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I don't see how $k < m$ can possibly occur in part (b).

Imagine $X \in R^{n, k}$ describes a matrix for the vectors $x^1$ to $x^K$.

$m = \dim( span(S)) \leq \min(n,k)$

Then if $k < m$ in part (b), then $k < m \leq \min(n,k)$. This inequality chain is impossible.

Imagine n = 100, k = 10.

Then, $10 < m \leq \min(100, 10)$

$10 < m \leq 10$

m cannot be equal to 10, because $10 < 10$ is impossible, and m cannot be less than 10, because $10 < m < 10$ is impossible.

What's going on?

  • 1
    $\begingroup$ You are confusing $k$ with $K$. Your $X \in R^{n, K}$ and $m \leq \min(n,K)$, while $k$ can be anything. Even $1$ if you just take the first vector, or $0$ if you take none. $\endgroup$ – Conifold Sep 16 at 5:27
  • $\begingroup$ @Conifold The proof is not complete to me. Why is it not possible that you must take linear combinations of $x_1, ..., x_K$ to generate the basis of size m? Why are we guaranteed to be able to just pick from $x_K$ without any linear combinations? $\endgroup$ – user3180 Sep 16 at 7:12
  • $\begingroup$ Because, by induction, you can pick $m$ linearly independent vectors by adding $m-k$ from the remaining ones. And any $m$ linearly independent vectors in a subspace of dimension $m$ are its basis. $\endgroup$ – Conifold Sep 16 at 7:42
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    $\begingroup$ I agree the problem might be the confusion between $k, K$. This is very bad pedagogy... $\endgroup$ – iarbel84 Sep 16 at 8:47

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