9.5: The Gram-Schmidt Orthogonalization procedure

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

• Isaiah Lankham, Mathematics Department at UC Davis
• Bruno Nachtergaele, Mathematics Department at UC Davis
• Anne Schilling, Mathematics Department at UC Davis

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