Linearly independent vs. orthogonal
A list of vectors can point in “different directions” in two senses. A set of linearly independent vectors have no redundant vectors: no vector in the list is a linear combination of the others. Orthogonal vectors are pairwise at right angles: every two different vectors in the list have inner product $0$. Linear independence is defined in every vector space, but orthogonality requires an inner product.
orthogonal and nonzero ⊂ linearly independent
Definitions
Missing definition: linearly_independent
Two vectors in the plane or in space are orthogonal if the vectors meet at a right angle, which happens exactly when their dot product is $0$.
Consider an inner product space $V$. Two vectors $u, v \in V$ are orthogonal if
\[\ip{u}{v} = 0.\]A list of vectors is orthogonal if every two different vectors in the list are orthogonal.
Examples
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The lines $y = x$ and $y = -x$ in the plane, with direction vectors $(1, 1)$ and $(1, -1)$
The direction vectors $(1, 1)$ and $(1, -1)$ of the lines $y = x$ and $y = -x$ are orthogonal, because their dot product is $1 - 1 = 0$.
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The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis in $\reals^3$
The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis have orthogonal direction vectors $(1, 0, 0)$ and $(0, 1, 0)$, so the lines are orthogonal.
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The vectors $(1, 0)$ and $(0, 1)$ in $\reals^2$
The vectors $(1, 0)$ and $(0, 1)$ are orthogonal, because their dot product is $1 \cdot 0 + 0 \cdot 1 = 0$.
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The vector $(1, 1)$ and the line $x + y = 1$ in the plane, whose direction vector is $(1, -1)$
The vector $(1, 1)$ is orthogonal to the direction vector $(1, -1)$ of the line $x + y = 1$, because their dot product is $1 - 1 = 0$.
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The functions $\sin$ and $\cos$ on $[-\pi, \pi]$, with the $L^2$ inner product $\ip{f}{g} = \int_{-\pi}^{\pi} f(x) g(x) \, dx$
The functions $\sin$ and $\cos$ are orthogonal in the $L^2$ inner product on $[-\pi, \pi]$, because $\int_{-\pi}^{\pi} \sin(x) \cos(x) \, dx = 0$, since the integrand is odd.
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The vectors $(1, 0)$ and $(0, 0)$ in $\reals^2$
The vectors $(1, 0)$ and $(0, 0)$ are orthogonal, because their dot product is $1 \cdot 0 + 0 \cdot 0 = 0$.
Non-examples
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The vectors $(1, 0)$ and $(2, 0)$ in $\reals^2$
The vectors $(1, 0)$ and $(2, 0)$ are not orthogonal, because their dot product is $1 \cdot 2 + 0 \cdot 0 = 2$.
Properties
- The zero vector is orthogonal to every vector.
- An orthogonal list of nonzero vectors is linearly independent.
- An orthogonal list of unit vectors is called orthonormal.
Examples
| Example | linearly_independent | orthogonal | Why |
|---|---|---|---|
| The vectors $(1, 0)$ and $(0, 1)$ in $\reals^2$ | ✓ | ✓ | |
Missing example: skew_basis_r2 | |||
| The vectors $(1, 0)$ and $(0, 0)$ in $\reals^2$ | ✗ | ✓ | The vectors $(1, 0)$ and $(0, 0)$ show why orthogonal vectors must be nonzero to be linearly independent: the zero vector is orthogonal to every vector, but every list containing the zero vector is linearly dependent. |
| The vectors $(1, 0)$ and $(2, 0)$ in $\reals^2$ | ✗ | ✗ | |
Notes
- The Gram–Schmidt process turns every linearly independent list into an orthogonal list with the same span.