# Tricky question, subspaces dimensions

• Jan 6th 2013, 09:32 AM
Stormey
Tricky question, subspaces dimensions
i need to disprove the following with an example:

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

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

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

• Jan 6th 2013, 11:19 AM
Deveno
Re: Tricky question, subspaces dimensions
suppose V = R3, and set:

U1 = span({(1,1,0)})

U2 = span({(1,0,0)})

U3 = span({(0,1,0)}).

then dim(U1+U2+U3) = 2

but dim(U1) + dim(U2) + dim(U3) - dim(U1∩U2) - dim(U1∩U3) - dim(U2∩U3) + dim(U1∩U2∩U3)

= 1 + 1 + 1 - 0 - 0 - 0 + 0 = 3
• Jan 6th 2013, 11:52 AM
johng
Re: Tricky question, subspaces dimensions
Presumably, this question arose from the true statement that $dim(U_1+U_2)=dim(U_1)+dim(U_2)-dim(U_1\cap U_2)$. When you try to extend this to 3 subspaces, you need to find $(U_1+U_2)\cap U_3$, which is definitely not equal to $(U_1\cap U_3)+(U_2\cap U_3)$. Basically, this is why Deveno's counter example above works. Notice though if $U_1\subset U_3$ or $U_2\subset U_3$, then $(U_1+U_2)\cap U_3=(U_1\cap U_3)+(U_2\cap U_3)$, and so in this case your original statement is true. (It boils down to the true statement about two subspaces.)
• Jan 6th 2013, 12:14 PM
Stormey
Re: Tricky question, subspaces dimensions
thanks man!

when i used Venn Diagram i always considered $dim(U_1+U_2+U_3)$ as being equal to $dim(V)$.
• Jan 6th 2013, 12:40 PM
Deveno
Re: Tricky question, subspaces dimensions
Quote:

Originally Posted by johng
Presumably, this question arose from the true statement that $dim(U_1+U_2)=dim(U_1)+dim(U_2)-dim(U_1\cap U_2)$. When you try to extend this to 3 subspaces, you need to find $(U_1+U_2)\cap U_3$, which is definitely not equal to $(U_1\cap U_3)+(U_2\cap U_3)$. Basically, this is why Deveno's counter example above works. Notice though if $U_1\subset U_3$ or $U_2\subset U_3$, then $(U_1+U_2)\cap U_3=(U_1\cap U_3)+(U_2\cap U_3)$, 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 $u_1 \in U_1, u_2 \in U_2$ and $u_1+u_2 \in U_3$, but that does not guarantee that $u_1,u_2 \in U_3$.