Thread: Finding dimension of sum of subspaces

U={(x,y,z,w):x-y+2w=0}

W={(a,b,a,0)}

are subspaces of R^4. Determine whether the sum U+W is direct and find dim(U+W).

I find it is not direct since, for example, (1,1,1,0) is a member of both.

Now for the second part I use theorem Dim(U+W)=dim(U)+dim(W)-dim('union'))

I know dim(U) and dim(W) as I earlier calculated basis for them but how do I find the dimension of the union?

Originally Posted by boromir

U={(x,y,z,w):x-y+2w=0}

W={(a,b,a,0)}

are subspaces of R^4. Determine whether the sum U+W is direct and find dim(U+W).

I find it is not direct since, for example, (1,1,1,0) is a member of both.

Now for the second part I use theorem Dim(U+W)=dim(U)+dim(W)-dim('union'))

I know dim(U) and dim(W) as I earlier calculated basis for them but how do I find the dimension of the union?

A basis for U is

and





and a Basis for W is





So the unions is these four basis vectors. Now just removed any dependant Vectors!

So you are saying the union of the bases immediately form a spanning set and then use the theorem that says it contains a basis? Well, doing simultaneous equations I get that {U1,U2,W1,W2} is a basis, so it is in fact equal to the whole space.

Originally Posted by boromir

Now for the second part I use theorem Dim(U+W)=dim(U)+dim(W)-dim('union'))

You got that slightly wrong. It should be Dim(U+W)=dim(U)+dim(W)-dim('intersection')). Now you know that the intersection contains (1,1,1,0), so its dimension is at least 1. If the dimension of the intersection is 2 then it would have to be the whole of W (since W is fairly obviously 2-dimensional).

From that, you should be able to pin down the dimension of , and then the theorem will tell you the dimension of

so how is related to dimensions of U and W? Does it have to be at most min{dim(V),dim(W)}?