Tricky question, subspaces dimensions

i need to disprove the following with an example:

Let $\displaystyle U_1,U_2,U_3$ be subspaces of $\displaystyle V$.

$\displaystyle dim(U_1+U_2+U_3)=dim(U_1)+dim(U_2)+dim(U_3)-dim(U_1\cap U_2)-dim(U_1\cap U_3)-dim(U_2\cap U_3)+dim(U_1\cap U_2\cap U_3)$

i used Venn Diagram and covered any possible variations of $\displaystyle U_1,U_2$ and $\displaystyle U_3$ but nothing seems to work.

Thanks in advanced!

Re: Tricky question, subspaces dimensions

suppose V = R^{3}, and set:

U_{1} = span({(1,1,0)})

U_{2} = span({(1,0,0)})

U_{3} = span({(0,1,0)}).

then dim(U_{1}+U_{2}+U_{3}) = 2

but dim(U_{1}) + dim(U_{2}) + dim(U_{3}) - dim(U_{1}∩U_{2}) - dim(U_{1}∩U_{3}) - dim(U_{2}∩U_{3}) + dim(U_{1}∩U_{2}∩U_{3})

= 1 + 1 + 1 - 0 - 0 - 0 + 0 = 3

Re: Tricky question, subspaces dimensions

Presumably, this question arose from the true statement that $\displaystyle dim(U_1+U_2)=dim(U_1)+dim(U_2)-dim(U_1\cap U_2)$. When you try to extend this to 3 subspaces, you need to find $\displaystyle (U_1+U_2)\cap U_3$, which is definitely not equal to $\displaystyle (U_1\cap U_3)+(U_2\cap U_3)$. Basically, this is why Deveno's counter example above works. Notice though if $\displaystyle U_1\subset U_3$ or $\displaystyle U_2\subset U_3$, then $\displaystyle (U_1+U_2)\cap U_3=(U_1\cap U_3)+(U_2\cap U_3)$, and so in this case your original statement is true. (It boils down to the true statement about two subspaces.)

Re: Tricky question, subspaces dimensions

thanks man!

when i used Venn Diagram i always considered $\displaystyle dim(U_1+U_2+U_3)$ as being equal to $\displaystyle dim(V)$.

Re: Tricky question, subspaces dimensions

Quote:

Originally Posted by

**johng** Presumably, this question arose from the true statement that $\displaystyle dim(U_1+U_2)=dim(U_1)+dim(U_2)-dim(U_1\cap U_2)$. When you try to extend this to 3 subspaces, you need to find $\displaystyle (U_1+U_2)\cap U_3$, which is definitely not equal to $\displaystyle (U_1\cap U_3)+(U_2\cap U_3)$. Basically, this is why Deveno's counter example above works. Notice though if $\displaystyle U_1\subset U_3$ or $\displaystyle U_2\subset U_3$, then $\displaystyle (U_1+U_2)\cap U_3=(U_1\cap U_3)+(U_2\cap U_3)$, and so in this case your original statement is true. (It boils down to the true statement about two subspaces.)

exactly so: we might have $\displaystyle u_1 \in U_1, u_2 \in U_2$ and $\displaystyle u_1+u_2 \in U_3$, but that does not guarantee that $\displaystyle u_1,u_2 \in U_3$.