{v}_3 = 2\mathbf{v}_1 - \mathbf{v}_2$. So span(T) = span{$\mathbf{v}_1, \mathbf{v}_2$}. For $(0,1,2)$ to be in the span, we need $(0,1,2) = c_1(1,2,0) + c_2(0,1,1)$, giving $c_1 = 0$, $2c_1 + c_2 = 1 \Rightarrow c_2 = 1$, $c_2 = 2$. This is inconsistent since $c_2$ cannot be both 1 and 2.
4. Let $A = \{(1,1,0), (1,0,1), (0,1,1)\}$ and $B = \{(2,1,1), (1,2,1), (1,1,2)\}$. Which statement about $\text{span}(A)$ and $\text{span}(B)$ is correct?
   Answer: $\text{span}(A) = \text{span}(B)$ and both equal $\mathbb{R}^3$ First, let's check if the vectors in A are linearly independent. The matrix $\begin{pmatrix} 1&1&0 \\ 1&0&1 \\ 0&1&1 \end{pmatrix}$ has determinant $1(0-1) - 1(1-0) + 0 = -1-1 = -2 ≠ 0$, so A spans $\mathbb{R}^3$. For B, the matrix $\begin{pmatrix} 2&1&1 \\ 1&2&1 \\ 1&1&2 \end{pmatrix}$ has determinant $2(4-1) - 1(2-1) + 1(1-2) = 6-1-1 = 4 ≠ 0$, so B also spans $\mathbb{R}^3$. Since both A and B span $\mathbb{R}^3$, we have $\text{span}(A) = \text{span}(B) = \mathbb{R}^3$.
5. Let $\{\vec{v}_1, \dots, \vec{v}_n\}$ be a generating set for a non-trivial vector space $V$. Suppose $\vec{u}$ is a vector in $V$. What must be true about the set $S' = \{\vec{v}_1, \dots, \vec{v}_n, \vec{u}\}$?
   Answer: $S'$ is a linearly dependent generating set for $V$. Since $\{\vec{v}_1, \dots, \vec{v}_n\}$ is a generating set for $V$, its span is $V$. Adding any vector $\vec{u}$ from $V$ to this set will not change the span, so $\mathrm{span}(S') = V$. Thus, $S'$ is also a generating set for $V$. Furthermore, because $\vec{u}$ is in $V = \mathrm{span}(\{\vec{v}_1, \dots, \vec{v}_n\})$, $\vec{u}$ can be written as a linear combination of $\{\vec{v}_1, \dots, \vec{v}_n\}$. By definition, this makes the set $S'$ linearly dependent. Therefore, $S'$ is a linearly dependent generating set for $V$.
6. Let $S_1 = \{\begin{pmatrix} 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 4 \end{pmatrix}\}$ and $S_2 = \{\begin{pmatrix} 3 \\ 6 \end{pmatrix}\}$ be sets of vectors in $\mathbb{R}^2$. Which statement correctly describes the relationship between their spans?
   Answer: $\mathrm{span}(S_1) = \mathrm{span}(S_2)$ In set $S_1$, the vector $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$ is a scalar multiple of $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ (specifically, 2 times the first vector). Therefore, the second vector is redundant, and $\mathrm{span}(S_1) = \mathrm{span}(\{\begin{pmatrix} 1 \\ 2 \end{pmatrix}\})$. The vector in $S_2$, $\begin{pmatrix} 3 \\ 6 \end{pmatrix}$, is also a scalar multiple of $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ (3 times). Thus, $\mathrm{span}(S_2)$ is the same set of vectors. Both sets span the same line through the origin in $\mathbb{R}^2$.
7. In the vector space $M_{2\times2}$ of $2 \times 2$ matrices, consider the subspace $W$ spanned by $S = \{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\}$. Which of the following matrices is not in $W$?
   Answer: $\begin{pmatrix} 2 & 2 \\ 2 & 0 \end{pmatrix}$ Any matrix in the subspace $W$ must be a linear combination of the two generating matrices. A general matrix in $W$ can be written as: $a\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} a & b \\ b & a \end{pmatrix}$. This means any matrix in $W$ must be symmetric (which is true for all options except D, which is trivially symmetric) and its diagonal elements must be equal. We check each option: A) Diagonals are both 3. In W. B) Diagonals are both 1. In W. C) Diagonals are 2 and 0. They are not equal. This matrix is not in W. D) The zero matrix is in every subspace. Here, $a=0$ and $b=0$.
8. For which value of the scalar $k$ does the set of vectors $S = \{\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ -2 \\ k \end{pmatrix}\}$ not form a generating set for $\mathbb{R}^3$?
   Answer: $k = 2$ A set of three vectors in $\mathbb{R}^3$ fails to be a generating set (i.e., does not span $\mathbb{R}^3$) if and only if the vectors are linearly dependent. This occurs when the matrix formed by these vectors as columns has a determinant of zero. The determinant of the matrix $A = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 2 & 1 & k \end{pmatrix}$ is calculated as $1(k - (-2)) - 0(...) + 2(0 - 2) = k + 2 - 4 = k - 2$. Setting the determinant to zero, we get $k - 2 = 0$, which gives $k = 2$. When $k=2$, the third vector is a linear combination of the first two ($2\vec{v}_1 - 2\vec{v}_2$), so the set is linearly dependent and spans only a plane.
9. Let $S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$ be a set of vectors in $\mathbb{R}^4$. Let $V = \mathrm{span}(S)$. Which of the following statements about $V$ must be true?
   Answer: $V$ cannot be all of $\mathbb{R}^4$. To span the entire space $\mathbb{R}^4$, a set must contain at least 4 vectors. Since the set $S$ only contains 3 vectors, its span, $V$, can have a dimension of at most 3. Therefore, $V$ cannot be the entire 4-dimensional space $\mathbb{R}^4$. The other options are not guaranteed: A) The dimension could be less than 3 if the vectors are linearly dependent. B) $S$ is a generating set for $V$ by definition, but it is only a basis if it is also linearly independent. D) If the three vectors are collinear, $V$ would be a line and could be spanned by a single non-zero vector.
10. Consider the vector space $\mathbb{P}_2$ of polynomials of degree at most 2. Let $W = \text{span}\{1 + x^2, x - x^2, 2 + x + x^2\}$. Which polynomial is NOT in $W$?
    Answer: $4 + 3x + 2x^2$ To determine if a polynomial is in W, we need to check if it can be written as $a(1+x^2) + b(x-x^2) + c(2+x+x^2) = (a+2c) + (b+c)x + (a-b+c)x^2$. For option C: $4+3x+2x^2$, we need $a+2c=4$, $b+c=3$, $a-b+c=2$. From the first two equations: $a=4-2c$ and $b=3-c$. Substituting into the third: $(4-2c)-(3-c)+c = 4-2c-3+c+c = 1 ≠ 2$. This is inconsistent, so $4+3x+2x^2 ∉ W$. We can verify the others work: A gives $a=1, b=-2, c=1$; B gives $a=-1, b=2, c=1$; D gives $a=0, b=1, c=2$.
11. Let $S = \{\vec{v}_1, \dots, \vec{v}_k\}$ be a set of vectors in a vector space $V$. If $\mathrm{span}(S) = V$ and the dimension of $V$ is $n$, which of the following statements is a necessary condition?
    Answer: $k \ge n$ A fundamental theorem of linear algebra states that any generating set for a vector space $V$ must contain at least as many vectors as the dimension of $V$. Therefore, if $\mathrm{dim}(V)=n$ and $\mathrm{span}(S)=V$, it must be true that $k \ge n$. Choice A is only true if $S$ is a basis. Choice C is not necessary; a generating set can be linearly dependent. Choice D is only true if $S$ is a basis (linearly independent), as linear dependence implies non-unique representations.
12. Let $\vec{u}$ and $\vec{v}$ be two non-zero, non-parallel vectors in $\mathbb{R}^3$. Let $\vec{w} = 2\vec{u} - 3\vec{v}$. Which geometric object best describes the set of all linear combinations of the vectors in $S = \{\vec{u}, \vec{v}, \vec{w}\}$?
    Answer: A plane through the origin The set of all linear combinations is the span of $S$, denoted $\mathrm{span}(S)$. Since $\vec{w}$ is a linear combination of $\vec{u}$ and $\vec{v}$, it is a redundant vector. Adding a redundant vector to a set does not change its span. Therefore, $\mathrm{span}(\{\vec{u}, \vec{v}, \vec{w}\}) = \mathrm{span}(\{\vec{u}, \vec{v}\})$. The span of two non-zero, non-parallel vectors is a plane through the origin.
13. Let $S = \{\vec{u}, \vec{v}\}$ and $T = \{\vec{u}, \vec{v}, \vec{u}+\vec{v}\}$ be two sets of vectors in a vector space $V$. Let $W_S = \mathrm{span}(S)$ and $W_T = \mathrm{span}(T)$. Which statement accurately describes the relationship between $W_S$ and $W_T$?
    Answer: $W_S = W_T$ The span of a set of vectors is the set of all possible linear combinations of those vectors. The vector $\vec{u}+\vec{v}$ is, by its form, a linear combination of $\vec{u}$ and $\vec{v}$. This means $\vec{u}+\vec{v}$ is already in $\mathrm{span}(\{\vec{u}, \vec{v}\})$. Adding a vector that is already in the span to a generating set does not change the span. Therefore, $\mathrm{span}(T) = \mathrm{span}(S)$, so $W_S = W_T$. This holds true whether or not $\vec{u}$ and $\vec{v}$ are linearly independent.
14. Let $S = \{1+x^2, x-x^2\}$ be a set of polynomials in $\mathcal{P}_2$, the space of polynomials of degree at most 2. Which of the following polynomials $p(x)$, when added to $S$, results in a set that does not generate $\mathcal{P}_2$?
    Answer: $p(x) = 1+2x-x^2$ The set will fail to generate $\mathcal{P}_2$ if the new polynomial $p(x)$ is in the span of the original set $S$. Using the standard basis $\{1, x, x^2\}$, the vectors for $S$ are $(1, 0, 1)$ and $(0, 1, -1)$. Any vector in their span has the form $a(1, 0, 1) + b(0, 1, -1) = (a, b, a-b)$. We test each option: A) For $p(x)=1$, the vector is $(1,0,0)$. This requires $a=1, b=0$, but then $a-b=1 \neq 0$. Not in the span. B) For $p(x)=x^2$, the vector is $(0,0,1)$. This requires $a=0, b=0$, but then $a-b=0 \neq 1$. Not in the span. C) For $p(x)=1+2x-x^2$, the vector is $(1,2,-1)$. This requires $a=1, b=2$. Then $a-b = 1-2 = -1$. This matches. Since $p(x)$ is a linear combination of the other two ($p(x) = 1(1+x^2) + 2(x-x^2)$), the set is linearly dependent and cannot span the 3-dimensional space $\mathcal{P}_2$. D) For $p(x)=2+x+3x^2$, the vector is $(2,1,3)$. This requires $a=2, b=1$, but then $a-b=1 \neq 3$. Not in the span.
15. Let $S = \{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$ be a set of vectors in $\mathbb{R}^3$. You are given that the system $A\vec{x} = \vec{b}$, where $A = [\vec{v}_1 \ \vec{v}_2 \ \vec{v}_3]$, has a unique solution for at least one vector $\vec{b} \in \mathbb{R}^3$. What can be concluded about $\mathrm{span}(S)$?
    Answer: $\mathrm{span}(S)$ is all of $\mathbb{R}^3$. The existence of a unique solution to $A\vec{x} = \vec{b}$ implies that the null space of matrix $A$ contains only the zero vector. For a square matrix $A$, this is equivalent to saying that $A$ is invertible. An invertible matrix has linearly independent columns. Therefore, the vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ are linearly independent. A set of three linearly independent vectors in $\mathbb{R}^3$ must form a basis for $\mathbb{R}^3$. By definition, a basis for a space spans that entire space. Thus, $\mathrm{span}(S) = \mathbb{R}^3$.

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