1. Let \(p_{n}(x)=x^{n}\) for \(x \in \mathbb{R}\) and let \(V=\operatorname{span}\left\{p_{0}, p_{1}, p_{2}, \ldots\right\}\). Then

Explanation for Question 1:
The set of polynomials \(p_{0}(x)=1\), \(p_{1}(x)=x\), \(p_{2}(x)=x^2\) and so on, forms a basis for the vector space of all polynomials. This is because any polynomial \(q(x)\) can be expressed as a finite linear combination \(q(x)=a_{0}+a_1x+a_2x^2+\cdots+a_nx^n\) where \(a_i\) are real numbers.
However, the exponential function \(e^x\) cannot be expressed as a finite linear combination of \(p_i\)’s, so it is not in the span of \(\{p_0, p_1, \ldots\}\). But \(e^x\) is a continuous function, so it belongs to the space of all continuous functions. Therefore, the vector space spanned by \(\{p_0, p_1, \ldots\}\) is a proper subset of the space of all continuous functions.

Solution

2. Which of the following are subspaces of the vector space \(\mathbb{R}^{3}\)?

Explanation for Question 2:
The set \(A = \{(x, y, z) \mid x+y=0\}\) can be rewritten as \(A = \{(x, -x, z) \mid x, z \in \mathbb{R}\}\). This set satisfies the properties of a vector space, as shown in the question.
Similarly, the set \(B = \{(x, y, z) \mid x-y=0\}\) also satisfies the properties of a vector space.
However, the set \(C = \{(x, y, z) \mid x+y=1\}\) does not contain the origin \((0, 0, 0)\), which is a requirement for a set to be a vector space. Therefore, \(C\) cannot be a vector space.
Similarly, the set \(D = \{(x, y, z) \mid x-y=1\}\) does not contain the origin and is also not a vector space.

Solution

3. For arbitrary not null subspaces \(U\), \(V\), and \(W\) of a finite dimensional vector space, which of the following hold:

Explanation for Question 3:
Answers (b) and (c) are correct.

Let:
\(U = \{(x, x) \mid x \in \mathbb{R}\}\)
\(V = \{(x, 0) \mid x \in \mathbb{R}\}\)
\(W = \{(0, y) \mid y \in \mathbb{R}\}\)
\(U \cap (V + W) = U\) and \((U \cap V) + (U \cap W) = \{0\}\).
Therefore, \(U \cap (V + W)\) not contained in \( (U \cap V) + (U \cap W)\).

Now, let:
\(U = \{(x, y, 0) \mid y \in \mathbb{R}\}\)
\(V = \{(0, y, z) \mid y, z \in \mathbb{R}\}\)
\(W = \{(x, 0, x) \mid x \in \mathbb{R}\}\)
\((U \cap V) + W = \{(0, y, 0) \mid y \in \mathbb{R}\} + W = \{(x, y, x) \mid x, y \in \mathbb{R}\}\)
And \((U + W) \cap (V + W) = \{(x, y, z) \mid x, y, z \in \mathbb{R}\} \cap \{(x, y, z) \mid x, y, z \in \mathbb{R}\} = \mathbb{R}^{3}\).
Therefore, \((U + W) \cap (V + W) \subset (U \cap V) + W\) is not true.

Let \(a \in (U \cap V) + (U \cap W)\):
\(\Rightarrow a = \alpha x + \beta y\) such that \(x \in U \cap V, y \in U \cap W\).
\(\Rightarrow x \in U, x \in V, y \in U, y \in W\).
\(\Rightarrow \alpha x + \beta y \in U\) and \( \alpha x + \beta y \in V + W\).
\(\Rightarrow a \in U \cap (V + W)\).
Therefore, \((U + V) + (U + W) \subset U \cap (V + W)\).

Let \(b \in (U \cap V) + W\):
\(\Rightarrow b = \alpha x + \beta y\) such that \(x \in U \cap V, y \in W\).
\(\Rightarrow x \in U, x \in V, y \in W\).
\(\Rightarrow \alpha x + \beta y \in U + W, \alpha x + \beta y \in V + W\).
\(\Rightarrow b \in (U + W) \cap (V + W)\).
Therefore, \((U \cap V) + W \subset (U + W) \cap (V + W)\).

Solution

4. Let \(V\) denote a vector space over a field \(F\) and with a basis \(B=\left\{e_{1}, e_{2}, \ldots, e_{n}\right\}\). Let \(x_{1}, x_{2}, \ldots, x_{n} \in F\)

Let

\(C=\left\{x_{1} e_{1}, x_{1} e_{1}+x_{2} e_{2}, \ldots, x_{1} e_{1}+x_{2} e_{2}+\ldots+x_{n} e_{n}\right\}\)

Then

Explanation for Question 4:
Answers (a), (b), (c), and (d) are correct.

First, define \(D = \left[\begin{array}{l}x_{1} e_{1} \\ x_{1} e_{1}+x_{2} e_{2} \\ \vdots \\ x_{1} e_{1}+x_{2} e_{2} \cdots+x_{n} e_{n}\end{array}\right] = \left[\begin{array}{ccccc}x_{1} & 0 & 0 & \cdots & 0 \\ x_{1} & x_{2} & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \\ x_{1} & x_{2} & x_{3} & \cdots & x_{n}\end{array}\right]\left[\begin{array}{c}e_{2} \\ e_{2} \\ \vdots \\ e_{n}\end{array}\right]\) & let \(C^{\prime} = \left[\begin{array}{ccccc}x_{1} & 0 & 0 & \cdots & 0 \\ x_{1} & x_{2} & 0 & \cdots & 0 \\ \vdots & & & & \\ x_{1} & x_{2} & x_{3} & \cdots & x_{n}\end{array}\right]\) & \(\left|C^{\prime}\right| = \prod_{i=1}^{n} x_{i}\)

Observe the following are equivalent:
a) \(C\) is a linearly independent set
b) \(\left|C^{\prime}\right| \neq 0\)
c) \(x_{i} \neq 0\) for all \(i=1,2, \ldots, n\)
d) \(C\) spans \(V\)

Solution

5. Consider the following row vectors :

\(a_{1}=(1,1,0,1,0,0), a_{2}=(1,1,0,0,1,0),\) \(a_{3}=(1,1,0,0,0,1), a_{4}=(1,0,1,1,0,0),\) \(a_{5}=(1,0,1,0,1,0), a_{6}=(1,0,1,0,0,1),\)

The dimension of the vector space spanned by these row vectors is

Explanation for Question 5:
The rank of the matrix \(D\) is 4, which means there are 4 linearly independent rows (vectors) in the matrix. This implies that the set \(C = \{a_1, a_2, \ldots, a_6\}\) has 4 linearly independent vectors.
Therefore, the dimension of the span of \(C\) is 4, which means the span of \(C\) is a 4-dimensional vector space.

Solution

6. Let \(\left\{v_{1}, \cdots, v_{n}\right\}\) be a linearly independent subset of a vector space \(V\) where \(n \geq 4\). Set \(w_{i j}=v_{i}-v_{j}\). Let \(W\) be the span of \(\left\{w_{i j} \mid 1 \leq i, j \leq n\right\}\). Then

Explanation for Answer 6:
Answer (a) and (c) are correct.
We define the set \(W\) as the span of vectors of the form \(v_i – v_j\) for \(1 \leq i, j \leq n\). This set is equal to the span of the given set \(S\).

Let \(W = \text{span}\{v_i – v_j : 1 \leq i, j \leq n\}\)
Clearly, \(W = W_1\) (since \(S\) is skew-symmetric)

\[ \begin{bmatrix} 0 & v_1 – v_2 & v_1 – v_3 & \cdots & v_1 – v_n \\ & 0 & v_2 – v_3 & \cdots & v_2 – v_n \\ & & 0 & \ddots & \vdots \\ & & & \ddots & v_{n-1} – v_n \\ & & & & 0 \end{bmatrix}_{n \times n} \]
\[ W_{12} + W_{23} = W_{13} \]
\(W_{23} + W_{34} = W_{24}\) and so on.
Therefore, all the upper diagonal elements will generate \(W\). These upper diagonal elements are linearly independent.
Hence, the set \(\{w_i : 1 \leq i < j \leq n\}\) is a linearly dependent subset of \(W\), and the dimension of \(W\) is \(\text{dim}(W) = n – 1\).

Solution

7. Let \(V\) be a 3-dimensional vector space over the field \(\mathbb{F}_{3}=\frac{\mathbb{Z}}{3 \mathbb{Z}}\) of 3 elements. The number of distinct 1-dimensional subspaces of \(V\) is

Explanation for Answer 7:
(a) is correct.
To find the number of distinct 1-dimensional subspaces of \(V\), we use the formula provided:
\(\frac{(p^n – 1)(p^n – p)\ldots(p^n – p^{k-1})}{(p^k – 1)(p^k – p)\ldots(p^k – p^{k-1})} \).
For the given values, \(p = 3\) and \(k = 1\), so the number of distinct 1-dimensional subspaces is \( \frac{3^{3}-1}{3-1} = 26\).

Solution

8. Let \(n\) be an integer, \(n \geq 3\), and let \(u_{1}, u_{2}, \ldots, u_{n}\) be \(n\) linearly independent elements in a vector space over \(\mathbb{R}\). Set \(u_{0}=0\) and \(u_{n+1}=u_{1}\). Define \(v_{i}=u_{i}+u_{i+1}\) and \(w_{i}=u_{i-1}+u_{i}\) for \(i=1,2, \ldots, n\). Then

Explanation for Answer 8:
Answers (b), (c), and (d) are correct.
We have \(v_1 = u_1 + u_2\), and similarly, \(v_2 = u_2 + u_3\), and so on up to \(v_n = u_n + u_1\).

\[ \begin{aligned} & v_1 = u_1 + u_2 \\ & v_2 = u_2 + u_3 \\ & \vdots \\ & v_n = u_n + u_1 \end{aligned} \]
Consider the linear combination \(\alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n = 0\).
This can be written as \((\alpha_1 + \alpha_n) u_1 + (\alpha_1 + \alpha_2) u_2 + \ldots + (\alpha_{n-1} + \alpha_n) u_1 = 0\).

Because \(u_1\)’s are linearly independent,
\[ \begin{aligned} & \alpha_1 + \alpha_n = 0 \\ & \alpha_1 + \alpha_2 = 0 \\ & \cdots \\ & \alpha_{n-1} + \alpha_n = 0 \end{aligned} \]
These equations lead to \(\alpha_1 = -\alpha_n\), \(\alpha_1 = -\alpha_2 = \alpha_3 = -\alpha_4 = \ldots\).
\(\alpha_i\) takes different values based on whether \(n\) is even or odd:
\[ \alpha_i = \begin{cases} \alpha_n; & \text{if } n \text{ is odd} \\ -\alpha_n; & \text{if } n \text{ is even} \end{cases} \]
Therefore, if \(n\) is odd,
\[ \begin{aligned} & \alpha_1 = -\alpha_n, \alpha_1 = \alpha_n \\ & \Rightarrow \alpha_1 = 0 \\ & \Rightarrow \alpha_i = 0 \text{ for all } i \end{aligned} \]
But if \(n\) is even,
\[ \alpha_1 = -\alpha_n \]
\(\Rightarrow\) all \(\alpha_i\)’s need not be zero.
Hence, \(v_i\)’s are linearly independent when \(n\) is odd, linearly dependent when \(n\) is even.
A similar argument can be made for \(w_i\)’s:
Consider the linear combination \(\alpha_1 w_1 + \alpha_2 w_2 + \ldots + \alpha_n w_n = 0\).
This can be written as \((\alpha_1 + \alpha_2) u_1 + (\alpha_2 + \alpha_3) u_2 + \ldots + \alpha_n u_n = 0\).
Because \(u_1\)’s are linearly independent,
\[ \begin{aligned} & \alpha_1 + \alpha_2 = 0 \\ & \alpha_2 + \alpha_3 = 0 \\ & \cdots \\ & \alpha_{n-1} + \alpha_n = 0 \\ & \alpha_n = 0 \\ & \alpha_i = 0 \text{ for all } i \end{aligned} \]
Therefore, \(w_i\)’s are linearly independent for all \(n\).

Solution

9. The dimension of the vector space of all symmetric matrices \(A=\left(a_{j k}\right)\) of order \(n \times n(n \geq 2)\) with real entries, \(a_{11}=0\) and trace zero is

Explanation for Answer 9:
(a) is correct.
The dimension of the vector space of all symmetric matrices of order \(n\) is \(\frac{n(n+1)}{2}\). Removing the constraints \(a_{11} = 0\) and \(\operatorname{trace}(A) = 0\) reduces the dimension by 2. So, the dimension of the subspace of symmetric matrices with \(a_{11} = 0\) and \(\operatorname{trace}(A) = 0\) is \(\frac{n(n+1)}{2} – 2\).

Solution

10. Let \(V_{1}, V_{2}\) be subspaces of a vector space \(V\), which of the following is necessarily a subspace of \(V\) ?

Explanation for Answer 10:
Answers (a) and (c) are correct.
Let \(V_1\) and \(V_2\) be subspaces of \(V\).
Then both the intersection \(V_1 \cap V_2\) and the sum \(V_1 + V_2\) are always subspaces.

However, consider the specific subspaces:
– \(V_1 = \{p(x) \in P_n(x) \mid p(0) = 0\}\)
– \(V_2 = \{p(x) \in P_n(x) \mid p(1) = 0\}\)
Clearly, both \(V_1\) and \(V_2\) are subspaces of \(V\).

Now, let’s choose \(p_1(x) = x \in V_1\) and \(p_2(x) = x – 1 \in V_2\).
If we consider \(p_1(x) + p_2(x)\), we get \(2x – 1\), which is not in either \(V_1\) or \(V_2\).

Therefore, \(V_1 \cup V_2\) is not a subspace of \(V\).

Additionally, if we look at \(V_1 \backslash V_2 = \{x \in V_1 : x \notin V_2\}\), it represents the set of vectors that are in \(V_1\) but not in \(V_2\). Since the zero vector does not belong to \(V_1 \backslash V_2\), it is not a subspace of \(V\) either.

Solution