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In this proof author tries to prove $m \leq n$ but then assumes he can do this process m times. How can we be sure about we don't run out of n? If this is not the case, why this proof is too long? Can't we just use the Linear Dependence Lemma to remove one span vector and add one linearly independent vector until we run out of linearly independent vectors? enter image description here

Sarvesh Ravichandran Iyer
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Ali
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  • Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community May 20 '23 at 08:36

2 Answers2

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Let's assume that $m > n$. In this case, after step $n$, we are left with the list of vectors $u_1, \ldots, u_n$, which by the argument in the proof must span $V$. Consider the vector $u_{n+1}$, which exists since $m > n$. On the one hand, the list $u_1, \ldots, u_{n+1}$ is assumed to be linearly independent. On the other hand, $u_{n+1} \in V = \operatorname{span}(u_1, \ldots, u_n)$, so the list $u_1, \ldots, u_{n+1}$ is linearly dependent. Contradiction.

Annie Carter
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  • how can you know u_1 to u_n spans V? Arguments says vectors w_1 to w_n spans V not u_1 to u_n. – Ali May 20 '23 at 11:38
  • The idea is that if you start with a set that spans $V$ and add a vector ($u_j$ at step $j$), the new set still spans $V$. If you then remove a vector (one of the $w_i$) that is linearly dependent on the others, the resulting set still spans $V$. In this way, starting with the spanning set ${w_1, \ldots, w_n}$, we alternate adding one $u_j$ and removing one $w_i$, until the set has been transformed to ${u_1, \ldots, u_n}$, and at each stage the set still spans $V$. – Annie Carter May 21 '23 at 09:10
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Suppose at step $j$ the new set of vectors obtained is denoted $B_j$,$\newcommand{\spn}{\text{span}}$ and $B_0$ is the set that contains all the $w$'s. The main idea behind this argument is that $\spn(B_j)=\spn(B_{j+1})$ for any $j$. This is guaranteed by the part (b) in the quoted Linear Dependence Lemma. Since $u_{j+1}\in\spn(B_0)=\spn(B_j)$, we are guaranteed that $\{u_{j+1}\}\cup B_{j}$ is linearly dependent.

Now, if the $n$ vectors in $B_0$ are used up but we still have some $u$'s not used, then $B_j$ contains full of $u$'s, so $\{u_{j+1}\}\cup B_j$ would be linearly independent. This is impossible.

Paprika7191
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