Linear Algebra: Proving Linear Dependence in Subspaces with Basis Intersection

[akima]

Let V be a 9 dimensional vector space and let U and W
be five dimensional subspaces of V with the bases Bu
and Bw respectively,
(a) show that if Bu intersect Bw is empty then
Bu union Bw is linearly dependen
(b)use part (a) to prove U intersect W is not
equal to the 0 vector
now i have already done part (a), now i have already
done part (a). can you please help me..

for part a.. this is what i have briefly...
we know Bu intersect Bw has nothing in common.
Since Bu and Bw is a basis we know that it is linearly independent.
Therefore Let bu be a linearly independent
subset of a vector space V and let Bw be a
vector in V that is not in Bu, then Bu union Bw
is linearly dependent iff Bw is in the span of Bu... (by a theorem)
by definition of a basis we know, span ( Bu) = v
therefore, the question now is if Bw is in V
which is true ( definition of a basis)... therefore
Bu union Bw is linearly dependent if Bu intersect Bw is nothing...

 

[akima]

my homework is due tomorrow... if some one can help me before then or even give me a clue.. it will be appreciated... :)

 

[ystael]

Your argument is severely confused.

for part a.. this is what i have briefly...
we know Bu intersect Bw has nothing in common.
Since Bu and Bw is a basis we know that it is linearly independent.
Therefore Let bu be a linearly independent
subset of a vector space V and let Bw be a
vector in V that is not in Bu

Here you have just defined $$B_U$$ for the second time, which won't fly. Either $$B_U$$ is a basis for $$U$$, or $$B_U$$ is an arbitrarily chosen linearly independent subset of $$V$$. The first is what the problem says.

then Bu union Bw
is linearly dependent iff Bw is in the span of Bu... (by a theorem)

No. For a single vector $$w$$, $$B_U \cup \{w\}$$ is linearly dependent if and only if $$w \in \mathop{\mathrm{span}} B_U$$. It is false that $$B_U \cup B_W$$ is linearly dependent iff $$B_W \subset \mathop{\mathrm{span}} B_U$$, because $$B_W$$ has more than one vector in it (try to construct some simple examples in $$\mathbb{R}^3$$ with $$B_U$$ and $$B_W$$ having two elements each).

by definition of a basis we know, span ( Bu) = v

No. $$B_U$$ is a basis for $$U$$, not $$V$$; $$\mathop{\mathrm{span}} B_U = U$$.

therefore, the question now is if Bw is in V
which is true ( definition of a basis)... therefore
Bu union Bw is linearly dependent if Bu intersect Bw is nothing...

It seems like you are thinking much too hard. The dimension of $$V$$ is 9. If $$B_U$$ and $$B_W$$ are disjoint, what is the size of the set $$B_U \cup B_W$$? Can this set be linearly independent?

 

[akima]

sorry about the late reply... I am kinda new to physics forum...
I am actually completely lost with this problem... well i would guess the Bu U Bw is linearly independent.. right? because any basis by definition is linearly independent... so wouldn't their union be independent as well? i don't know.. um just confused abou this problem..
if you could help me out that would be really great :)