# Linear Subspace is Convex Set

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## Theorem

Let $V$ be a vector space over $\R$ or $\C$, and let $L$ be a linear subspace of $V$.

Then $L$ is a convex set.

## Proof

Let $x, y \in L, t \in \left[{0 .. 1}\right]$ be arbitrary.

Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that

- $tx + \left({1-t}\right)y \in L$

Hence $L$ is convex.

$\blacksquare$

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*: $\S I.2$