# Thread: Direct Sum of Dimensions

1. ## Direct Sum of Dimensions

I have a question here on direct sums which states:

Let U and V be subspaces of R5 such that U + V = R5, dimU = 2, and dimV = 3. Show that R5 = U is a direct sum of V.

This is how I approached it:

R5 = U is a direct sum of V
dim(U+V)= 5 = dimR5

If U is contained in V, dimU = dimV => U=V

Therefore: U+V is in R5
R5 = U is a direct sum of V.

Does that make any sense? Could someone please review this and tell me if I am correct, or atleast steer me on the right path.

2. $\dim(U+V)=\dim(U)+\dim(V)-\dim(U\cap V),$ that leads $\dim(U\cap V)=0,$ so $U\oplus V=\mathbb R^5.$

3. Originally Posted by GreenDay14
I have a question here on direct sums which states:

Let U and V be subspaces of R5 such that U + V = R5, dimU = 2, and dimV = 3. Show that R5 = U is a direct sum of V.

This is how I approached it:

R5 = U is a direct sum of V
dim(U+V)= 5 = dimR5

If U is contained in V, dimU = dimV => U=V

Therefore: U+V is in R5
R5 = U is a direct sum of V.

Does that make any sense? Could someone please review this and tell me if I am correct, or atleast steer me on the right path.
I'm not really sure what you tried to do there, but you only need to show that $U \cap V= \{ 0 \}$