Exercise I.C.6

Question

Let $V$ be a vector space over a field $\mathbb{K}$, and let $U,W$ be finite-dimensional subspaces of $V$. Prove that $U+W$ and $U\cap W$ are finite-dimensional subspaces of $V$, and
$$\dim(U + W) + \dim(U \cap W) = \dim(U) + \dim(W)$$

Toğrul Topal · Accepted Answer

\begin{itemize}
    \item Pick an arbitray $a,b \in U + W$ and $c \in \mathbb{K}$. Then $a = u + w$ and $b=u' + w'$ for some $u,u' \in U$ and $w,w' \in W$. We then have $ca + b = cu + cw + u' + w' = (cu + u') + (cw + w')$. But $cu + u' \in U$ and $cw +w' \in W$, and so $ca+b \in U + W$. Thus, $U+W$ is a vector subspace of $V$.
    \item Let $a,b \in U \cap W$ and $c \in \mathbb{K}$. Then $a, b \in U$ and $a,b \in W$, and so by the closure of both under linear combinations we have $ca+b \in U$ and $ca+b \in W$; thus $ca + b \in U \cap W$.
    \item Let $\{v_1,\ldots,v_p\}$ be a basis for $U \cap W$. By extension theorem (Proposition I.11)
      \begin{itemize}
      \item Since $U \cap W \subseteq U$ we can find a basis $\{v_1,\ldots,v_p, u_1,\ldots,u_q\}$ of $U$
      \item Since $U \cap W \subseteq W$ we can find a basis $\{v_1,\ldots,v_p, w_1,\ldots,w_r\}$ of $W$
      \end{itemize}
      Now we demonstrate that the set
      $$\{v_1,\ldots,v_n,u_1,\ldots,u_q,w_1,\ldots,w_r\} $$
      is a basis for $U+W$.
      \begin{itemize}
      \item (Spanning) Let $x \in U+W$ be arbitrary, i.e., $x = u + w$ for some $u \in U$ and $w\in W$. By the definition of bases of $U$ and $W$ we can write
        \begin{align*}
          x &= u + w = \left(a_1v_1 + \ldots + a_pv_p + a_{p+1}u_1 + \ldots + a_{p+q}u_qight) + \left(b_1v_1 + \ldots + b_pv_p + b_{p+1}w_1 + \ldots + b_{p+r}w_right)\
            &= (a_1+b_1)v_1 + \ldots + (a_p+b_p)v_p + a_{p+1}u_1 + \ldots + a_{p+q}u_q + b_{p+1}w_1 + \ldots + b_{p+r}w_r
        \end{align*}
      \item (Independence) Let $a_1,\ldots,a_p,a_{p+1},\ldots,a_{p+q},a_{p+q+1},\ldots,a_{p+q+r}$ be such that
        $$a_1v_1 + \ldots + a_pv_p + a_{p+1}u_1 + \ldots + a_{p+q}u_q + a_{p+q+1}w_1 + \ldots + a_{p+q+r}w_r = 0$$
        Then we notice two things:
        \begin{align*}
          (1) \quad & a_1v_1 + \ldots + a_pv_p + a_{p+1}u_1 + \ldots + a_{p+q}u_q = (-a_{p+q+1})w_1 + \ldots + (-a_{p+q+r})w_r\
          (2) \quad & a_1v_1 + \ldots + a_pv_p + a_{p+q+1}w_1 + \ldots + a_{p+q+r}w_r = (-a_{p+1})u_1 + \ldots + (-a_{p+q})u_q
        \end{align*}
        In other words,
        \begin{enumerate}
        \item[(1)] $(-a_{p+q+1})w_q + \ldots + (-a_{p+q+r})w_r \in U \cap W$; thus, $(-a_{p+q+1})w_1 + \ldots + (-a_{p+q+r})w_r = c_1v_1 + \ldots + c_pv_p$ for some $c_1,\ldots,c_p$. But this is is a linear combination of basis elements of $W$ which results in a sum zero; thus, all coefficients, including $a_{p+q+1},\ldots,a_{p+q+r}$ must be zero.
        \item[(2)] Similarly, $(-a_{p+1})u_1 + \ldots + (-a_{p+q})u_q \in U \cap W$; thus, $(-a_{p+1})u_1 + \ldots + (-a_{p+q})u_r = c_1v_1 + \ldots + c_pv_p$ for some $c_1,\ldots,c_p$. But this is is a linear combination of basis elements of $U$ which results in a sum zero; thus, all coefficients, including $a_{p+1},\ldots,a_{p+r}$ must be zero.
        \item[(3)] Thus, $a_{p+1},\ldots,a_{p+r},a_{p+r+1},\ldots,a_{p+r+q}$ are all zero which leaves us with $a_1v_1 + \ldots + a_pv_p = 0$. By the basis assumption, $a_1,\ldots,a_p$ must be zero as well, and we are done.
        \end{enumerate}
      \end{itemize}
    \end{itemize}

Exercise I.C.6

Answers

Comments

Exercise I.C.6

Answers

Comments

Add answer