9.5: The Gram-Schmidt Orthogonalization procedure

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Page ID: 260

$Schilling.jpg$
Contributed by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling
Professor (Mathematics) at University of California, Davis

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We now come to a fundamentally important algorithm, which is called the Gram-Schmidt orthogonalization procedure. This algorithm makes it possible to construct, for each list of linearly independent vectors (resp. basis), a corresponding orthonormal list (resp. orthonormal basis).

Theorem 9.5.1

If $(v_1,\ldots,v_m) $ is a list of linearly independent vectors in $V$, then there exists an orthonormal list $(e_1,\ldots,e_m) $ such that

\[ \Span(v_1,\ldots,v_k) = \Span(e_1,\ldots,e_k), \quad \text{for all $k=1,\ldots,m$.} \label{9.5.1} \]

Proof

The proof is constructive, that is, we will actually construct vectors $e_1,\ldots,e_m $ having the desired properties. Since $(v_1,\ldots,v_m) $ is linearly independent, $v_k\neq 0 $ for each $k=1,2,\ldots,m$. Set $e_1=\frac{v_1}{\norm{v_1}}$. Then $e_{1} $ is a vector of norm 1 and satisfies Equation (9.5.1) for $k=1$. Next, set

\[\begin{equation*}
e_2 = \frac{v_2 - \inner{v_2}{e_1} e_1}{\norm{v_2 - \inner{v_2}{e_1} e_1}}.
\end{equation*} \]

This is, in fact, the normalized version of the orthogonal decomposition Equation(9.3.1)~\eqref{eq:orthogonal decomp}. I.e.,

\[\begin{equation*}
w = v_2 - \inner{v_2}{e_1}e_1,
\end{equation*} \]

where $w\bot e_1$. Note that $\norm{e_2}=1 $ and $\Span(e_1,e_2)=\Span(v_1,v_2)$.

Now, suppose that $e_1,\ldots,e_{k-1} $ have been constructed such that $(e_1,\ldots,e_{k-1})$ is an orthonormal list and $\Span(v_1,\ldots,v_{k-1}) = \Span(e_1,\ldots,e_{k-1})$. Then define
\[\begin{equation*}
e_k = \frac{v_k - \inner{v_k}{e_1}e_1 - \inner{v_k}{e_2} e_2 - \cdots -\inner{v_k}{e_{k-1}} e_{k-1}}
{\norm{v_k - \inner{v_k}{e_1}e_1 - \inner{v_k}{e_2} e_2 - \cdots -\inner{v_k}{e_{k-1}} e_{k-1}}}.
\end{equation*} \]

Since $(v_1,\ldots,v_k) $ is linearly independent, we know that $v_k\not\in \Span(v_1,\ldots,v_{k-1})$. Hence, we also know that $v_k\not\in \Span(e_1,\ldots,e_{k-1})$. It follows that the norm in the definition of $e_k $ is not zero, and so $e_k $ is well-defined (i.e., we are not dividing by zero). Note that a vector divided by its norm has norm 1 so that $\norm{e_k}=1$. Furthermore,

\[\begin{equation*}
\begin{split}
\inner{e_k}{e_i} &= \left\langle \frac{v_k - \inner{v_k}{e_1}e_1 - \inner{v_k}{e_2} e_2 - \cdots
-\inner{v_k}{e_{k-1}} e_{k-1}}
{\norm{v_k - \inner{v_k}{e_1}e_1 - \inner{v_k}{e_2} e_2 - \cdots -\inner{v_k}{e_{k-1}} e_{k-1}}},
e_i \right\rangle \\
&= \frac{\inner{v_k}{e_i} - \inner{v_k}{e_i}}
{\norm{v_k - \inner{v_k}{e_1}e_1 - \inner{v_k}{e_2} e_2 - \cdots -\inner{v_k}{e_{k-1}} e_{k-1}}}=0,
\end{split}
\end{equation*} \]

for each $1\le i<k$. Hence, $(e_1,\ldots,e_k) $ is orthonormal.

$\square$

From the definition of $e_k$, we see that $v_k\in \Span(e_1,\ldots,e_k) $ so that $\Span(v_1,\ldots,v_k) \subset \Span(e_1,\ldots,e_k)$. Since both lists $(e_1,\ldots,e_k) $ and $(v_1,\ldots,v_k) $ are linearly independent, they must span subspaces of the same dimension and therefore are the same subspace. Hence Equation (9.5.1) holds.

Example $\PageIndex{2}$

Take $v_1=(1,1,0) $ and $v_2=(2,1,1) $ in $\mathbb{R}^3$. The list $(v_1,v_2) $ is linearly independent (as you should verify!). To illustrate the Gram-Schmidt procedure, we begin by setting
\[\begin{equation*}
e_1 = \frac{v_1}{\norm{v_1}} = \frac{1}{\sqrt{2}} (1,1,0).
\end{equation*} \]
Next, set
\[\begin{equation*}
e_2 = \frac{v_2 - \inner{v_2}{e_1}e_1}{\norm{v_2 - \inner{v_2}{e_1}e_1}}.
\end{equation*} \]
The inner product $\inner{v_2}{e_1}=\frac{1}{\sqrt{2}}\inner{(1,1,0)}{(2,1,1)}=\frac{3}{\sqrt{2}}$,
so
\[\begin{equation*}
u_2 = v_2 - \inner{v_2}{e_1}e_1 = (2,1,1) - \frac{3}{2}(1,1,0) = \frac{1}{2}(1,-1,2).
\end{equation*} \]
Calculating the norm of $u_2$, we obtain $\norm{u_2}=\sqrt{\frac{1}{4}(1+1+4)} = \frac{\sqrt{6}}{2}$.
Hence, normalizing this vector, we obtain
\[\begin{equation*}
e_2 = \frac{u_2}{\norm{u_2}} = \frac{1}{\sqrt{6}}(1,-1,2).
\end{equation*} \]
The list $(e_1,e_2) $ is therefore orthonormal and has the same span as $(v_1,v_2)$.

Corollary 9.5.3.

Every finite-dimensional inner product space has an orthonormal basis.

Proof

Let $(v_1,\ldots,v_n) $ be any basis for $V$. This list is linearly independent and spans $V$. Apply the Gram-Schmidt procedure to this list to obtain an orthonormal list $(e_1,\ldots,e_n)$, which still spans $V $ by construction. By Proposition9.4.2~\ref{prop:orth li}, this list is linearly independent and hence a basis of $V$.

Corollary 9.5.4.

Every orthonormal list of vectors in $V $ can be extended to an orthonormal basis of $V$.

Proof

Let $(e_1,\ldots,e_m) $ be an orthonormal list of vectors in $V$. By Proposition9.4.2~\ref{prop:orth li}, this list is linearly independent and hence can be extended to a basis $(e_1,\ldots,e_m,v_1,\ldots,v_k) $ of $V $ by the Basis Extension Theorem. Now apply the Gram-Schmidt procedure to obtain a new orthonormal basis $(e_1,\ldots,e_m,f_1,\ldots,f_k)$. The first $m $ vectors do not change since they already are orthonormal. The list still spans $V $ and is linearly independent by Proposition9.4.2~\ref{prop:orth li} and therefore forms a basis.

Recall Theorem7.5.3~\ref{thm:ComplexLinearMapsUpperTriangularWrtSomeBasis}: given an operator $T \in \mathcal{L}(V, V) $ on a complex vector space $V$, there exists a basis $B $ for $V$ such that the matrix $M(T)$ of $T$ with respect to $B $ is upper triangular. We would like to extend this result to require the additional property of orthonormality.

Corollary 9.5.5

Let $V $ be an inner product space over $\mathbb{F} $ and $T\in\mathcal{L}(V,V)$. If $T $ is upper-triangular with respect to some basis, then $T $ is upper-triangular with respect to some orthonormal basis.

Proof

Let $(v_1,\ldots,v_n) $ be a basis of $V $ with respect to which $T $ is upper-triangular. Apply the Gram-Schmidt procedure to obtain an orthonormal basis $(e_1,\ldots,e_n)$, and note that

\[ \Span(e_1,\ldots,e_k) = \Span(v_1,\ldots,v_k), \quad \text{for all $1\le k\le n$.} \]

We proved before that $T $ is upper-triangular with respect to a basis $(v_1,\ldots,v_n) $ if and only if $\Span(v_1,\ldots,v_k) $ is invariant under $T $ for each $1\le k\le n$. Since these spans are unchanged by the Gram-Schmidt procedure, $T $ is still upper triangular for the corresponding orthonormal basis.

Contributors

Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.