Orthonormal Bases and the Gram-Schmidt Process

The prior subsection suggests that projecting onto the line spanned by ${\vec {s}}$ decomposes a vector ${\vec {v}}$ into two parts

$\displaystyle {\vec {v}}={\mbox{proj}}_{[{\vec {s}}\,]}({\vec {v}})\,+\,\left({\vec {v}}-{\mbox{proj}}_{[{\vec {s}}\,]}({\vec {v}})\right)$

that are orthogonal and so are "not interacting". We will now develop that suggestion.

Definition 2.1

Vectors ${\vec {v}}_{1},\dots ,{\vec {v}}_{k}\in \mathbb {R} ^{n}$ are mutually orthogonal when any two are orthogonal: if $i\neq j$ then the dot product ${\vec {v}}_{i}\cdot {\vec {v}}_{j}$ is zero.

Theorem 2.2

If the vectors in a set $\{{\vec {v}}_{1},\dots ,{\vec {v}}_{k}\}\subset \mathbb {R} ^{n}$ are mutually orthogonal and nonzero then that set is linearly independent.

Proof: Consider a linear relationship $c_{1}{\vec {v}}_{1}+c_{2}{\vec {v}}_{2}+\dots +c_{k}{\vec {v}}_{k}={\vec {0}}$. If $i\in [1..k]$ then taking the dot product of ${\vec {v}}_{i}$ with both sides of the equation

${\begin{array}{rl}{\vec {v}}_{i}\cdot (c_{1}{\vec {v}}_{1}+c_{2}{\vec {v}}_{2}+\dots +c_{k}{\vec {v}}_{k})&={\vec {v}}_{i}\cdot {\vec {0}}\\c_{i}\cdot ({\vec {v}}_{i}\cdot {\vec {v}}_{i})&=0\end{array}}$

shows, since ${\vec {v}}_{i}$ is nonzero, that $c_{i}$ is zero.

Corollary 2.3

If the vectors in a size $k$ subset of a $k$ dimensional space are mutually orthogonal and nonzero then that set is a basis for the space.

Proof: Any linearly independent size $k$ subset of a $k$ dimensional space is a basis.

Of course, the converse of Corollary 2.3 does not hold— not every basis of every subspace of $\mathbb {R} ^{n}$ is made of mutually orthogonal vectors. However, we can get the partial converse that for every subspace of $\mathbb {R} ^{n}$ there is at least one basis consisting of mutually orthogonal vectors.

Example 2.4: The members ${\vec {\beta }}_{1}$ and ${\vec {\beta }}_{2}$ of this basis for $\mathbb {R} ^{2}$ are not orthogonal.

$\displaystyle B=\left\langle {\begin{pmatrix}4\\2\end{pmatrix}},{\begin{pmatrix}1\\3\end{pmatrix}}\right\rangle$

However, we can derive from $B$ a new basis for the same space that does have mutually orthogonal members. For the first member of the new basis we simply use ${\vec {\beta }}_{1}$.

${\vec {\kappa }}_{1}={\begin{pmatrix}4\\2\end{pmatrix}}$

For the second member of the new basis, we take away from ${\vec {\beta }}_{2}$ its part in the direction of ${\vec {\kappa }}_{1}$,

$\displaystyle {\vec {\kappa }}_{2}={\begin{pmatrix}1\\3\end{pmatrix}}-{\mbox{proj}}_{\scriptstyle [{\vec {\kappa }}_{1}]}({\begin{pmatrix}1\\3\end{pmatrix}})={\begin{pmatrix}1\\3\end{pmatrix}}-{\begin{pmatrix}2\\1\end{pmatrix}}={\begin{pmatrix}-1\\2\end{pmatrix}}$

which leaves the part, ${\vec {\kappa }}_{2}$ pictured above, of ${\vec {\beta }}_{2}$ that is orthogonal to ${\vec {\kappa }}_{1}$ (it is orthogonal by the definition of the projection onto the span of ${\vec {\kappa }}_{1}$). Note that, by the corollary, $\{{\vec {\kappa }}_{1},{\vec {\kappa }}_{2}\}$ is a basis for $\mathbb {R} ^{2}$.

Definition 2.5:

An orthogonal basis for a vector space is a basis of mutually orthogonal vectors.

Example 2.6: To turn this basis for $\mathbb {R} ^{3}$

$\left\langle {\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}0\\2\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\3\end{pmatrix}}\right\rangle$

into an orthogonal basis, we take the first vector as it is given.

${\vec {\kappa }}_{1}={\begin{pmatrix}1\\1\\1\end{pmatrix}}$

We get ${\vec {\kappa }}_{2}$ by starting with the given second vector ${\vec {\beta }}_{2}$ and subtracting away the part of it in the direction of ${\vec {\kappa }}_{1}$.

${\vec {\kappa }}_{2}={\begin{pmatrix}0\\2\\0\end{pmatrix}}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({\begin{pmatrix}0\\2\\0\end{pmatrix}})={\begin{pmatrix}0\\2\\0\end{pmatrix}}-{\begin{pmatrix}2/3\\2/3\\2/3\end{pmatrix}}={\begin{pmatrix}-2/3\\4/3\\-2/3\end{pmatrix}}$

Finally, we get ${\vec {\kappa }}_{3}$ by taking the third given vector and subtracting the part of it in the direction of ${\vec {\kappa }}_{1}$, and also the part of it in the direction of ${\vec {\kappa }}_{2}$.

${\vec {\kappa }}_{3}={\begin{pmatrix}1\\0\\3\end{pmatrix}}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({\begin{pmatrix}1\\0\\3\end{pmatrix}})-{\mbox{proj}}_{[{\vec {\kappa }}_{2}]}({\begin{pmatrix}1\\0\\3\end{pmatrix}})={\begin{pmatrix}-1\\0\\1\end{pmatrix}}$

Again the corollary gives that

$\left\langle {\begin{pmatrix}1\\1\\1\end{pmatrix}},{\begin{pmatrix}-2/3\\4/3\\-2/3\end{pmatrix}},{\begin{pmatrix}-1\\0\\1\end{pmatrix}}\right\rangle$

is a basis for the space.

The next result verifies that the process used in those examples works with any basis for any subspace of an $\mathbb {R} ^{n}$ (we are restricted to $\mathbb {R} ^{n}$ only because we have not given a definition of orthogonality for other vector spaces).

Theorem 2.7 (Gram-Schmidt orthogonalization):

If $\left\langle {\vec {\beta }}_{1},\ldots {\vec {\beta }}_{k}\right\rangle$

is a basis for a subspace of $\mathbb {R} ^{n}$ then, where

${\begin{array}{rl}{\vec {\kappa }}_{1}&={\vec {\beta }}_{1}\\{\vec {\kappa }}_{2}&={\vec {\beta }}_{2}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {\beta }}_{2}})\\{\vec {\kappa }}_{3}&={\vec {\beta }}_{3}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {\beta }}_{3}})-{\mbox{proj}}_{[{\vec {\kappa }}_{2}]}({{\vec {\beta }}_{3}})\\&\vdots \\{\vec {\kappa }}_{k}&={\vec {\beta }}_{k}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {\beta }}_{k}})-\cdots -{\mbox{proj}}_{[{\vec {\kappa }}_{k-1}]}({{\vec {\beta }}_{k}})\end{array}}$

the ${\vec {\kappa }}\,$'s form an orthogonal basis for the same subspace.

Proof: We will use induction to check that each ${\vec {\kappa }}_{i}$ is nonzero, is in the span of $\left\langle {\vec {\beta }}_{1},\ldots {\vec {\beta }}_{i}\right\rangle$ and is orthogonal to all preceding vectors: ${\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{i}=\cdots ={\vec {\kappa }}_{i-1}\cdot {\vec {\kappa }}_{i}=0$. With those, and with Corollary 2.3, we will have that $\left\langle {\vec {\kappa }}_{1},\ldots {\vec {\kappa }}_{k}\right\rangle$ is a basis for the same space as $\left\langle {\vec {\beta }}_{1},\ldots {\vec {\beta }}_{k}\right\rangle$.

We shall cover the cases up to $i=3$, which give the sense of the argument. Completing the details is Problem 15.

The $i=1$ case is trivial— setting ${\vec {\kappa }}_{1}$ equal to ${\vec {\beta }}_{1}$ makes it a nonzero vector since ${\vec {\beta }}_{1}$ is a member of a basis, it is obviously in the desired span, and the "orthogonal to all preceding vectors" condition is vacuously met.

For the $i=2$ case, expand the definition of ${\vec {\kappa }}_{2}$.

${\vec {\kappa }}_{2}={\vec {\beta }}_{2}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {\beta }}_{2}})={\vec {\beta }}_{2}-{\frac {{\vec {\beta }}_{2}\cdot {\vec {\kappa }}_{1}}{{\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{1}}}\cdot {\vec {\kappa }}_{1}={\vec {\beta }}_{2}-{\frac {{\vec {\beta }}_{2}\cdot {\vec {\kappa }}_{1}}{{\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{1}}}\cdot {\vec {\beta }}_{1}$

This expansion shows that ${\vec {\kappa }}_{2}$ is nonzero or else this would be a non-trivial linear dependence among the ${\vec {\beta }}$'s (it is nontrivial because the coefficient of ${\vec {\beta }}_{2}$ is $1$) and also shows that ${\vec {\kappa }}_{2}$ is in the desired span. Finally, ${\vec {\kappa }}_{2}$ is orthogonal to the only preceding vector

${\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{2}={\vec {\kappa }}_{1}\cdot ({\vec {\beta }}_{2}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {\beta }}_{2}}))=0$

because this projection is orthogonal.

The $i=3$ case is the same as the $i=2$ case except for one detail. As in the $i=2$ case, expanding the definition

${\begin{array}{rl}{\vec {\kappa }}_{3}&={\vec {\beta }}_{3}-{\frac {{\vec {\beta }}_{3}\cdot {\vec {\kappa }}_{1}}{{\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{1}}}\cdot {\vec {\kappa }}_{1}-{\frac {{\vec {\beta }}_{3}\cdot {\vec {\kappa }}_{2}}{{\vec {\kappa }}_{2}\cdot {\vec {\kappa }}_{2}}}\cdot {\vec {\kappa }}_{2}\\&={\vec {\beta }}_{3}-{\frac {{\vec {\beta }}_{3}\cdot {\vec {\kappa }}_{1}}{{\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{1}}}\cdot {\vec {\beta }}_{1}-{\frac {{\vec {\beta }}_{3}\cdot {\vec {\kappa }}_{2}}{{\vec {\kappa }}_{2}\cdot {\vec {\kappa }}_{2}}}\cdot {\bigl (}{\vec {\beta }}_{2}-{\frac {{\vec {\beta }}_{2}\cdot {\vec {\kappa }}_{1}}{{\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{1}}}\cdot {\vec {\beta }}_{1}{\bigr )}\end{array}}$

shows that ${\vec {\kappa }}_{3}$ is nonzero and is in the span. A calculation shows that ${\vec {\kappa }}_{3}$ is orthogonal to the preceding vector ${\vec {\kappa }}_{1}$.

${\begin{array}{rl}{\vec {\kappa }}_{1}\cdot {\vec {\kappa }}_{3}&={\vec {\kappa }}_{1}\cdot {\bigl (}{\vec {\beta }}_{3}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {\beta }}_{3}})-{\mbox{proj}}_{[{\vec {\kappa }}_{2}]}({{\vec {\beta }}_{3}}){\bigr )}\\&={\vec {\kappa }}_{1}\cdot {\bigl (}{\vec {\beta }}_{3}-{\mbox{proj}}_{[{\vec {\kappa }}_{1}]}({{\vec {\beta }}_{3}}){\bigr )}-{\vec {\kappa }}_{1}\cdot {\mbox{proj}}_{[{\vec {\kappa }}_{2}]}({{\vec {\beta }}_{3}})\\&=0\end{array}}$

(Here's the difference from the $i=2$ case— the second line has two kinds of terms. The first term is zero because this projection is orthogonal, as in the $i=2$ case. The second term is zero because ${\vec {\kappa }}_{1}$ is orthogonal to ${\vec {\kappa }}_{2}$ and so is orthogonal to any vector in the line spanned by ${\vec {\kappa }}_{2}$.) The check that ${\vec {\kappa }}_{3}$ is also orthogonal to the other preceding vector ${\vec {\kappa }}_{2}$ is similar.

Beyond having the vectors in the basis be orthogonal, we can do more; we can arrange for each vector to have length one by dividing each by its own length (we can normalize the lengths).

Example 2.8: Normalizing the length of each vector in the orthogonal basis of Example 2.6 produces this orthonormal basis.

$\left\langle {\begin{pmatrix}1/{\sqrt {3}}\\1/{\sqrt {3}}\\1/{\sqrt {3}}\end{pmatrix}},{\begin{pmatrix}-1/{\sqrt {6}}\\2/{\sqrt {6}}\\-1/{\sqrt {6}}\end{pmatrix}},{\begin{pmatrix}-1/{\sqrt {2}}\\0\\1/{\sqrt {2}}\end{pmatrix}}\right\rangle$

Besides its intuitive appeal, and its analogy with the standard basis ${\mathcal {E}}_{n}$ for $\mathbb {R} ^{n}$, an orthonormal basis also simplifies some computations.

Licensing

Content obtained and/or adapted from:

• Gram-Schmidt Orthogonalization, Wikibooks: Linear Algebra under a CC BY-SA license