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The columns of $A$ are $n$ vectors from $\mathbb{R^m}$. If they're linearly independent, what's $rank(A)$? If they span $\mathbb{R^m},$ what's $rank(A)$? If they're a basis for $\mathbb{R^m},$ what's $rank(A)$?
Solution: $n$-independent columns $\Large{\color{red}{[}}$ implies rank $= n$ $\Large{{\color{red}{]}}}$.
Columns span $\mathbb{R^m}$ $\Large{\color{red}{[}}$ implies rank $= m$ $\Large{{\color{red}{]}}}$ .
Columns are basis for $\mathbb{R^m}$ implies rank = $m = n. \qquad \square$

P124: The column space consists of all linear combinations of the columns.
The combinations are all possible vectors $\mathbf{Ax}$.
P144: The rank of a matrix is its number of pivots.
P171: A set of vectors spans a space if their linear combinations fill the space.

$\Large{1.}$ I'm trying to complete the steps in this (terse) solution. Since the $n$ columns are linearly independent, thus $\left[\vec{\text{col 1 of A}} \ldots \vec{\text{col n of A}} \right]_{m \times n}\mathbf{x = 0} \implies \sum_{1 \le i \le n}x_i(\vec{\text{col i of A}}) = \mathbf{0} \implies \mathbf{x = 0}.$
How and why does this imply $rank(A) = n$?

$\Large{2.}$ Are there more intuitive, lucid arguments/proofs than #$1$?

$\Large{3.}$ Since the columns that span $\mathbb{R^m}$ might be linearly dependent, shouldn't this be rank $\le m$ ?
I then considered $L = \left[c \quad kc \right]_{1 \times 2}$ where $m = 2, n = 2$ and
by inspection, the number of pivots $ = 1$. For all $k \neq 0$, this is just the straight line $x_2 = (-1/k)x_1,$ which isn't all of $\mathbb{R^2}$. So $L$ doesn't span $\mathbb{R^2}$ ?

This question precedes dimension, dimensions/theorems of the 4 subspaces, Orthogonality, Determinants, eigenvalues and eigenvectors, and linear transformations, all of which should please be omitted.

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  • $\begingroup$ Re: 1: If the columns are independent, then the dimension of the column space of $A$ is $n$. The rank is equal to the dimension of the column space. Depending on the course you're in, that may be all they expect to have as the answer... Have you talked about column spaces? $\endgroup$ – apnorton Nov 22 '13 at 15:06
  • $\begingroup$ @anorton: Thank you for your comment, by virtue of which I just added the defn of column space which we did cover. Rank hasn't been discussed as the dimension of the column space. Please let me know if you need more info. $\endgroup$ – Greek - Area 51 Proposal Nov 22 '13 at 15:26
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Re:1,2 : If you apply eliminations (on a matrix with n independent columns) to get reduced row echelon form (rref) then the result will be matrix with n pivots in n columns. Since rank equates with no. of pivots, in this case it'll be n.

Re:3 : It's not mentioned that if either n$\leq$m or n$\geq$m. Since, given n vectors are in $R^m$ and are linearly independent, n=m follows. Otherwise, if n < m then, they can't span whole of $R^m$ but just part of it. Moreover, if n > m then they can't be linearly independent which contradicts (1).
In your example you took linearly dependent columns ($L = [c\hspace1ex kc]$) that's why it has single pivot and rank=1. If you choose ($L = [c\hspace1ex d]$) where c and d are independent then you have two pivots and L has rank=2. Thus, n=m follows.

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