Direct sum of a subspace and its orthogonal complement

Question

Can someone give me an geometric and intuitive explanation for why this Theorem is right? Direct sum of a subspace and its orthogonal complement Suppose U is a finite-dimensional subspace of V. Then $V=U⊕U^⊥$

Thank You

Answer 1

This has a geometric picture in $\mathbb{R}^2$ and $\mathbb{R}^3$.

In $\mathbb{R}^2$, you can think of $U$ as a line through the origin, and $U^\perp$ as the line through the origin perpendicular to it. For example, $U$ could be the line spanned by $(1,1)$ and $U^\perp$ the line spanned by $(-1,1)$. To get to any vector in $\mathbb{R}^2$, we can first go along some vector in $U$, then add a vector in $U^\perp$, just like the parallelogram law for the sum of two vectors, and there is only one way to do this.

In $\mathbb{R}^3$, you can think of $U$ as a plane through the origin, and $U^\perp$ as the line through the origin that’s normal to $U$. For example, $U$ might be the $xy$-plane and $U^\perp$ the $z$-axis. To get to any point in $\mathbb{R}^3$, there is a unique way to write it as the sum of some vector in the $xy$-plane and some vector in the $z$-axis. We can first look at its “shadow” in the $xy$-plane, i.e. a point $(x,y,0)$, which corresponds to the $U$ component, and then add to this the “height” component $(0,0,z)$, which corresponds to the $U^\perp$ component.