--- Commentary: $\textbf{Exercise 4.}$ (a) Let $U$ be the subspace of $\mathbb{C}^5$ defined by $U = {(z_1, z_2, z_3, z_4, z_5) \in \mathbb{C}^5 : 6z_1 = z_2 \text{ and } z_3 + 2z_4 + 3z_5 = 0}.$ Find a basis of $U$. (b) Extend the basis in (a) to a basis of $\mathbb{C}^5$. (c) Find a subspace $W$ of $\mathbb{C}^5$ such that $\mathbb{C}^5 = U \oplus W$. $\textbf{Solution 4.}$ (a) The list $(1, 6, 0, 0, 0)$, $(0, 0, -2, 1, 0)$, $(0, 0, -3, 0, 1)$ consists of vectors in $U$. It is easy to see from the definition of linear independence that this list is linearly independent. If $(z_1, z_2, z_3, z_4, z_5) \in U$, then $(z_1, z_2, z_3, z_4, z_5) = z_1(1, 6, 0, 0, 0) + z_4(0, 0, -2, 1, 0) + z_5(0, 0, -3, 0, 1).$ Thus the list $(1, 6, 0, 0, 0)$, $(0, 0, -2, 1, 0)$, $(0, 0, -3, 0, 1)$ spans $U$. Because $(1, 6, 0, 0, 0)$, $(0, 0, -2, 1, 0)$, $(0, 0, -3, 0, 1)$ is linearly independent and spans $U$, this list is a basis of $U$. (b) The list $(1, 6, 0, 0, 0), (0, 0, -2, 1, 0), (0, 0, -3, 0, 1), (1, 0, 0, 0, 0), (0, 0, 1, 0, 0)$ is a basis of $\mathbb{C}^5$, as is easy to verify. (c) Using the idea of the proof of 2.33 and the answer above to (b), we see that taking $W = {(\alpha, 0, \beta, 0, 0) : \alpha, \beta \in \mathbb{C}}$ gives a subspace $W$ such that $\mathbb{C}^5 = U \oplus W$. $\textit{Commentary:}$ This exercise is similar to Exercise 3, but takes place in $\mathbb{C}^5$ instead of $\mathbb{R}^5$ and has slightly different defining equations for the subspace $U$. In part (a), we start by noting that $(1, 6, 0, 0, 0)$, $(0, 0, -2, 1, 0)$, and $(0, 0, -3, 0, 1)$ are in $U$. The first vector corresponds to $z_1 = 1, z_4 = z_5 = 0$, the second to $z_1 = z_5 = 0, z_4 = 1$, and the third to $z_1 = z_4 = 0, z_5 = 1$. These vectors are linearly independent because they are non-zero and orthogonal (when viewed as vectors in $\mathbb{R}^{10}$). Then, we show that every vector in $U$ can be written as a linear combination of these three vectors, proving that they form a basis of $U$. In part (b), we extend this basis to a basis of $\mathbb{C}^5$ by adding $(1, 0, 0, 0, 0)$ and $(0, 0, 1, 0, 0)$, which introduce the missing dimensions. In part (c), we again choose $W$ to be the subspace spanned by the vectors we added in part (b), ensuring that $\mathbb{C}^5 = U \oplus W$. $\textit{Examples:}$ 1. In $\mathbb{C}^4$, let $U = {(z, w, iw, iz) : z, w \in \mathbb{C}}$. A basis for $U$ is $(1, 0, 0, i)$ and $(0, 1, i, 0)$. This can be extended to a basis of $\mathbb{C}^4$ by adding $(i, 0, 0, -1)$ and $(0, i, -1, 0)$. The subspace $W = {(ai, 0, 0, -a) + (0, bi, -b, 0) : a, b \in \mathbb{C}}$ satisfies $\mathbb{C}^4 = U \oplus W$. 2. In the space of complex polynomials $\mathbb{C}[z]$, let $U = {p(z) : p(i) = p(-i)}$. A basis for $U$ is $1, z^2, z^4, \ldots$. This can be extended to a basis of $\mathbb{C}[z]$ by adding $z, z^3, z^5, \ldots$. The subspace $W = {a_1z + a_3z^3 + a_5z^5 + \ldots : a_k \in \mathbb{C}}$ satisfies $\mathbb{C}[z] = U \oplus W$. These examples demonstrate the same principles in a complex vector space and a complex polynomial space. The steps are to find a basis for the subspace, extend it to a basis of the whole space, and identify a complementary subspace for the direct sum decomposition.