On the dimension of the algebraic sum of subspaces

Theorem 1.1

Let U be a finite-dimensional vector space over an arbitrary field F . Let U 1 , … , U n be linear subspaces of U. Then,

where

for i = 1 , … , n .

Lemma 2.1

Let U be a finite-dimensional vector space over an arbitrary field  F and let U 1 , … , U n be the subspaces of U. The following conditions are equivalent:

1. the algebraic sum U 1 + … + U n is the direct sum U 1 ⊕ … ⊕ U n ;

2. for all 1 ≤ i ≤ n , there is U i ∩ U i * = 0 .

Proof

The implication from (a) to (b) is straightforward. For the opposite direction, let us assume (b) and show that any vector u ∈ U 1 + … + U n has a unique presentation as

Suppose that we have two such presentations

Then, for any 1 ≤ i ≤ n , we have

By (b), this implies u i = u i ′ for all i = 1 , … , n , and we are done.□

Proof of Theorem 1.1

Case 1. Assume that

We claim that under this assumption, all algebraic sums appearing in formula (1) are direct sums. Of course, it suffices to show that it is the case for the sum of all involved subspaces, so we claim that

We want to use Lemma 2.1, so let

for some 1 ≤ i ≤ n . Let j be an index different from i . Then, u i ∈ U j * because U i ⊂ U j * for all j ≠ i . Since also u i was taken as an element of U i * , we conclude that

It follows that the right-hand side of (1) is

and we are done in this case.

Case 2. Let W ≔ U 1 * ∩ … ∩ U n * be now arbitrary. The idea is to reduce the situation to Case 1, taking quotients of all involved spaces by W . In order to control the dimensions, we show that

The containment W ⊂ U i * follows by the definition of W , and of course, every U i * is contained in U 1 + … + U n .

Let π : U → U ⁄ W be the quotient map. Then,

1. dim ( π ( U 1 + … + U n ) ) = dim ( U 1 + … + U n ) − dim ( W ) and;

2. dim ( π ( U i * ) ) = dim ( U i * ) − dim ( W ) .

Moreover, formula (1) holds for subspaces π ( U 1 ) , … , π ( U n ) in U ⁄ W because by construction,

Indeed, let y ∈ π ( U 1 * ) ∩ … ∩ π ( U n * ) . So there are elements x i ∈ U i * with π ( x i ) = y for i = 1 , … , n . Hence, x i − x j ∈ W for all i , j ∈ { 1 , … , n } . But then x i = ( x i − x j ) + x j ∈ U j * for all 1 ≤ i , j ≤ n , or equivalently, x i ∈ U 1 * ∩ … ∩ U n * = W . Thus, y = π ( x i ) = 0 in U ⁄ W .

Using (a) and (b) as mentioned earlier, it is now elementary to check that formula (1) holds for the spaces before taking the quotient.□

Corollary 2.2

Let U be a finite-dimensional vector space over an arbitrary field F . Let U 1 , … , U n be linear subspaces of U. Then,

Proof

The idea is to use formula (2) for spaces U i and U i * . We have

Since U i + U i * = ∑ i = 1 n U i , adding equation (5.i) for i = 1 , … , n , we obtain

Now, subtracting (1) on both sides does the job.□