Examples

Gram-Schmidt illustration for 3 vectors

Gram-Schmidt illustration for 3 vectors

Problem: Use the Gram-Schmidt process to find an orthogonal basis for the following span:

$${\mathbf{v}}_1=\left(\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right),{\mathbf{v}}_2=\left(\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right),{\mathbf{v}}_3=\left(\begin{array}{c}1\\ 0\\ 1\\ 1\end{array}\right).$$

Solution: We first define ${\mathbf{w}}_1={\mathbf{v}}_1$. Then compute ${\mathrm{proj}}_{{\mathbf{w}}_1}({\mathbf{v}}_2)$:

$${\mathrm{proj}}_{{\mathbf{w}}_1}({\mathbf{v}}_2)=\frac{{\mathbf{v}}_2\cdot {\mathbf{w}}_1}{{\mathbf{w}}_1\cdot {\mathbf{w}}_1}{\mathbf{w}}_1=\frac{1}{2}{\mathbf{w}}_1,$$

and therefore ${\mathbf{w}}_2={\mathbf{v}}_2-\frac{1}{2}{\mathbf{w}}_1=\left(\begin{array}{c}-1/2\\ 1/2\\ 1\\ 0\end{array}\right)$. At this point we have $⟨{\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3⟩=⟨{\mathbf{w}}_1,{\mathbf{w}}_2,{\mathbf{v}}_3⟩$. Since we are just trying to preserve the span, it is convenient to change the definition of ${\mathbf{w}}_2$ to $\left(\begin{array}{c}-1\\ 1\\ 2\\ 0\end{array}\right)$ by multiplying it by $2$. This won’t change the span, and we eliminate the fractions.

Finally compute ${\mathrm{proj}}_{V_2}({\mathbf{v}}_3)$. Since $V_2=⟨{\mathbf{w}}_1,{\mathbf{w}}_2⟩$, and this is actually an orthogonal basis for $V_2$, we can use the formula for $\mathrm{proj}$ that splits it into components according to the orthogonal basis:

$$\begin{array}{rl}{\mathrm{proj}}_{V_2}({\mathbf{v}}_3)& ={\mathrm{proj}}_{{\mathbf{w}}_1}({\mathbf{v}}_3)+{\mathrm{proj}}_{{\mathbf{w}}_2}({\mathbf{v}}_3)\\ & =\frac{{\mathbf{v}}_3\cdot {\mathbf{w}}_1}{{\mathbf{w}}_1\cdot {\mathbf{w}}_1}{\mathbf{w}}_1+\frac{{\mathbf{v}}_3\cdot {\mathbf{w}}_2}{{\mathbf{w}}_2\cdot {\mathbf{w}}_2}{\mathbf{w}}_2\\ & =\frac{1}{2}{\mathbf{w}}_1+\frac{1}{6}{\mathbf{w}}_2=\left(\begin{array}{c}1/3\\ 2/3\\ 1/3\\ 0\end{array}\right).\end{array}$$

Therefore ${\mathbf{w}}_3=\left(\begin{array}{c}2/3\\ -2/3\\ 2/3\\ 1\end{array}\right)$, and we can scale this up as with ${\mathbf{w}}_2$ and use instead $\left(\begin{array}{c}2\\ -2\\ 2\\ 3\end{array}\right)$. So we have an orthogonal basis:

$$⟨\left(\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}-1\\ 1\\ 2\\ 0\end{array}\right),\left(\begin{array}{c}2\\ -2\\ 2\\ 3\end{array}\right)⟩.$$

Computing distance of point to subspace

Computing distance of point to subspace

Problem: Find the closest point on the subspace given by the span

$$U=⟨\left(\begin{array}{c}0\\ 2\\ 3\\ -1\end{array}\right),\left(\begin{array}{c}1\\ -2\\ 1\\ -1\end{array}\right),\left(\begin{array}{c}2\\ 0\\ 1\\ 3\end{array}\right)⟩$$

to the vector $\mathbf{x}=(-1,-3,2,1)$.

Solution: Label the vectors in the span ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$, in order. Notice that they are already pairwise orthogonal, so we can skip Gram-Schmidt. Then compute ${\mathrm{proj}}_U(\mathbf{x})$:

$$\begin{array}{rl}{\mathrm{proj}}_U(\mathbf{x})& ={\mathrm{proj}}_{{\mathbf{v}}_1}(\mathbf{x})+{\mathrm{proj}}_{{\mathbf{v}}_2}(\mathbf{x})+{\mathrm{proj}}_{{\mathbf{v}}_3}(\mathbf{x})\\ & =\frac{\mathbf{x}\cdot {\mathbf{v}}_1}{{\mathbf{v}}_1\cdot {\mathbf{v}}_1}{\mathbf{v}}_1+\frac{\mathbf{x}\cdot {\mathbf{v}}_2}{{\mathbf{v}}_2\cdot {\mathbf{v}}_2}{\mathbf{v}}_2+\frac{\mathbf{x}\cdot {\mathbf{v}}_3}{{\mathbf{v}}_3\cdot {\mathbf{v}}_3}{\mathbf{v}}_3\\ & =\frac{-1}{14}{\mathbf{v}}_1+\frac{6}{7}{\mathbf{v}}_2+\frac{3}{14}{\mathbf{v}}_3=\left(\begin{array}{c}9/7\\ -13/7\\ 6/7\\ -1/7\end{array}\right).\end{array}$$

Therefore

$$\mathbf{x}-{\mathrm{proj}}_U(\mathbf{x})=\left(\begin{array}{c}-16/7\\ -8/7\\ 8/7\\ 8/7\end{array}\right)$$

and the length of this vector is $\sqrt{\frac{448}{49}}$, which is our answer.