I have a question about the following proof of the statement that for every subspace $U$ of a finite dimensional vector space $V$ there is a subspace $W$ of $V$ such that $V = U \oplus W$.


Because $V$ is finite-dimensional, so is $U$. Thus there is a basis $(u_1,\ldots,u_m)$ of U. Of course $(u_1,\ldots,u_m)$ is a linearly independent list of vectors in $V$, and thus it can be extended to a basis $(u_1,\ldots,u_m, w_1, \ldots,w_n)$ of $V$. Let $W = \text{span}(w_1,\ldots,w_n).$

To prove that $V = U \oplus W$, we need to show that $$V = U + W \text{ and } U \cap W = \{0\}.$$

To prove the first equation, suppose that $v \in V$. Then, because the list $(u_1,\ldots,u_m, w_1, \ldots,w_n)$ spans $V$, there exist scalars $a_1,\ldots,a_m,b_1,\ldots,b_n \in \mathbb{F}$ such that $$v = \underbrace{a_1u_1 + \dotsb + a_mu_m} + \underbrace{b_1w_1 + \dotsb + b_nw_n}.$$

In other words, we have $v = u + w$, where $u \in U$ and $w \in W$ are defined as above. Thus $v \in U + W$.

I understand everything up until this point, the next part of the proof is the portion that I wish to clarify.

To show that $U \cap W = \{0\}$, suppose $v \in U \cap W$. Then there exist scalars $a_1,\ldots,a_m,b_1,...,b_n \in \mathbb{F}$ such that $$v = a_1u_1 + \dotsb+a_mu_m = b_1w_1 + \dotsb + b_nw_n. $$

Halt! See the above. How does the notion that $v \in U\cap W$ lead to the equality $v = a_1u_1 + \dotsb+a_mu_m = b_1w_1 + \dotsb + b_nw_n$? Or how does the containment of $v$ in $U$ and $W$ yield the equality $v = a_1u_1 + \dotsb+a_mu_m = b_1w_1 + \dotsb + b_nw_n$?

  • 1
    $\begingroup$ Note that $v\in U\cap W$ means containment of $v$ both in $U$ and in $W$, but you say "either ... or" instead. $\endgroup$ Oct 3 '13 at 6:31
  • $\begingroup$ I was thinking of $v \in U \cup W$ for some reason. Thanks for pointing that out. $\endgroup$
    – St Vincent
    Oct 3 '13 at 18:05
  • $\begingroup$ remaining part: as $v=a_1u_1+...+a_mu_m= b_1w_1+...+b_nw_n$. Thus we have, $a_1u_1+...+a_mu_m-b_1w_1-...-b_nw_n=0$. But, $\{u_1,...,u_m,w_1,...,w_n\}$ is basis of $V$ and hence linearly independent. Hence each $a_i, b_j=0$ for $i=1,...,m$ and $j=1,..n$. This implies $v=0$ and thus $U\cap W=\{0\}$. Is am I correct? $\endgroup$ Mar 17 '19 at 12:46

Since $v$ is in $U$, it can be written as a linear combination of basis elements of $U$ , and the same goes for $W$.

While you're at it, and since this wasn't too thrilling, you can go ahead and prove that

$({\rm i})$ $U\oplus W=V$

$(\rm ii)$ If $B'$ is a basis of $U$ and $B''$ a basis of $W$ then $B=B'\cup B''$ is a basis of $V$

are equivalent statements.

  • $\begingroup$ Wow I feel dumb, Thanks! +1 $\endgroup$
    – St Vincent
    Oct 3 '13 at 6:02
  • $\begingroup$ For the second part, do you mean $B = B' \cup B''$? $\endgroup$
    – St Vincent
    Oct 3 '13 at 6:15

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