Recall that the sum of spans of single vectors being a direct sum is equivalent to the linear independence. The following is an extension of the observation, which complements this result about spanning vectors in a sum.

Let` H`_{1} + `H`_{2} + ... + ` H_{k}` be a direct sum of subspaces of

Proof For the case *k* = 3 and `B`_{1} = {**v**_{11}, **v**_{12}}, `B`_{2} = {**v**_{21}, **v**_{22}}, `B`_{3} = {**v**_{31}, **v**_{32}, **v**_{33}} are linearly independent vectors in **H**_{1}, **H**_{2}, **H**_{3}, we will prove that `B` = {**v**_{11}, **v**_{12}, **v**_{21}, **v**_{22}, **v**_{31}, **v**_{32}, **v**_{33}} is linearly independent. The general case is similar.

Let a linear combination of vectors of `B` be trivial:

(`c`_{11}`v`_{11} + `c`_{12}`v`_{12}) + (`c`_{21}`v`_{21} + `c`_{22}`v`_{22}) + (`c`_{31}`v`_{31} + `c`_{32}`v`_{32} + `c`_{33}`v`_{33}) = `v`_{1} + `v`_{2} + `v`_{3} = **0**.

We need to conclude that all the seven coefficients are zero. Since

`v`_{1} = `c`_{11}`v`_{11} + `c`_{12}`v`_{12} ∈ **H**_{1}, `v`_{2} = `c`_{21}`v`_{21} + `c`_{22}`v`_{22} ∈ **H**_{2}, `v`_{3} = `c`_{31}`v`_{31} + `c`_{32}`v`_{32} + `c`_{33}`v`_{33} ∈ **H**_{3},

the fact that **H**_{1} + **H**_{2} + **H**_{3} is a direct sum implies that `v`_{1} = `v`_{2} = `v`_{3} = **0**. Then by `c`_{11}`v`_{11} + `c`_{12}`v`_{12} = `v`_{1} = **0** and `B`_{1} = {**v**_{11}, **v**_{12}} being linearly independent, we have `c`_{11} = `c`_{12} = 0. Similarly, we have `c`_{21} = `c`_{22} = 0 and `c`_{31} = `c`_{32} = `c`_{33} = 0.

In the special case `B _{i}` is a basis of

`H`_{1} + `H`_{2} + ... + ` H_{k}` is a direct sum ⇔

We remark that by this exercise, we have `dim`(`H`_{1} + `H`_{2} + ... + ` H_{k}`) ≤

Proof For the ⇐ direction, we again take a basis `B _{i}` of

Now for the special case *k* = 3 and `B`_{1} = {**v**_{11}, **v**_{12}}, `B`_{2} = {**v**_{21}, **v**_{22}}, `B`_{3} = {**v**_{31}, **v**_{32}, **v**_{33}}, the linear independence of `B` means that

(`c`_{11}`v`_{11} + `c`_{12}`v`_{12}) + (`c`_{21}`v`_{21} + `c`_{22}`v`_{22}) + (`c`_{31}`v`_{31} + `c`_{32}`v`_{32} + `c`_{33}`v`_{33}) = **0**

implies all seven coefficients are zero. Equivalently, for the arbitrary (becase `B`_{1}, `B`_{2}, `B`_{3} are bases) vectors

`v`_{1} = `c`_{11}`v`_{11} + `c`_{12}`v`_{12} ∈ **H**_{1}, `v`_{2} = `c`_{21}`v`_{21} + `c`_{22}`v`_{22} ∈ **H**_{2}, `v`_{3} = `c`_{31}`v`_{31} + `c`_{32}`v`_{32} + `c`_{33}`v`_{33} ∈ **H**_{3},

`v`_{1} + `v`_{2} + `v`_{3} = **0** implies `v`_{1} = `v`_{2} = `v`_{3} = **0**. This means exactly that `H`_{1} + `H`_{2} + `H`_{3} is a direct sum.