Re: rank([X;Y]) < rank(X) + rank(Y)
- From: "John D'Errico" <woodchips@xxxxxxxxxxxxxxxx>
- Date: Fri, 17 Oct 2008 10:20:17 +0000 (UTC)
Greg Heath <heath@xxxxxxxxxxxxxxxx> wrote in message <1c2d7246-a730-4ba0-92b8-0b807b18cfe5@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
Thanks, John. I have been determined to do a similar
construction using 13 + 31 original vectors that create
nonorthogonal bases. Hence the reason for this thread.
I purposely excluded orthogonalization because I wanted
to understand what lessons original vectors could tell
me from a physical point of view.
Kudos aside, maybe my basic problem is that I really
don't understand the terminology (dumb engineer etc).
spanX contains all possible linear combinations of the
17 X row vectors.
spanY contains all possible linear combinations of the
60 Y row vectors
spanZ contains all possible linear combinations of the
77 Z row vectors.
Therefore spanZ not only contains spanX and spanY,
but it also contains all possible linear combinations
of X and Y row vector mixtures. Consequently, I
concluded that
spanZ > spanX + spanY
Maybe that is the wrong terminology for what i just
described. If so, and my logic makes sense, what is
the correct terminology?
Greg- Hide quoted text -
- Show quoted text -
The problem is that span(X) is a set of vectors.
It is an entire subspace, a set of infinitely
many points that comprise a subspace of the
vectors in a vector space of dimension 45. The
dimension of this subspace was determined
numerically by the function rank to be 13. The
number 13 here is the number of basis vectors
that will completely describe any possible point
in that subspace as a linear combination of those
basis vectors.
Likewise, the span(Y) is a subspace of dimension
31, again as determined by rank. But remember,
that these subspaces are sets. So when you
compare their "sizes", as determined by their
rank, you must think in terms of a venn diagram.
I think the problem is that you are thinking of
these "mixed linear combinations", and thinking
that they are somehow not described, or that
we are somehow missing them. But mixed
linear combinations is not something that has
a mathematical definition. Lets go back to the
tiny example I gave long ago. I'll reduce it to
only 4 dimensions.
X = [1 0 0 0;1 1 1 1];
Y = [0 1 0 0;1 1 1 1];
The second vector is the same in each, but if
you prefer it does not need to be. I could have
as easily set
Y = [0 1 0 0;1 2 1 1];
See that the vector [1 1 1 1] is in the span of
both X and Y.
Clearly rank(X) = rank(Y) = 2. But, rank([X;Y])
is just 3.
What is span(X)? Any point in the row space
of X can be written as a linear combination of
the two orthonormal vectors:
[1 0 0 0]
and
[0 1 1 1]/sqrt(3)
In fact, orth gives a different pair of vectors.
orth(X')'
ans =
-0.6011 -0.4614 -0.4614 -0.4614
-0.79917 0.34705 0.34705 0.34705
No matter, since these two are just rotated from
the pair that I suggested. The choice is truly
arbitrary. Span(X) is a subspace of dimension 2
in our 4-d vector space. Likewise, span(Y) is
another subspace of dimension 2. But there is
some overlap. They share a smaller subspace
between them.
Now, what happens when we form [X;Y]? What
is the rank of the
X = [1 0 0 0;1 1 1 1];
Y = [0 1 0 0;1 2 1 1];
rank([X;Y])
ans =
3
What does it mean to be in the row space of
the combined array? A vector will be in that
row space if it can be written as some linear
combination of any of the rows of that array.
However, you can never represent the vector
[0 0 1 -1]
in any as a linear combination of those rows.
And it does not make any sense to talk about
"mixed linear combinations", since if a vector
lies in the span([X;Y]), all this means is it can
be represented as SOME linear combination of
the 3 basis vectors that span that subspace.
Any such "mixed linear combination" is already
included in that description.
Does this make sense?
John
.
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