linear space - considerations about dimensions

Question

$X$ is linear space and $U$, $V$ are subspaces of $X$ and $\dim U= 4$, $\dim V=6$, $\dim X = 10$. Then:
a. $X = U \oplus V$,
b. $\dim(U \cap V)\geq 2$,
c. $\dim(U+V) = \dim(X) \implies U \cap V = \{0\}$.

a. It is not true. Take $X = \operatorname{span}(e_1, \dotsc, e_{10})$, $U = \operatorname{span}(e_1, \dotsc, e_4)$ and $V = \operatorname{span}(e_1, \dotsc, e_6)$. Then $(0,0,0,0,0,0,0,0,0,1) \in X$, but $(0,0,0,0,0,0,0,0,0,1)\notin U \oplus V$.

b. I don't know how to solve it. Can you help me?

c. Is not true: If $X = U = V = \operatorname{span}(e_1, \dotsc, e_{10})$, then $U + V = X$.

What about correctness of a and c?

Jendrik Stelzner · Accepted Answer · 2016-08-06T23:31:58.460

0

Your counterexample for c doesn’t work, because in this case $\dim U = \dim V = 10$, but it is given that $\dim U = 4$ and $\dim V = 6$.

Your counterexample for a works, but instead of $U \oplus V$ you need to write $U + V$, because the sum is not direct.

Hint for c: This are straightforward application of a dimension formula, which relates $\dim(U + V)$ and $\dim(U \cap V)$.

Hint for b: Try to construct a counterexample, similarly to a. You can also use the dimension formula mentioned above to translate the inequality $\dim(U \cap V) \geq 2$ into an equivalent inequality concerning $\dim(U+V)$, which may be more intuitive to you.

edited Aug 06 '16 at 23:31

answered Aug 06 '16 at 23:16

Jendrik Stelzner

17,081

Thanks to relation c is true, yeah ? Tomorrow I will consider b. – Aug 06 '16 at 23:33
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Yes, statement c is true. – Jendrik Stelzner Aug 06 '16 at 23:37
For b. $U=span(e_1,...,e_4)$, $V=(e_5,...,e_{10})$ $U+V=X$. $\dim U\cap V) = 0$. Ok ? – Aug 07 '16 at 13:03
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Yes, this works (it is also the counterexample I tought about). – Jendrik Stelzner Aug 07 '16 at 16:15

linear space - considerations about dimensions

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