On 8/25/06, Stefan van der Walt <[EMAIL PROTECTED]> wrote:
Matrix rank has nothing to do with numpy rank. Numpy rank is simply the number of indices required to address an element of an ndarray. I always thought a better name for the Numpy rank would be dimensionality, but like everything else one gets used to the numpy jargon, it only needs to be defined someplace for what it is.
Chuck
On Thu, Aug 24, 2006 at 11:10:24PM -0400, Sasha wrote:
> I would welcome an effort to make the glossary more novice friendly,
> but not at the expense of oversimplifying things.
>
> BTW, do you think "Rank ... (2) number of orthogonal dimensions of a
> matrix" is clear? Considering that matrix is defined a "an array of
> rank 2"? Is "rank" in linear algebra sense common enough in numpy
> documentation to be included in the glossary?
>
> For comparison, here are a few alternative formulations of matrix rank
> definition:
>
> "The rank of a matrix or a linear map is the dimension of the image of
> the matrix or the linear map, corresponding to the number of linearly
> independent rows or columns of the matrix, or to the number of nonzero
> singular values of the map."
> <http://mathworld.wolfram.com/MatrixRank.html >
>
> "In linear algebra, the column rank (row rank respectively) of a
> matrix A with entries in some field is defined to be the maximal
> number of columns (rows respectively) of A which are linearly
> independent."
> <http://en.wikipedia.org/wiki/Rank_(linear_algebra)>
I prefer the last definition. Introductory algebra courses teach the
term "linearly independent" before "orthogonal" (IIRC). As for
"linear map", it has other names, too, and doesn't (in my mind)
clarify the definition of rank in this context.
Matrix rank has nothing to do with numpy rank. Numpy rank is simply the number of indices required to address an element of an ndarray. I always thought a better name for the Numpy rank would be dimensionality, but like everything else one gets used to the numpy jargon, it only needs to be defined someplace for what it is.
Chuck
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