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If a nonzero continuous real function on an interval is orthogonal to every polynomial of degree below $n$, it has at least $n$ distinct interior zeros. If it had only finitely many zeros and fewer than $n$ sign changes, multiply it by the product of linear factors at its sign-changing zeros. The product has constant sign and is nonzero on an open interval, so its integral cannot vanish. Orthogonality to this degree-below-$n$ polynomial contradicts that conclusion. Infinitely many interior zeros already satisfy the assertion. Orthogonal polynomials of respective degrees $0,\dots ,n-1$ form a basis of the needed polynomial space.

An orthonormal basis is a basis whose vectors are pairwise orthogonal and each have norm one.

Rotate an orthonormal basis counterclockwise by $\theta$ while keeping the vector $v$ fixed. The new basis vectors, written in the old basis, are the columns of $P=R_{\theta }$. The change-of-basis matrix convention therefore givesFor example, the fixed vector with old coordinate vector $(2,0{)}^T$ has new coordinate vector $(1,-\sqrt{3}{)}^T$ after the basis turns by $\pi /3$ radians. Its negative second coordinate means that the arrow lies below the new first axis. By contrast, an active planar rotation through that same angle moves the arrow to old coordinates $(1,\sqrt{3}{)}^T$. The sign difference follows from which object moves. The passive component formulas are $x^{\prime }=x\cos \theta +y\sin \theta$, $y^{\prime }=-x\sin \theta +y\cos \theta$.^[ref]

If coordinate vectors are instead written as rows, transposing the active equation gives $(v^{\prime }{)}^T=v^T{R}_{\theta }^T$. Switching to rows changes the side and the transpose; it does not by itself turn an active planar rotation into passive rotation of coordinates.