Intersection Condition for Direct Sum of Subspaces

This article needs to be tidied. In particular: Please study house rules, and other pages to get a feel for how pages are structured. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code.

This article needs to be linked to other articles. In particular: throughout You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.

Theorem

Let $U$ and $W$ be subspaces of a vector space $V$.

Then $U + W$ is a direct sum if and only if $U \cap W = 0$.

Proof

We must first prove that if $U + W$ is a direct sum, then:

$U \cap W = 0$

Let $U + W$ be a direct sum.

Let $\mathbf v \in U \cap W$ be an arbitrary vector in $U \cap W$.

Then:

$\mathbf 0 = \mathbf v + \paren {-\mathbf v}$

where:

$\mathbf v \in U$

$-\mathbf v \in W$

This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Also see

• Two-Step Vector Subspace Test
• Null Space is Subspace

Sources

• 2015: Sheldon Axler: Linear Algebra Done Right (3rd ed.)