Skip to content

11.13 Finite-Dimensional Vector Spaces

  1. Check whether the following sets of vectors in \(\mathbb{R}^3\) are L.I. (linearly independent). If not, find a non-trivial linear relation between the vectors.

    1. \(a = \{(1, 0, -1), (1, 2, 1), (0, -3, 2)\}\)

      We consider \(a(1, 0, -1) + b(1, 2, 1) + c(0, -3, 2) = (0, 0, 0)\).

      \(\begin{aligned}\begin{pmatrix}1 & 1 & 0\\ 0 & 2 & -3\\ -1 & 1 & 2\end{pmatrix}\begin{pmatrix}a\\ b\\ c\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}\end{aligned}\)

      Since the matrix can be reduced to a identity matrix, thus it has only the trivial solution \((a, b, c) = (0, 0, 0)\) 2. \(b = \{(1, 0, -1), (1, -2, 1), (2, -2, 0)\}\)

      \(A = \begin{pmatrix} 1 & 1 & 2 \\ 0 & -2 & -2 \\ -1 & 1 & 0 \end{pmatrix}\)​\(\rightarrow\begin{pmatrix}1 & 1 & 2\\ 0 & -2 & -2\\ 0 & 0 & 0\end{pmatrix}\)

      \(\Rightarrow A\) is not identity matrix \(\Rightarrow\)there exists non-trivial solution, then it is not linear independent

      A non-trivial relation is given by:

      \(1 \cdot (1, 0, -1) + 1 \cdot (1, -2, 1) + (-1) \cdot (2, -2, 0) = (0, 0, 0)\)

  2. Prove the following:

    1. Every subset of a linear independent set is linear independent

      Let \(\{v_1, \dots, v_n\}\) be a L.I. (linearly independent) set and consider \(\{w_1, \dots, w_k\} \subseteq \{v_1, \dots, v_n\}\) a subset.

      Since \(\{v_1, \dots, v_n\}\) is L.I., if we have \(a_1 v_1 + \dots + a_n v_n = 0\), then \(a_1 = \dots = a_n = 0\).

      Now, if we take \(c_1 w_1 + \dots + c_k w_k = 0\), we can consider

      \(\{w_1, \dots, w_k, w_{k+1}, \dots, w_n\} = \{v_1, \dots, v_n\}\) (adding the vectors we are missing).

      \(\Rightarrow 0 = c_1 w_1 + \dots + c_k w_k + 0 \cdot w_{k+1} + \dots + 0 \cdot w_n\)

      By the linear independence of \(\{v_1, \dots, v_n\}\), it follows that \(c_1 = \dots = c_k = 0\). 2. Every set that contains a linear dependent subset is linear dependent

      Let \(\{v_1, \dots, v_n\}\) be L.D. (linearly dependent) and consider a set that contains \(\{v_1, \dots, v_n\}\), say \(\{v_1, \dots, v_n, w_{n+1}, \dots, w_k\}\).

      Since \(\{v_1, \dots, v_n\}\) is L.D., there exist \(a_1, \dots, a_n \in F\), not all zero, such that \(a_1 v_1 + \dots + a_n v_n = 0\).

      Then, \(0 = a_1 v_1 + \dots + a_n v_n + 0 \cdot w_{n+1} + 0 \cdot w_{n+2} + \dots + 0 \cdot w_k\)

      Therefore, \(\{v_1, \dots, v_n, w_{n+1}, \dots, w_k\}\) is L.D. 3. Every set that contains the \(0\) vector is linear dependent

      Consider \(\{0, v_1, \dots, v_n\}\), a set that includes the \(0\) vector. Take \(1 \in F\), and then we have \(1 \cdot 0 + 0 \cdot v_1 + \dots + 0 \cdot v_n = 0\).

      So we have a non-trivial linear combination equal to \(0\) \(\Rightarrow \{0, v_1, \dots, v_n\}\) is L.D.

  3. Suppose \(\{v_1, v_2, v_3, v_4\}\) spans \(V\). Prove that \(\{v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4\}\) also spans \(V\).

    Let \(w \in V\). Since \(\{v_1, v_2, v_3, v_4\}\) spans \(V\), there exist \(a_1, a_2, a_3, a_4 \in F\) such that \(w = a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4\).

    We would like to find \(b_1, b_2, b_3, b_4 \in F\) such that \(w = b_1 (v_1 - v_2) + b_2 (v_2 - v_3) + b_3 (v_3 - v_4) + b_4 v_4\).
    ​\(w = b_1 v_1 + (b_2 - b_1) v_2 + (b_3 - b_2) v_3 + (b_4 - b_3) v_4\)

    It is enough to consider the following equations: \(\begin{cases}b_1=a_1,\\ b_2-b_1=a_2,\\ b_3-b_2=a_3,\\ b_4-b_3=a_4.\end{cases}\)

    Solving, we find: \(\begin{aligned}b_1 & =a_1,\\ b_2 & =a_2+a_1,\\ b_3 & =a_3+a_2+a_1,\\ b_4 & =a_4+a_3+a_2+a_1.\end{aligned}\)​

    Thus, we've written \(w\) as a linear combination of \(\{v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4\}\).

    \(\Rightarrow\) These vectors span \(V\).

  4. Definition: A basis of a vector space \(V\) is a list of vectors in \(V\) that are L.I. (linearly independent) and that span \(V\).

    Theorem: \(\{v_1, \dots, v_n\}\) is a basis for \(V\) \(\iff\) every \(v \in V\) can be written uniquely in the form \(a_1 v_1 + \dots + a_n v_n = v\) for \(a_1, \dots, a_n \in F\).

    Find all complex vector spaces that have exactly one basis.

    For example, \(\mathbb{R}^3\) has the canonical basis \(\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}\), but there exist other bases, for example, what we saw in (1).

    We claim that the only vector space with only one basis is the trivial vector space \(\{0\}\).

    It is clear that \(\{0\}\) has only one basis

    Let \(V \neq \{0\}\) be a non-trivial vector space. Let's see that it has more than one basis.

    • Finite-dimensional \(V\)​ : Suppose \(\{v_1, v_2, \dots, v_n\}\) is a basis for \(V\).

    Let \(c \in C\), \(c \neq 0\), \(c \neq 1\). Let's see that \(\{c v_1, c v_2, \dots, c v_n\}\) is also a basis for \(V\).

    • \(\{c v_1, \dots, c v_n\}\) is L.I. : If we have \(a_1 (c v_1) + \dots + a_n (c v_n) = 0\), then
      ​\(\Rightarrow (a_1 c) v_1 + \dots + (a_n c) v_n = 0\).
      Since \(\{v_1, \dots, v_n\}\) is L.I., it follows that \(a_i c = 0\) for each \(i = 1, \dots, n\).
      ​\(\Rightarrow a_i = 0\) for all \(i\) (since \(c \neq 0\)).

    • \(\{c v_1, \dots, c v_n\}\) spans \(V\)​ . Since \(\{v_1, \dots, v_n\}\) spans \(V\), if \(v \in V\), we can write \(v = a_1 v_1 + \dots + a_n v_n\) for some \(a_i \in F\).

      \(\Rightarrow v = \frac{a_1}{c} (c v_1) + \dots + \frac{a_n}{c} (c v_n)\), so \(\{c v_1, \dots, c v_n\}\) spans \(V\).

    Therefore, \(\{c v_1, \dots, c v_n\}\) is another basis for \(V\). * Infinite-dimensional \(V\)​ : Now \(\{v_1, v_2, \dots\}\) is a basis for \(V\).

    \(\Rightarrow \{c v_1, c v_2, \dots\}\) for \(c \neq 0\), \(c \neq 1\), is also a basis by the exact same steps as before.

  5. Consider the following vectors in \(\mathbb{R}^4\):

    \(\alpha_1 = (-1, 0, 1, 2), \quad \alpha_2 = (3, 4, -2, 5), \quad \alpha_3 = (1, 4, 0, 9)\)

    Find a system of homogeneous linear equations for which the space of solutions is exactly the subspace spanned by \(\alpha_1, \alpha_2, \alpha_3\).

    Let's try to "simplify" the space \(\langle \alpha_1, \alpha_2, \alpha_3 \rangle\)​

    \(\begin{pmatrix}-1 & 0 & 1 & 2\\ 3 & 4 & -2 & 5\\ 1 & 4 & 0 & 9\end{pmatrix}\), then \(\begin{pmatrix}1 & 0 & -1 & -2\\ 0 & 1 & \frac14 & \frac{11}{4}\\ 0 & 0 & 0 & 0\end{pmatrix}\)​

    Thus, any element in \(\langle \alpha_1, \alpha_2, \alpha_3 \rangle\) is in the form \((x, y, z, w) = a(1, 0, -1, -2) + b(0, 1, \frac{1}{4}, \frac{11}{4})\). We find the system of equations that determine \(\langle \alpha_1, \alpha_2, \alpha_3 \rangle\) as: \(\begin{cases}z=-x+\frac{y}{4}\\ w=-2x+\frac{11y}{4}\end{cases}\)​

  6. Let \(U = \{(x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5 : x_1 = 3x_2, \, x_3 = 7x_4\}\) be a subspace of \(\mathbb{R}^5\).

    1. Find a basis for \(U\).

      We have that \((x_1, x_2, x_3, x_4, x_5) \in U\) can be written as: \((x_1, x_2, x_3, x_4, x_5) = (3x_2, x_2, 7x_4, x_4, x_5)\)

      \(= x_2 (3, 1, 0, 0, 0) + x_4 (0, 0, 7, 1, 0) + x_5 (0, 0, 0, 0, 1)\).

      So, \(\{(3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1)\}\) is a basis for \(U\). 2. Extend the basis found in (1) to a basis of \(\R^5\)

      We take \((x, y, z, w, u) \in \mathbb{R}^5\). With the vectors in the basis for \(U\), we can generate \(y, w,\) and \(u\), so we need to be able to generate \(x\) and \(z\).

      So we take \((1, 0, 0, 0, 0)\) and \((0, 0, 1, 0, 0)\).

      And our basis will be:
      ​\(\{(1, 0, 0, 0, 0), (0, 0, 1, 0, 0), (3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1)\}\).

      (Exercise: check that it is a basis) 3. Find a subspace \(W\) of \(\mathbb{R}^5\) such that \(\mathbb{R}^5 = U \oplus W\).

      By definition, if \(U \oplus W = \mathbb{R}^5\), then we must have:

      1. \(U + W = \mathbb{R}^5\)
      2. \(U \cap W = \{0\}\)

      Since \(U = \langle v_1, v_2, v_3 \rangle\) and \(\mathbb{R}^5 = \langle v_1, v_2, v_3, (1, 0, 0, 0, 0), (0, 0, 1, 0, 0) \rangle\), we can take:

      \(W = \langle (1, 0, 0, 0, 0), (0, 0, 1, 0, 0) \rangle\)

      This choice ensures that \(\mathbb{R}^5 = U \oplus W\), because every vector \(v \in \mathbb{R}^5\) can be written uniquely as \(v = u + w\) for \(u \in U\) and \(w \in W\).

‍