Every subspace has a partner made of all the vectors at right angles to it. In three dimensions a line and the plane perpendicular to it are partners, and every vector splits into a part in each. For a matrix, the solutions of the homogeneous system are exactly the partner of its row space.

A flagpole stands at right angles to the ground. Every step you take splits into how far you climbed along the pole and how far you moved across the ground, and neither part tells you anything about the other.

In the plane, the orthogonal complement of the line $W$ spanned by $\mathbf{w}=(2,1)$ is the line spanned by $\mathbf{n}=(-1,2)$. Every vector of one is perpendicular to every vector of the other, the two meet only at $\mathbf{0}$, and $\dim W+\dim W^{\perp}=2$.

The orthogonal complement of a subspace $W$ of $\mathbb{R}^n$ is the set $W^{\perp}$ of vectors orthogonal to every vector of $W$. It is a subspace, and to test a vector it is enough to check it against a spanning list of , because the dot product is linear. and meet only at , since a vector orthogonal to itself is zero.

• $\operatorname{Nul}A=(\operatorname{Row}A)^{\perp}$, the theorem below, and applied to $A^{T}$: .

Entry $i$ of $A\mathbf{x}$ is row $i$ of $A$ dotted with $\mathbf{x}$. So $A\mathbf{x}=\mathbf{0}$ says exactly that is orthogonal to every row, and so to every combination of rows. Rank–nullity then counts: is for , and every subspace is the row space of the matrix whose rows are a basis of it.

Entry $i$ of $A\mathbf{x}$ is row $i$ of $A$, $\mathbf{r}_i$, dotted with .

A vector orthogonal to each row is orthogonal to every combination of rows, because the dot product is linear.

The two sets are the same set.

Rank–nullity; with $W=\operatorname{Row}A$ this reads $\dim W+\dim W^{\perp}=n$.

Put a spanning list of $W$ in as the rows of a matrix. $W^{\perp}$ is its null space: the vectors whose dot product with every row is $0$.

• $W^{\perp}$: all vectors orthogonal to $W$, found as the null space of a matrix with a spanning list of $W$ as rows.
• $\dim W+\dim W^{\perp}=n$, and .

Great work! You now know how to:

• find an orthogonal complement as a null space;
• use $\dim W+\dim W^{\perp}=n$ and $(W^{\perp})^{\perp}=W$;