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 :)