Re: Linear Algebra


Arturo Magidin wrote:
In article <1160687083.787621.98000@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
jennifer <scrilla_12_1999@xxxxxxxxx> wrote:

Arturo Magidin wrote:
In article <1160686651.389119.179710@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
jennifer <scrilla_12_1999@xxxxxxxxx> wrote:

[...]

Let V denote a 10-dimensional vector space, and let U and W denote the
subspaces of V, where dim U= 8 and dim W= 9.
Prove that there are only two possible values of the dimension of UnW.

[...]

So... What do you think are the "two values" that dim(U/\W) "can
have"?

Dim(U/\W) = 7 or Dim(U/\W) = 8, right?


Yes. Now, you have proven that Dim(U/\W) >= 7.

Now, why is it that Dim(U/\W) <= 8?


Because we have only 8 vectors in U/\W, from the example you gave.

No. No. No. No. No. No. No. No.

First, in ->my<- example there are plenty more than "8 vectors". There
are uncountably many different vectors.

Second, you need to prove it IN GENERAL. Look at your original
problem!

Sigh. U/\W is certainly a subspace of V. But is it a subspace of
something ->smaller<-, as well?

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

It is a subspace of U so it's dimension has to be less than or equal to
dimU because if we have a list of vectors in UnW, it can be extended to
form a basis in U.

.

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