Note 1
Solutions

Here are the solutions to the practice problems at the end of each note.

1.1 Solutions for Note 1

1.

It suffices to show that \(\alpha (x, y, z) + \beta (x', y', z') \in S\) for any scalars \(\alpha \) and \(\beta \). With standard vector addition and scalar multiplication, \(\alpha (x, y, z) + \beta (x', y', z') = (0, \alpha y + \beta y', \alpha z + \beta z') \in S\). Thus, the set is a vector space.

2.

Let \(f\) and \(g\) be functions in \(\mathbb {R}\) such that \(f'(0) = 0\) and \(g'(0) = 0\). Notice that \((f + g)'(0) = f'(0) + g'(0) = 0\) and \(f + g \in S\). Moreover, for any scalars \(\alpha \), \((\alpha f)'(0) = \alpha f'(0) = 0\). Thus \(\alpha f \in S\) and \(S\) is a vector space.

3.

First, for any scalars \(\alpha \), notice that \(\alpha (x_1, \ldots , x_n) = (\alpha x_1, \ldots , \alpha x_n)\). If each coefficient is divided by \(\alpha \neq 0\), the scalar multiplication holds. The case where \(\alpha = 0\) naturally holds. Continuing, notice that \((x_1, \ldots , x_n) + (x_1', \ldots , x_n') = (x_1 + x_1', \ldots , x_n + x_n') \in S\) due to the property of addition. Thus, \(S\) is a vector space.

4.

Let \(V = W_1 \cap \cdots \cap W_n\) where \(W_i\) for integer \(i \in [1, n]\) is a subspace of \(V\). For \(u, v \in V\), notice that since \(u, v \in W_i\) for all \(i\) and \(\alpha u + \beta v \in W_i\) for any scalars \(\alpha \) and \(\beta \) by construction, \(\alpha u + \beta v \in V\) and the intersection of subspaces of a vector space is also a subspace of the vector space.

5.

First and foremost, it is evident that \(e^{a_1 x}\) is linearly independent since \(e^{a_1 x} > 0\) for all \(a_1\) and \(x\). Therefore, it suffices to show that if the set is linearly independent for \(n - 1\), then it is linearly independent for \(n\). Consider the following equations for some \(c_k\). \begin{align*} \sum _{k=1}^n c_k e^{a_k x} &= 0 \\ \sum _{k=1}^n c_k a_k e^{a_k x} &= 0 \\ \sum _{k=1}^n c_k a_n e^{a_k x} &= 0 \end{align*}

Subtracting the third equation, which is obtained by multiplying \(a_n\) in both sides, to the second equation obtained by differentiating both sides, \[ \sum _{k=1}^n c_k (a_k - a_n) e^{a_k x} = \sum _{k=1}^{n-1} c_k (a_k - a_n) e^{a_k x} = 0 \] is obtained. Because \(a_k \neq a_n\) for integer \(k \in [1, n-1]\) and \(\{ e^{a_1 x}, \ldots , e^{a_{n-1} x} \}\) is linearly independent \(c_1 = \cdots = c_{n-1} = 0\). In other words, \(c_n e^{a_n x} = 0\) and \(c_n = 0\). Therefore, \(\{ e^{a_1 x}, \ldots , e^{a_n x} \}\) is linearly independent by induction.

6.

By definition, note that \(C = A + B\). Therefore, it suffices to show that \(A \cap B = \{ 0 \}\). Notice that \(A\) and \(B\) can be written as \(A = \Span \{ (1, 1, 1) \}\) and \(B = \Span \{ (20, 15, 12) \}\). Because \(\lambda (1, 1, 1) = (20, 15, 12)\) if and only if \(\lambda = 0\), \(A \cap B = \{ 0 \}\) and \(C = A \oplus B\).

7.

Consider the following equations for some scalars \(k_1, k_2, k_3\). \begin{align*} k_1 (a + b) + k_2 (b + c) + k_3 (c + a) &= 0 \\ a (k_3 + k_1) + b (k_1 + k_2) + c (k_2 + k_3) &= 0 \end{align*}

The linear combinations of \(\{ a + b, b + c, c + a \}\) equal zero if and only if the linear combinations of \(a\), \(b\), and \(c\) equal zero. Moreover, from the equation above, \(k_1 = -k_2 = k_3 = -k_1\). In other words, \(k_1 = k_2 = k_3 = 0\) and \(\{ a + b, b + c, c + a \}\) is linearly independent.

8.

Let \(f\) and \(g\) be polynomials such that \(f(0) = g(0) = 0\). Therefore, \((f + g)(0) = 0\) and \(f + g\) is in the set. Moreover, \((\alpha f)(0) = 0\) for any scalars \(\alpha \). Therefore, the set of all polynomials in \(\mathbb {P}^3\) that includes the origin is a subspace of \(\mathbb {P}^3\).